Ex 5.13 – Continuity and Differentiability | ML Aggarwal Class 12 Solutions

Solution: We begin by differentiating the function: \[ \frac{d}{dx} \left( x^{20} \right) = 20x^{19}. \] Next, we differentiate the result to obtain the second derivative: \[ \frac{d^2}{dx^2} \left( x^{20} \right) = \frac{d}{dx} \left( 20x^{19} \right) = 20 \cdot 19x^{18} = 380x^{18}. \]

Solution: First, differentiate the function: \[ \frac{d}{dx}(x^{2} + 3x + 2) = 2x + 3. \] Next, differentiate the result to obtain the second derivative: \[ \frac{d^2}{dx^2}(x^{2} + 3x + 2) = \frac{d}{dx}(2x + 3) = 2. \]

Solution: First, differentiate the function: \[ \frac{d}{dx}\left(x^{3}-5x^{2}+3x+4\right) = 3x^{2}-10x+3. \] Next, differentiate the first derivative: \[ \frac{d^2}{dx^2}\left(x^{3}-5x^{2}+3x+4\right) = \frac{d}{dx}\left(3x^{2}-10x+3\right) = 6x-10. \]

Solution: Differentiate the function: \[ \frac{d}{dx}\left(x^{3}+\tan x\right) = 3x^{2}+\sec^{2}x. \] Then, differentiate again to obtain the second derivative. For the first term: \[ \frac{d}{dx}\left(3x^{2}\right) = 6x. \] For the second term, recall that \[ \frac{d}{dx}\left(\sec^{2}x\right) = 2\sec^{2}x\tan x. \] Thus, the second derivative is: \[ \frac{d^{2}}{dx^{2}}\left(x^{3}+\tan x\right) = 6x+2\sec^{2}x\tan x. \]

Solution: First, differentiate the function: \[ \frac{d}{dx}\left(\log x\right) = \frac{1}{x}. \] Next, differentiate the result: \[ \frac{d^2}{dx^2}\left(\log x\right) = \frac{d}{dx}\left(\frac{1}{x}\right) = -\frac{1}{x^2}. \]

Solution: First, differentiate the function: \[ \frac{d}{dx}\left(\tan^{-1} x\right)=\frac{1}{1+x^{2}}. \] Next, differentiate the first derivative using the chain rule: \[ \frac{d^{2}}{dx^{2}}\left(\tan^{-1} x\right)=\frac{d}{dx}\left(\frac{1}{1+x^{2}}\right)=-\frac{2x}{(1+x^{2})^{2}}. \]

Solution: Given: \[ y=\log (x-2). \] First, differentiate \( y \) with respect to \( x \): \[ \frac{dy}{dx}=\frac{1}{x-2}. \] Next, differentiate \( \frac{dy}{dx} \) to obtain the second derivative: \[ \frac{d^{2}y}{dx^{2}}=\frac{d}{dx}\left(\frac{1}{x-2}\right)=-\frac{1}{(x-2)^{2}}. \]

Solution: Given: \[ y=\cot x. \] First, differentiate \( y \) with respect to \( x \): \[ \frac{dy}{dx}=-\mathrm{cosec}^{2} x. \] Next, differentiate the first derivative: \[ \frac{d^{2}y}{dx^{2}}=\frac{d}{dx}\left(-\mathrm{cosec}^{2} x\right) =2\mathrm{cosec}^{2} x \cot x. \] Now, evaluate at \( x=\frac{\pi}{4} \). We have: \[ \mathrm{cosec}\left(\frac{\pi}{4}\right)=\frac{1}{\sin\frac{\pi}{4}}=\frac{1}{\frac{\sqrt{2}}{2}}=\sqrt{2}, \quad \mathrm{cosec}^{2}\left(\frac{\pi}{4}\right)=2, \] and \[ \cot\left(\frac{\pi}{4}\right)=1. \] Therefore, \[ \frac{d^{2}y}{dx^{2}}\Bigg|_{x=\frac{\pi}{4}}=2\cdot 2\cdot 1=4. \]

Solution: Let \[ y=\sin^{-1} x. \] Then the first derivative is \[ \frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}. \] Writing the derivative in exponent form, \[ \frac{dy}{dx}=(1-x^2)^{-\frac{1}{2}}, \] we differentiate to obtain the second derivative: \[ \frac{d^2y}{dx^2}=\frac{d}{dx}\left((1-x^2)^{-\frac{1}{2}}\right) = -\frac{1}{2}(1-x^2)^{-\frac{3}{2}}(-2x) = \frac{x}{(1-x^2)^{\frac{3}{2}}}. \]

Solution: We start by differentiating the function using the product rule: \[ y = x \cos x \quad \Rightarrow \quad y’ = \frac{d}{dx}(x)\cos x + x\,\frac{d}{dx}(\cos x) = \cos x – x \sin x. \] Next, we differentiate \( y’ \) to obtain the second derivative: \[ y” = \frac{d}{dx}\left(\cos x – x \sin x\right) = -\sin x – \left(\sin x + x \cos x\right) = -2 \sin x – x \cos x. \]

Solution: First, we differentiate using the product rule. For the function \[ y = x \sin 2x, \] we have: \[ y’ = \frac{d}{dx}(x)\sin 2x + x\,\frac{d}{dx}(\sin 2x) = \sin 2x + x\cdot(2\cos 2x) = \sin 2x + 2x\cos 2x. \] Next, we differentiate \( y’ \) to obtain the second derivative: \[ y” = \frac{d}{dx}\left(\sin 2x + 2x\cos 2x\right). \] Differentiating term-by-term: \[ \frac{d}{dx}(\sin 2x) = 2\cos 2x, \] and applying the product rule to \(2x\cos 2x\): \[ \frac{d}{dx}(2x\cos 2x) = 2\cos 2x + 2x\cdot(-2\sin 2x) = 2\cos 2x – 4x\sin 2x. \] Thus, combining the results: \[ y” = 2\cos 2x + 2\cos 2x – 4x\sin 2x = 4\cos 2x – 4x\sin 2x. \]

Solution: We start by differentiating the given function using the product rule. \[ y = e^{x}\sin 5x. \] First derivative: \[ y’ = \frac{d}{dx}\left(e^{x}\right)\sin 5x + e^{x}\frac{d}{dx}\left(\sin 5x\right) = e^{x}\sin 5x + e^{x}\cdot 5\cos 5x = e^{x}\left(\sin 5x + 5\cos 5x\right). \] Next, we differentiate \( y’ \) to find the second derivative: \[ y” = \frac{d}{dx}\left(e^{x}\left(\sin 5x + 5\cos 5x\right)\right). \] Using the product rule again: \[ y” = e^{x}\left(\sin 5x + 5\cos 5x\right) + e^{x}\left(5\cos 5x – 25\sin 5x\right). \] Combining like terms, we obtain: \[ y” = e^{x}\left[\sin 5x + 5\cos 5x + 5\cos 5x – 25\sin 5x\right] = e^{x}\left(-24\sin 5x + 10\cos 5x\right). \] This can be factored as: \[ y” = 2e^{x}\left(5\cos 5x – 12\sin 5x\right). \]

Solution: First, differentiate using the product rule. Given: \[ y = e^{2x}\sin 3x. \] The first derivative is: \[ y’ = \frac{d}{dx}\left(e^{2x}\right)\sin 3x + e^{2x}\frac{d}{dx}\left(\sin 3x\right) = 2e^{2x}\sin 3x + e^{2x}\cdot 3\cos 3x. \] Thus, we can write: \[ y’ = e^{2x}\left(2\sin 3x + 3\cos 3x\right). \] Next, differentiate \( y’ \) to find the second derivative: \[ y” = \frac{d}{dx}\left[e^{2x}\left(2\sin 3x + 3\cos 3x\right)\right]. \] Applying the product rule again: \[ y” = \frac{d}{dx}\left(e^{2x}\right)\left(2\sin 3x + 3\cos 3x\right) + e^{2x}\frac{d}{dx}\left(2\sin 3x + 3\cos 3x\right). \] Compute each derivative: \[ \frac{d}{dx}\left(e^{2x}\right)= 2e^{2x}, \] \[ \frac{d}{dx}\left(2\sin 3x\right) = 6\cos 3x,\quad \frac{d}{dx}\left(3\cos 3x\right) = -9\sin 3x. \] Therefore: \[ y” = 2e^{2x}\left(2\sin 3x + 3\cos 3x\right) + e^{2x}\left(6\cos 3x – 9\sin 3x\right). \] Combine like terms: \[ y” = e^{2x}\left[4\sin 3x + 6\cos 3x + 6\cos 3x – 9\sin 3x\right] = e^{2x}\left[-5\sin 3x + 12\cos 3x\right]. \]

Solution: We are given \[ y = \text{log}(\text{log}\, x). \] First, we differentiate \( y \) with respect to \( x \). Using the chain rule: \[ y’ = \frac{1}{\text{log}\, x} \cdot \frac{d}{dx}(\text{log}\, x) = \frac{1}{\text{log}\, x} \cdot \frac{1}{x} = \frac{1}{x\,\text{log}\, x}. \] Next, we differentiate \( y’ \) to find the second derivative. Write: \[ y’ = \frac{1}{x\,\text{log}\, x}. \] Differentiating using the product rule (or quotient rule), we have: Let \[ f(x) = x^{-1} \quad \text{and} \quad g(x) = (\text{log}\, x)^{-1}. \] Then, \[ f'(x) = -x^{-2}, \] and by the chain rule, \[ g'(x) = -\frac{1}{(\text{log}\, x)^2}\cdot\frac{1}{x} = -\frac{1}{x\,(\text{log}\, x)^2}. \] Applying the product rule: \[ y” = f'(x)g(x) + f(x)g'(x) = \left(-\frac{1}{x^2}\right)\frac{1}{\text{log}\, x} + \frac{1}{x}\left(-\frac{1}{x\,(\text{log}\, x)^2}\right). \] Simplify the expression: \[ y” = -\frac{1}{x^2\,\text{log}\, x} – \frac{1}{x^2\,(\text{log}\, x)^2} = -\frac{\text{log}\, x + 1}{x^2\,(\text{log}\, x)^2}. \]

Solution: We begin by rewriting the function as: \[ y = \log x \cdot x^{-1}. \] Differentiating using the product rule: \[ y’ = \frac{d}{dx}(\log x)\cdot x^{-1} + \log x\cdot \frac{d}{dx}(x^{-1}) = \frac{1}{x}\cdot x^{-1} – \log x\cdot x^{-2} = \frac{1-\log x}{x^2}. \] Next, differentiate \( y’ \) to obtain the second derivative. Express \( y’ \) as: \[ y’ = (1-\log x)x^{-2}. \] Applying the product rule: \[ y” = \frac{d}{dx}(1-\log x)\cdot x^{-2} + (1-\log x)\cdot \frac{d}{dx}(x^{-2}). \] We have: \[ \frac{d}{dx}(1-\log x) = -\frac{1}{x} \quad \text{and} \quad \frac{d}{dx}(x^{-2}) = -2x^{-3}. \] Thus, \[ y” = \left(-\frac{1}{x}\right)x^{-2} + (1-\log x)\cdot\left(-2x^{-3}\right) = -\frac{1}{x^3} – \frac{2(1-\log x)}{x^3}. \] Simplify the numerator: \[ y” = -\frac{1+2-2\log x}{x^3} = \frac{2\log x-3}{x^3}. \]

Solution: We are given: \[ y = x^2\,\text{log}|\cos x|. \] First, differentiate using the product rule: \[ y’ = \frac{d}{dx}\left(x^2\right) \text{log}|\cos x| + x^2\,\frac{d}{dx}\left(\text{log}|\cos x|\right). \] Since \[ \frac{d}{dx}(x^2)=2x, \] and noting that \[ \frac{d}{dx}\left(\text{log}|\cos x|\right) = \frac{-\sin x}{\cos x}=-\tan x, \] we obtain: \[ y’ = 2x\,\text{log}|\cos x| – x^2\tan x. \] Next, differentiate \( y’ \) to find \( y” \). Differentiate term by term: For the first term: \[ \frac{d}{dx}\left(2x\,\text{log}|\cos x|\right) = 2\,\text{log}|\cos x| + 2x\left(-\tan x\right) = 2\,\text{log}|\cos x| – 2x\tan x. \] For the second term: \[ \frac{d}{dx}\left(-x^2\tan x\right) = -\left(2x\tan x + x^2\sec^2 x\right). \] Combining these results: \[ y” = \left(2\,\text{log}|\cos x| – 2x\tan x\right) – \left(2x\tan x + x^2\sec^2 x\right) = 2\,\text{log}|\cos x| – 4x\tan x – x^2\sec^2 x. \]

Solution: First, we rewrite the function using the hint: \[ y = \frac{2x+1}{2x+3} = 1 – \frac{2}{2x+3}. \] Now, differentiate \( y \) with respect to \( x \). **Step 1: First Derivative** Since \[ y = 1 – 2(2x+3)^{-1}, \] we differentiate: \[ y’ = -2 \cdot \left(-1\right)(2x+3)^{-2}\cdot 2 = \frac{4}{(2x+3)^2}. \] **Step 2: Second Derivative** Differentiate \( y’ = 4(2x+3)^{-2} \): \[ y” = 4 \cdot (-2)(2x+3)^{-3}\cdot 2 = -\frac{16}{(2x+3)^3}. \]

Solution: We start with \[ y = \sec(ax) = \frac{1}{\cos(ax)}. \] First, differentiate \( y \) with respect to \( x \) using the chain rule: \[ y’ = a\,\sec(ax)\tan(ax). \] Next, differentiate \( y’ \) to obtain the second derivative. Write: \[ y’ = a\,\sec(ax)\tan(ax). \] Differentiating using the product rule: \[ y” = a\,\frac{d}{dx}\Big[\sec(ax)\tan(ax)\Big] = a\left[\sec(ax)\tan(ax)\cdot a\tan(ax) + \sec(ax)\cdot a\sec^2(ax)\right]. \] This simplifies to: \[ y” = a^2\,\sec(ax)\left[\tan^2(ax) + \sec^2(ax)\right]. \] Recall that \( \tan^2(ax) = \sec^2(ax)-1 \). Hence, \[ \tan^2(ax) + \sec^2(ax) = (\sec^2(ax)-1) + \sec^2(ax)=2\sec^2(ax)-1. \] Therefore, the second derivative becomes: \[ y” = a^2\,\sec(ax)\left(2\sec^2(ax)-1\right). \]

Solution: Let \[ y=\cot (1-2 x). \] First, we differentiate using the chain rule. Recall that \[ \frac{d}{dx}\left(\cot u\right)=-\mathrm{cosec}^{2}u\cdot\frac{du}{dx}, \] where \( u=1-2x \) and \( \frac{du}{dx}=-2 \). Thus, the first derivative is \[ \frac{dy}{dx}=-\mathrm{cosec}^{2}(1-2x)\cdot(-2)=2\,\mathrm{cosec}^{2}(1-2x). \] Next, we differentiate the first derivative. We differentiate \[ \frac{dy}{dx}=2\,\mathrm{cosec}^{2}(1-2x) \] with respect to \( x \). Again, applying the chain rule along with the derivative \[ \frac{d}{du}(\mathrm{cosec}^{2} u)=-2\,\mathrm{cosec}^{2}u\cot u, \] and noting that \( \frac{du}{dx}=-2 \), we have: \[ \frac{d^{2}y}{dx^{2}}=2\cdot\frac{d}{dx}\left(\mathrm{cosec}^{2}(1-2x)\right) =2\cdot\left[-2\,\mathrm{cosec}^{2}(1-2x)\cot(1-2x)\cdot(-2)\right]. \] Simplify the constants: \[ 2\cdot\left[-2\cdot(-2)\,\mathrm{cosec}^{2}(1-2x)\cot(1-2x)\right] =2\cdot(4\,\mathrm{cosec}^{2}(1-2x)\cot(1-2x)) =8\,\mathrm{cosec}^{2}(1-2x)\cot(1-2x). \]

Solution: We start by expressing the function using the product-to-sum formula: \[ \sin 3x \cos 5x = \frac{1}{2}\Big[\sin(3x+5x) + \sin(3x-5x)\Big] = \frac{1}{2}\Big[\sin 8x + \sin(-2x)\Big] = \frac{1}{2}\Big[\sin 8x – \sin 2x\Big]. \] Thus, we have: \[ y = \frac{1}{2}\left[\sin 8x – \sin 2x\right]. \] **Step 1: First Derivative** Differentiating term by term: \[ y’ = \frac{1}{2}\left[8\cos 8x – 2\cos 2x\right] = 4\cos 8x – \cos 2x. \] **Step 2: Second Derivative** Differentiate \( y’ \) term by term: \[ y” = \frac{d}{dx}\left(4\cos 8x\right) – \frac{d}{dx}\left(\cos 2x\right) = -32\sin 8x + 2\sin 2x. \]

Solution: We start by using the identity: \[ \sin^{3} x = \frac{1}{4}\Bigl(3\sin x-\sin 3x\Bigr). \] Differentiating with respect to \( x \), the first derivative is: \[ \frac{d}{dx}\Bigl(\sin^{3} x\Bigr)=\frac{1}{4}\Bigl(3\cos x-3\cos 3x\Bigr) =\frac{3}{4}\Bigl(\cos x-\cos 3x\Bigr). \] Differentiating once more to get the second derivative: \[ \frac{d^{2}}{dx^{2}}\Bigl(\sin^{3} x\Bigr)=\frac{3}{4}\Bigl(-\sin x+3\sin 3x\Bigr) =\frac{3}{4}\Bigl(3\sin 3x-\sin x\Bigr). \]

Solution: We begin with \[ y = \cos\left(2x^2-1\right). \] **Step 1: First Derivative** Using the chain rule, we differentiate: \[ y’ = -\sin\left(2x^2-1\right)\cdot \frac{d}{dx}\left(2x^2-1\right) = -\sin\left(2x^2-1\right)\cdot 4x = -4x\,\sin\left(2x^2-1\right). \] **Step 2: Second Derivative** Differentiate \( y’ = -4x\,\sin\left(2x^2-1\right) \) using the product rule: \[ y” = \frac{d}{dx}\left(-4x\right)\sin\left(2x^2-1\right) + (-4x)\frac{d}{dx}\left(\sin\left(2x^2-1\right)\right). \] We have: \[ \frac{d}{dx}\left(-4x\right) = -4, \] and \[ \frac{d}{dx}\left(\sin\left(2x^2-1\right)\right) = \cos\left(2x^2-1\right)\cdot \frac{d}{dx}\left(2x^2-1\right) = \cos\left(2x^2-1\right)\cdot 4x. \] Thus, \[ y” = -4\,\sin\left(2x^2-1\right) + (-4x)\cdot\left(4x\,\cos\left(2x^2-1\right)\right) = -4\,\sin\left(2x^2-1\right) – 16x^2\,\cos\left(2x^2-1\right). \]

Solution: We have \[ y=\sqrt{1-x^2}=(1-x^2)^{\frac{1}{2}}. \] Differentiating with respect to \( x \): \[ \frac{dy}{dx}=\frac{1}{2}(1-x^2)^{-\frac{1}{2}}(-2x)=-\frac{x}{\sqrt{1-x^2}}. \] Differentiating again, we write \[ \frac{dy}{dx}=-x(1-x^2)^{-\frac{1}{2}}. \] Using the product rule with \( u=-x \) and \( v=(1-x^2)^{-\frac{1}{2}} \): \[ u’=-1, \quad v’=\frac{d}{dx}\left((1-x^2)^{-\frac{1}{2}}\right) = -\frac{1}{2}(1-x^2)^{-\frac{3}{2}}(-2x) = x(1-x^2)^{-\frac{3}{2}}. \] Thus, \[ \frac{d^2y}{dx^2}=u’v+uv’ =-1\cdot(1-x^2)^{-\frac{1}{2}}+(-x)\cdot\left[x(1-x^2)^{-\frac{3}{2}}\right]. \] Simplifying: \[ \frac{d^2y}{dx^2}=-\frac{1}{(1-x^2)^{\frac{1}{2}}}-\frac{x^2}{(1-x^2)^{\frac{3}{2}}} = -\frac{(1-x^2)+x^2}{(1-x^2)^{\frac{3}{2}}} = -\frac{1}{(1-x^2)^{\frac{3}{2}}}. \]

Solution: Given \[ y=\cos^{-1}x, \] we have \[ \frac{dy}{dx}=-\frac{1}{\sqrt{1-x^2}}. \] Since \( y=\cos^{-1}x \) implies \( x=\cos y \) and \( \sqrt{1-x^2}=\sin y \) (because \(0\le y\le\pi\)), it follows that \[ \frac{dy}{dx}=-\frac{1}{\sin y}. \] Differentiating with respect to \( x \) using the chain rule: \[ \frac{d^2y}{dx^2}=\frac{d}{dy}\left(-\frac{1}{\sin y}\right)\frac{dy}{dx}. \] First, differentiate with respect to \( y \): \[ \frac{d}{dy}\left(-\frac{1}{\sin y}\right)=\frac{\cos y}{\sin^2 y}. \] Then, substituting \(\frac{dy}{dx}=-\frac{1}{\sin y}\): \[ \frac{d^2y}{dx^2}=\frac{\cos y}{\sin^2 y}\left(-\frac{1}{\sin y}\right) = -\frac{\cos y}{\sin^3 y}. \] Recognizing that \[ \frac{\cos y}{\sin^3 y}=\cot y \, \mathrm{cosec}^2 y, \] we obtain \[ \frac{d^2y}{dx^2}=-\mathrm{cosec}^2 y\, \cot y. \]

Solution: We can rewrite the function as: \[ y=\log\left(\frac{x^2}{e^x}\right)=\log(x^2)-\log(e^x)=2\log x-x. \] Differentiating with respect to \( x \): \[ \frac{dy}{dx}=\frac{2}{x}-1. \] Differentiating again: \[ \frac{d^2y}{dx^2}=-\frac{2}{x^2}. \]

Solution: Given \[ y=\cot x, \] we differentiate to obtain the first derivative: \[ \frac{dy}{dx}=-\mathrm{cosec}^2 x. \] Differentiating again for the second derivative: \[ \frac{d^2y}{dx^2}=-\frac{d}{dx}\left(\mathrm{cosec}^2 x\right). \] Using the chain rule and knowing that \[ \frac{d}{dx}(\mathrm{cosec} x)=-\mathrm{cosec} x\,\cot x, \] we have \[ \frac{d}{dx}\left(\mathrm{cosec}^2 x\right)=2\,\mathrm{cosec} x\left(-\mathrm{cosec} x\,\cot x\right) = -2\,\mathrm{cosec}^2 x\,\cot x. \] Hence, \[ \frac{d^2y}{dx^2}=-\left(-2\,\mathrm{cosec}^2 x\,\cot x\right) = 2\,\mathrm{cosec}^2 x\,\cot x. \] Next, calculate \[ 2y\frac{dy}{dx}=2\cot x\left(-\mathrm{cosec}^2 x\right) = -2\,\mathrm{cosec}^2 x\,\cot x. \] Adding the two results: \[ \frac{d^2y}{dx^2}+2y\frac{dy}{dx} = 2\,\mathrm{cosec}^2 x\,\cot x – 2\,\mathrm{cosec}^2 x\,\cot x = 0. \]

Solution: Given \[ y=5\cos x-3\sin x, \] we differentiate to obtain the first derivative: \[ \frac{dy}{dx}=-5\sin x-3\cos x. \] Differentiating again, the second derivative is: \[ \frac{d^2y}{dx^2}=-5\cos x+3\sin x. \] Now, adding \( y \) to the second derivative: \[ \frac{d^2y}{dx^2}+y=(-5\cos x+3\sin x)+(5\cos x-3\sin x)=0. \] Thus, we have proved that \[ \frac{d^2 y}{dx^2}+y=0. \]

Solution: We are given \[ y=x+\tan x. \] First, differentiate with respect to \( x \): \[ \frac{dy}{dx}=1+\frac{d}{dx}(\tan x)=1+\sec^2 x. \] Next, differentiate the first derivative to find the second derivative: \[ \frac{d^2y}{dx^2}=\frac{d}{dx}(1+\sec^2 x)=0+2\sec^2 x\,\frac{d}{dx}(\sec x). \] Since \[ \frac{d}{dx}(\sec x)=\sec x\,\tan x, \] it follows that \[ \frac{d^2y}{dx^2}=2\sec^2 x\cdot\sec x\,\tan x=2\sec^3 x\,\tan x. \] However, a simpler approach is to differentiate \(\sec^2 x\) directly: \[ \frac{d}{dx}(\sec^2 x)=2\sec x\cdot \sec x\,\tan x=2\sec^2 x\,\tan x. \] Thus, we have \[ \frac{d^2y}{dx^2}=2\sec^2 x\,\tan x. \] Now, consider the expression \[ \cos^2 x\cdot \frac{d^2y}{dx^2}-2y+2x. \] Substitute the value of \(\frac{d^2y}{dx^2}\) and \(y\): \[ \cos^2 x\cdot (2\sec^2 x\,\tan x)-2(x+\tan x)+2x. \] Note that \(\cos^2 x\,\sec^2 x=1\), so \[ 2\tan x-2x-2\tan x+2x=0. \] This confirms that \[ \cos^2 x\cdot \frac{d^2y}{dx^2}-2y+2x=0. \]

Solution: We start with \[ y=\tan x+\sec x. \] Expressing in terms of sine and cosine: \[ y=\frac{\sin x}{\cos x}+\frac{1}{\cos x}=\frac{1+\sin x}{\cos x}. \] Multiplying numerator and denominator by \(1-\sin x\): \[ y=\frac{(1+\sin x)(1-\sin x)}{\cos x(1-\sin x)} =\frac{1-\sin^2 x}{\cos x(1-\sin x)} =\frac{\cos^2 x}{\cos x(1-\sin x)} =\frac{\cos x}{1-\sin x}. \] Now, differentiate \( y=\frac{\cos x}{1-\sin x} \). Let \[ u=\cos x,\quad v=1-\sin x. \] Then, \( u’=-\sin x \) and \( v’=-\cos x \). Using the quotient rule: \[ \frac{dy}{dx}=\frac{u’v-u v’}{v^2} =\frac{(-\sin x)(1-\sin x)-\cos x(-\cos x)}{(1-\sin x)^2} =\frac{-\sin x(1-\sin x)+\cos^2 x}{(1-\sin x)^2}. \] Simplify the numerator using \(\cos^2 x=1-\sin^2 x\): \[ -\sin x(1-\sin x)+\cos^2 x =-\sin x+\sin^2 x+1-\sin^2 x =1-\sin x. \] Thus, \[ \frac{dy}{dx}=\frac{1-\sin x}{(1-\sin x)^2} =\frac{1}{1-\sin x}. \] Next, differentiate \(\frac{dy}{dx}\) with respect to \( x \): \[ \frac{dy}{dx}=(1-\sin x)^{-1}. \] Differentiating using the chain rule: \[ \frac{d^2y}{dx^2}=-1\cdot(1-\sin x)^{-2}\cdot(-\cos x) =\frac{\cos x}{(1-\sin x)^2}. \]

Solution: We start with \[ y=\sec x-\tan x. \] Differentiate to obtain the first derivative: \[ \frac{dy}{dx}=\frac{d}{dx}(\sec x)-\frac{d}{dx}(\tan x) =\sec x\,\tan x-\sec^2 x. \] Next, differentiate \(\frac{dy}{dx}\) to get the second derivative. Differentiating \(\sec x\,\tan x\) using the product rule: \[ \frac{d}{dx}(\sec x\,\tan x) =\frac{d}{dx}(\sec x)\cdot\tan x+\sec x\cdot\frac{d}{dx}(\tan x) =\sec x\,\tan x\cdot\tan x+\sec x\,\sec^2 x =\sec x\,\tan^2 x+\sec^3 x. \] Differentiating \(-\sec^2 x\): \[ \frac{d}{dx}(-\sec^2 x) =-2\sec x\,\frac{d}{dx}(\sec x) =-2\sec x\,( \sec x\,\tan x) =-2\sec^2 x\,\tan x. \] Hence, the second derivative is: \[ \frac{d^2y}{dx^2}=\sec x\,\tan^2 x+\sec^3 x-2\sec^2 x\,\tan x. \] Multiplying by \(\cos x\): \[ \cos x\cdot\frac{d^2y}{dx^2} =\cos x\left(\sec x\,\tan^2 x+\sec^3 x-2\sec^2 x\,\tan x\right). \] Noting that \(\cos x\,\sec x=1\), \(\cos x\,\sec^2 x=\sec x\), and \(\cos x\,\sec^3 x=\sec^2 x\), we have: \[ \cos x\cdot\frac{d^2y}{dx^2} =\tan^2 x+\sec^2 x-2\sec x\,\tan x. \] Now, observe that: \[ y^2=(\sec x-\tan x)^2=\sec^2 x-2\sec x\,\tan x+\tan^2 x. \] Thus, we conclude: \[ \cos x\cdot\frac{d^2y}{dx^2}=y^2. \]

Solution: We are given \[ y=x+\cot x. \] First, differentiate \( y \) with respect to \( x \): \[ \frac{dy}{dx}=1+\frac{d}{dx}(\cot x) =1-\mathrm{cosec}^2 x. \] Next, differentiate the first derivative to obtain the second derivative. Since \[ \frac{d}{dx}(-\mathrm{cosec}^2 x)=2\,\mathrm{cosec}^2 x\,\cot x, \] it follows that \[ \frac{d^2y}{dx^2}=2\,\mathrm{cosec}^2 x\,\cot x. \] Now, consider the expression: \[ \sin^2 x\cdot \frac{d^2y}{dx^2}-2y+2x. \] Substitute \(\frac{d^2y}{dx^2}\) and \(y\): \[ \sin^2 x\cdot \left(2\,\mathrm{cosec}^2 x\,\cot x\right)-2(x+\cot x)+2x. \] Since \(\sin^2 x\cdot \mathrm{cosec}^2 x=1\), the expression simplifies to: \[ 2\,\cot x-2x-2\,\cot x+2x=0. \] Thus, we have proved that: \[ \sin^2 x\cdot \frac{d^2y}{dx^2}-2y+2x=0. \]

Solution: We are given \[ y = a e^{mx} + b e^{-mx}. \] Differentiating with respect to \( x \): \[ y_1 = \frac{dy}{dx} = a m e^{mx} – b m e^{-mx}. \] Differentiating again: \[ y_2 = \frac{d^2y}{dx^2} = a m^2 e^{mx} + b m^2 e^{-mx}. \] Factor out \( m^2 \): \[ y_2 = m^2 (a e^{mx} + b e^{-mx}) = m^2 y. \] Therefore, \[ y_2 – m^2 y = 0. \]

Solution: We are given \[ y = 500 e^{7x} + 600 e^{-7x}. \] Differentiating with respect to \( x \): \[ y’ = 500 \cdot 7 e^{7x} – 600 \cdot 7 e^{-7x} = 3500 e^{7x} – 4200 e^{-7x}. \] Differentiating again to find the second derivative: \[ y” = 3500 \cdot 7 e^{7x} + 4200 \cdot 7 e^{-7x} = 24500 e^{7x} + 29400 e^{-7x}. \] Notice that: \[ 24500 e^{7x} = 49 \cdot 500 e^{7x} \quad \text{and} \quad 29400 e^{-7x} = 49 \cdot 600 e^{-7x}. \] Hence, we can write: \[ y” = 49 \left( 500 e^{7x} + 600 e^{-7x} \right) = 49 y. \]

Solution: Given \[ y = a e^{2x} + b e^{-x}, \] we differentiate with respect to \( x \) to find the first derivative: \[ y’ = \frac{d}{dx}(a e^{2x} + b e^{-x}) = 2a e^{2x} – b e^{-x}. \] Differentiating again to obtain the second derivative: \[ y” = \frac{d}{dx}(2a e^{2x} – b e^{-x}) = 4a e^{2x} + b e^{-x}. \] Now, consider the expression: \[ y” – y’ – 2y. \] Substitute the expressions for \( y \), \( y’ \), and \( y” \): \[ y” – y’ – 2y = \left(4a e^{2x} + b e^{-x}\right) – \left(2a e^{2x} – b e^{-x}\right) – 2\left(a e^{2x} + b e^{-x}\right). \] Simplify step-by-step: \[ = 4a e^{2x} + b e^{-x} – 2a e^{2x} + b e^{-x} – 2a e^{2x} – 2b e^{-x}, \] \[ = \left(4a e^{2x} – 2a e^{2x} – 2a e^{2x}\right) + \left(b e^{-x} + b e^{-x} – 2b e^{-x}\right), \] \[ = 0 + 0 = 0. \] Hence, we have shown that: \[ \frac{d^2 y}{dx^2} – \frac{dy}{dx} – 2y = 0. \]

Solution: We start with \[ y = \mathrm{A} \sin x + \mathrm{B} \cos x. \] Differentiating with respect to \( x \), we obtain the first derivative: \[ y’ = \mathrm{A} \cos x – \mathrm{B} \sin x. \] Differentiating again gives the second derivative: \[ y” = -\mathrm{A} \sin x – \mathrm{B} \cos x. \] Now, adding \( y \) to \( y” \) we have: \[ y” + y = \left(-\mathrm{A} \sin x – \mathrm{B} \cos x\right) + \left(\mathrm{A} \sin x + \mathrm{B} \cos x\right) = 0. \]

Solution: We start with \[ y = \mathrm{A} \cos nx + \mathrm{B} \sin nx. \] Differentiating with respect to \( x \) gives: \[ y’ = -\mathrm{A} n \sin nx + \mathrm{B} n \cos nx. \] Differentiating once more to obtain the second derivative: \[ y” = -\mathrm{A} n^2 \cos nx – \mathrm{B} n^2 \sin nx. \] This can be written as: \[ y” = -n^2\left(\mathrm{A} \cos nx + \mathrm{B} \sin nx\right) = -n^2 y. \] Rearranging the equation, we get: \[ y” + n^2 y = 0. \]

Solution: Consider the function \[ y=\tan^{-1} x. \] Differentiating, we have \[ \frac{dy}{dx}=\frac{1}{1+x^2}. \] Differentiating once more yields \[ \frac{d^2 y}{dx^2}=-\frac{2x}{(1+x^2)^2}. \] Multiplying the second derivative by \(1+x^2\) and adding \(2x\) times the first derivative, we obtain: \[ \left(1+x^{2}\right)\frac{d^2 y}{dx^2}+2x\frac{dy}{dx} = \left(1+x^2\right)\left(-\frac{2x}{(1+x^2)^2}\right)+2x\left(\frac{1}{1+x^2}\right). \] Simplifying, \[ = -\frac{2x}{1+x^2}+ \frac{2x}{1+x^2} = 0. \]

Solution: Consider the function \[ y=\sin^{-1} x. \] Differentiating with respect to \( x \), we have \[ \frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}. \] Differentiating once more, we obtain \[ \frac{d^2 y}{dx^2}=\frac{d}{dx}\left(\frac{1}{\sqrt{1-x^2}}\right) =\frac{x}{\left(1-x^2\right)^{3/2}}. \] Now, multiplying the second derivative by \(1-x^2\) and subtracting \(x\) times the first derivative, we get: \[ \left(1-x^2\right)\frac{d^2 y}{dx^2}-x\frac{dy}{dx} = \left(1-x^2\right)\left(\frac{x}{\left(1-x^2\right)^{3/2}}\right) -x\left(\frac{1}{\sqrt{1-x^2}}\right). \] Simplifying, \[ =\frac{x}{\sqrt{1-x^2}}-\frac{x}{\sqrt{1-x^2}}=0. \]

Solution: Consider \[ y=\sin (\log x). \] First, differentiate with respect to \( x \): \[ \frac{dy}{dx}=\cos (\log x)\cdot \frac{1}{x}=\frac{\cos (\log x)}{x}. \] Next, differentiate again: \[ \frac{d^{2}y}{dx^{2}}=\frac{d}{dx}\left(\frac{\cos (\log x)}{x}\right). \] Using the quotient rule, \[ \frac{d^{2}y}{dx^{2}}=\frac{-\sin (\log x)\cdot\frac{1}{x}\cdot x-\cos (\log x)\cdot 1}{x^{2}} =\frac{-\sin (\log x)-\cos (\log x)}{x^{2}}. \] However, on careful re-evaluation of the differentiation, we note that the derivative of \(\cos (\log x)\) is \[ -\sin (\log x)\cdot \frac{1}{x}, \] so applying the quotient rule correctly: \[ \frac{d^{2}y}{dx^{2}} = \frac{x\cdot\left(-\sin (\log x)\cdot \frac{1}{x}\right)-\cos (\log x)}{x^{2}} = \frac{-\sin (\log x)-\cos (\log x)}{x^{2}}. \] Multiplying by \(x^2\) and adding \(x\frac{dy}{dx}+y\), we get: \[ x^{2}\frac{d^{2}y}{dx^{2}}+x\frac{dy}{dx}+y = \left[-\sin (\log x)-\cos (\log x)\right] + \cos (\log x)+\sin (\log x)=0. \]

Solution: Consider \[ y=3\cos (\log x)+4\sin (\log x). \] Differentiating with respect to \( x \), we have \[ \frac{dy}{dx}=\frac{-3\sin (\log x)+4\cos (\log x)}{x}. \] Differentiating again using the quotient rule yields \[ \frac{d^{2}y}{dx^{2}}=\frac{x\left[\frac{-3\cos (\log x)-4\sin (\log x)}{x}\right]-\left(-3\sin (\log x)+4\cos (\log x)\right)}{x^{2}}, \] which simplifies to \[ \frac{d^{2}y}{dx^{2}}=\frac{-3\cos (\log x)-4\sin (\log x)+3\sin (\log x)-4\cos (\log x)}{x^{2}} =\frac{-7\cos (\log x)-\sin (\log x)}{x^{2}}. \] Now, multiply the second derivative by \( x^2 \): \[ x^2\frac{d^{2}y}{dx^{2}}=-7\cos (\log x)-\sin (\log x). \] Also, multiplying the first derivative by \( x \) gives: \[ x\frac{dy}{dx}=-3\sin (\log x)+4\cos (\log x). \] Adding these results to \( y \) we obtain: \[ x^2\frac{d^{2}y}{dx^{2}}+x\frac{dy}{dx}+y = \left[-7\cos (\log x)-\sin (\log x)\right] +\left[-3\sin (\log x)+4\cos (\log x)\right] +\left[3\cos (\log x)+4\sin (\log x)\right]. \] Grouping like terms, the coefficients of \(\cos (\log x)\) add up as: \[ -7+4+3=0, \] and those of \(\sin (\log x)\) add up as: \[ -1-3+4=0. \] Hence, \[ x^2\frac{d^{2}y}{dx^{2}}+x\frac{dy}{dx}+y=0. \]

Solution: Consider the function \[ y=x\cos x. \] Differentiating with respect to \( x \): \[ \frac{dy}{dx}=\cos x – x\sin x. \] Differentiating again, we have: \[ \frac{d^{2}y}{dx^{2}}=-\sin x – \left(\sin x + x\cos x\right) = -2\sin x – x\cos x. \] Now, compute each term of the given expression: \[ x^2\frac{d^{2}y}{dx^{2}}=x^2\left(-2\sin x – x\cos x\right) = -2x^2\sin x – x^3\cos x, \] \[ -2x\frac{dy}{dx}=-2x\left(\cos x – x\sin x\right) = -2x\cos x + 2x^2\sin x, \] \[ \left(2+x^2\right)y=\left(2+x^2\right)(x\cos x) = 2x\cos x + x^3\cos x. \] Adding these three results: \[ \begin{aligned} &\quad\, x^2\frac{d^{2}y}{dx^{2}}-2x\frac{dy}{dx}+\left(2+x^2\right)y\\[1mm] &=\left[-2x^2\sin x – x^3\cos x\right] +\left[-2x\cos x + 2x^2\sin x\right] +\left[2x\cos x + x^3\cos x\right]\\[1mm] &=\left(-2x^2\sin x+2x^2\sin x\right) +\left(-x^3\cos x+x^3\cos x\right) +\left(-2x\cos x+2x\cos x\right)\\[1mm] &=0. \end{aligned} \] Hence, the given differential equation is verified.

Solution: Let \[ u = \cos^{-1} x \quad \text{so that} \quad y = u^2. \] Then, \[ \frac{du}{dx} = -\frac{1}{\sqrt{1-x^2}}, \] and by the chain rule the first derivative is \[ y_{1} = \frac{dy}{dx} = 2u\,\frac{du}{dx} = -\frac{2\cos^{-1} x}{\sqrt{1-x^2}}. \] Differentiating \( y_{1} = -2\cos^{-1} x\,(1-x^2)^{-1/2} \) with respect to \( x \) using the product rule: \[ y_{2} = \frac{d}{dx}\left(-2\cos^{-1} x\,(1-x^2)^{-1/2}\right) = -2\left[\frac{d}{dx}(\cos^{-1} x)\,(1-x^2)^{-1/2} + \cos^{-1} x\,\frac{d}{dx}\left((1-x^2)^{-1/2}\right)\right]. \] We already have \[ \frac{d}{dx}(\cos^{-1} x) = -\frac{1}{\sqrt{1-x^2}}, \] and differentiating \((1-x^2)^{-1/2}\) yields \[ \frac{d}{dx}\left((1-x^2)^{-1/2}\right) = \frac{x}{(1-x^2)^{3/2}}. \] Therefore, \[ y_{2} = -2\left[-\frac{1}{\sqrt{1-x^2}}\,(1-x^2)^{-1/2} + \cos^{-1} x\,\frac{x}{(1-x^2)^{3/2}}\right]. \] Simplifying the first term: \[ -\frac{1}{\sqrt{1-x^2}}\,(1-x^2)^{-1/2} = -\frac{1}{1-x^2}, \] so that \[ y_{2} = -2\left[-\frac{1}{1-x^2} + \frac{x\cos^{-1} x}{(1-x^2)^{3/2}}\right] = 2\left[\frac{1}{1-x^2} – \frac{x\cos^{-1} x}{(1-x^2)^{3/2}}\right]. \] Multiply \( y_{2} \) by \( (1-x^2) \): \[ (1-x^2)y_{2} = 2\left[1 – \frac{x\cos^{-1} x}{\sqrt{1-x^2}}\right]. \] Next, compute the term involving \( y_{1} \): \[ -xy_{1} = -x\left(-\frac{2\cos^{-1} x}{\sqrt{1-x^2}}\right) = \frac{2x\cos^{-1} x}{\sqrt{1-x^2}}. \] Adding these two expressions, we have \[ \left(1-x^{2}\right)y_{2} – x y_{1} = 2\left[1 – \frac{x\cos^{-1} x}{\sqrt{1-x^2}}\right] + \frac{2x\cos^{-1} x}{\sqrt{1-x^2}} = 2. \]

Solution: Let \[ u = \tan^{-1} x \quad \text{so that} \quad y = u^2. \] Then, the first derivative is \[ \frac{dy}{dx} = 2u\,\frac{du}{dx} = 2\tan^{-1} x \cdot \frac{1}{1+x^2} = \frac{2\tan^{-1} x}{1+x^2}. \] Differentiating \(\frac{dy}{dx} = 2\tan^{-1} x\,(1+x^2)^{-1}\) using the product rule, we have: \[ \frac{d^2y}{dx^2} = 2\left[\frac{d}{dx}\left(\tan^{-1} x\right)(1+x^2)^{-1} + \tan^{-1} x\,\frac{d}{dx}\left((1+x^2)^{-1}\right)\right]. \] We know that \[ \frac{d}{dx}\left(\tan^{-1} x\right) = \frac{1}{1+x^2}, \] and \[ \frac{d}{dx}\left((1+x^2)^{-1}\right) = -\frac{2x}{(1+x^2)^2}. \] Thus, \[ \frac{d^2y}{dx^2} = 2\left[\frac{1}{1+x^2}\,(1+x^2)^{-1} – \tan^{-1} x\,\frac{2x}{(1+x^2)^2}\right] = 2\left[\frac{1}{(1+x^2)^2} – \frac{2x\,\tan^{-1} x}{(1+x^2)^2}\right]. \] Hence, \[ \frac{d^2y}{dx^2} = \frac{2 – 4x\,\tan^{-1} x}{(1+x^2)^2}. \] Multiplying by \(\left(x^{2}+1\right)^{2}\), we obtain: \[ \left(x^{2}+1\right)^{2} \frac{d^2y}{dx^2} = 2 – 4x\,\tan^{-1} x. \] Next, we have: \[ 2x\left(x^{2}+1\right) \frac{dy}{dx} = 2x\left(x^{2}+1\right) \cdot \frac{2\tan^{-1} x}{1+x^2} = 4x\,\tan^{-1} x. \] Adding these two expressions gives: \[ \left(x^{2}+1\right)^{2} \frac{d^2y}{dx^2} + 2x\left(x^{2}+1\right) \frac{dy}{dx} = \left(2 – 4x\,\tan^{-1} x\right) + 4x\,\tan^{-1} x = 2. \]

Solution: We begin with \[ y=\log\left(x+\sqrt{x^{2}+1}\right). \] Recognize that \[ y=\sinh^{-1}x, \] and it is well-known that \[ \frac{d}{dx}\left(\sinh^{-1}x\right)=\frac{1}{\sqrt{x^{2}+1}}. \] Hence, \[ \frac{dy}{dx}=\frac{1}{\sqrt{x^{2}+1}}. \] Differentiating \( \frac{dy}{dx}=\left(x^{2}+1\right)^{-1/2} \) with respect to \( x \): \[ \frac{d^{2}y}{dx^{2}}=\frac{d}{dx}\left[\left(x^{2}+1\right)^{-1/2}\right] = -\frac{1}{2}\left(x^{2}+1\right)^{-3/2}\cdot 2x = -\frac{x}{\left(x^{2}+1\right)^{3/2}}. \] Now, compute \[ \left(x^{2}+1\right)\frac{d^{2}y}{dx^{2}}+x\frac{dy}{dx} = \left(x^{2}+1\right)\left(-\frac{x}{\left(x^{2}+1\right)^{3/2}}\right) + x\left(\frac{1}{\sqrt{x^{2}+1}}\right). \] Simplify the first term: \[ \left(x^{2}+1\right)\left(-\frac{x}{\left(x^{2}+1\right)^{3/2}}\right) = -\frac{x}{\sqrt{x^{2}+1}}. \] The second term is: \[ x\left(\frac{1}{\sqrt{x^{2}+1}}\right) = \frac{x}{\sqrt{x^{2}+1}}. \] Adding these, we obtain: \[ -\frac{x}{\sqrt{x^{2}+1}}+\frac{x}{\sqrt{x^{2}+1}}=0. \]

Solution: We start with \[ y=\log\left(x+\sqrt{x^{2}+a^{2}}\right). \] First, differentiate with respect to \( x \): \[ y_{1}=\frac{dy}{dx}=\frac{1}{x+\sqrt{x^2+a^2}}\left(1+\frac{x}{\sqrt{x^2+a^2}}\right). \] Note that \[ 1+\frac{x}{\sqrt{x^2+a^2}}=\frac{x+\sqrt{x^2+a^2}}{\sqrt{x^2+a^2}}, \] so that \[ y_{1}=\frac{1}{\sqrt{x^2+a^2}}. \] Next, differentiate \( y_{1}=\left(x^{2}+a^{2}\right)^{-1/2} \) to obtain \( y_{2} \): \[ y_{2}=\frac{d}{dx}\left(\left(x^{2}+a^{2}\right)^{-1/2}\right) = -\frac{1}{2}\left(x^{2}+a^{2}\right)^{-3/2}\cdot 2x = -\frac{x}{\left(x^{2}+a^{2}\right)^{3/2}}. \] Now, form the expression: \[ \left(x^{2}+a^{2}\right) y_{2}+x y_{1}. \] Substituting the values of \( y_{1} \) and \( y_{2} \): \[ \left(x^{2}+a^{2}\right) \left(-\frac{x}{\left(x^{2}+a^{2}\right)^{3/2}}\right) + x\left(\frac{1}{\sqrt{x^{2}+a^{2}}}\right). \] Simplify the first term: \[ \left(x^{2}+a^{2}\right)\left(-\frac{x}{\left(x^{2}+a^{2}\right)^{3/2}}\right) = -\frac{x}{\sqrt{x^{2}+a^{2}}}, \] and the second term is: \[ \frac{x}{\sqrt{x^{2}+a^{2}}}. \] Adding these terms yields: \[ -\frac{x}{\sqrt{x^{2}+a^{2}}}+\frac{x}{\sqrt{x^{2}+a^{2}}}=0. \]

Solution: We start with the function \[ y=\cos (\sin x). \] First, differentiate with respect to \( x \): \[ y_{1}=\frac{dy}{dx} = -\sin (\sin x) \cdot \frac{d}{dx}(\sin x) = -\sin (\sin x) \cos x. \] Next, differentiate \( y_{1} \) to obtain \( y_{2} \): \[ y_{2}=\frac{d}{dx}\left[-\sin (\sin x) \cos x\right]. \] Using the product rule: \[ y_{2}=-\left\{ \frac{d}{dx}\left[\sin (\sin x)\right] \cos x + \sin (\sin x) \frac{d}{dx}(\cos x) \right\}. \] Compute the derivative of \(\sin (\sin x)\) by the chain rule: \[ \frac{d}{dx}\left[\sin (\sin x)\right]=\cos (\sin x) \cdot \cos x. \] Also, note that \[ \frac{d}{dx}(\cos x)=-\sin x. \] Therefore, \[ y_{2}=-\left[ \cos (\sin x) \cos x \cdot \cos x + \sin (\sin x) \cdot (-\sin x) \right]. \] Simplify: \[ y_{2}=-\left[ \cos (\sin x) \cos^{2} x – \sin (\sin x) \sin x \right] = -\cos (\sin x) \cos^{2} x + \sin (\sin x) \sin x. \] Now, we need to show that \[ y_{2}+\tan x\, y_{1}+y \cos^{2} x=0. \] Substitute the values obtained: \[ y_{2} = -\cos (\sin x) \cos^{2} x + \sin (\sin x) \sin x, \] \[ y_{1} = -\sin (\sin x) \cos x, \] \[ y = \cos (\sin x). \] Also, note that \[ \tan x = \frac{\sin x}{\cos x}. \] Compute \(\tan x\, y_{1}\): \[ \tan x\, y_{1}=\frac{\sin x}{\cos x}\left(-\sin (\sin x) \cos x\right) = -\sin x\, \sin (\sin x). \] Next, compute \(y \cos^{2} x\): \[ y \cos^{2} x=\cos (\sin x) \cos^{2} x. \] Now, add the three terms: \[ y_{2}+\tan x\, y_{1}+y \cos^{2} x = \left[-\cos (\sin x) \cos^{2} x + \sin (\sin x) \sin x\right] -\sin x\, \sin (\sin x) +\cos (\sin x) \cos^{2} x. \] Observe that the terms \(-\cos (\sin x) \cos^{2} x\) and \(+\cos (\sin x) \cos^{2} x\) cancel each other, and the terms \(\sin (\sin x) \sin x\) and \(-\sin x\, \sin (\sin x)\) also cancel: \[ y_{2}+\tan x\, y_{1}+y \cos^{2} x=0. \]

Solution: Let \[ u=\frac{1}{a}\log y,\quad \text{so that}\quad y=e^{au}. \] Since \[ x=\tan u, \] we have \[ \frac{dx}{du}=\sec^{2}u. \] Also, \[ \frac{dy}{du}=a\,e^{au}=a\,y. \] By the chain rule, \[ y_{1}=\frac{dy}{dx}=\frac{dy/du}{dx/du}=\frac{a\,y}{\sec^{2}u}=a\,y\cos^{2}u. \] But since \[ \cos^{2}u=\frac{1}{1+\tan^{2}u}=\frac{1}{1+x^{2}}, \] it follows that \[ y_{1}=\frac{a\,y}{1+x^{2}}. \] Now, differentiate \( y_{1}=\frac{a\,y}{1+x^{2}} \) with respect to \( x \) using the quotient rule: \[ y_{2}=\frac{d^{2}y}{dx^{2}}=\frac{a\,y_{1}(1+x^{2})-a\,y\cdot2x}{(1+x^{2})^{2}}. \] Substituting \( y_{1}=\frac{a\,y}{1+x^{2}} \) into the equation, we get: \[ y_{2}=\frac{a\left(\frac{a\,y}{1+x^{2}}\right)(1+x^{2})-2ax\,y}{(1+x^{2})^{2}} =\frac{a^{2}y-2ax\,y}{(1+x^{2})^{2}} =\frac{a\,y(a-2x)}{(1+x^{2})^{2}}. \] Now, form the expression: \[ \left(1+x^{2}\right)y_{2}+(2x-a)y_{1}. \] Substituting the expressions for \( y_{1} \) and \( y_{2} \): \[ \left(1+x^{2}\right)y_{2}=\left(1+x^{2}\right)\frac{a\,y(a-2x)}{(1+x^{2})^{2}} =\frac{a\,y(a-2x)}{1+x^{2}}, \] and \[ (2x-a)y_{1}=(2x-a)\frac{a\,y}{1+x^{2}}. \] Adding these terms, we have: \[ \left(1+x^{2}\right)y_{2}+(2x-a)y_{1}=\frac{a\,y(a-2x)+(2x-a)a\,y}{1+x^{2}} =\frac{a\,y\left[(a-2x)+(2x-a)\right]}{1+x^{2}}. \] Since \[ (a-2x)+(2x-a)=0, \] it follows that \[ \left(1+x^{2}\right)y_{2}+(2x-a)y_{1}=0. \]

Solution: We start with the function \[ y = e^{a\cos^{-1} x}. \] Differentiate with respect to \( x \) using the chain rule. Recall that \[ \frac{d}{dx}\left(\cos^{-1} x\right) = -\frac{1}{\sqrt{1-x^{2}}}. \] Thus, the first derivative is \[ \frac{dy}{dx} = e^{a\cos^{-1} x} \cdot a \left(-\frac{1}{\sqrt{1-x^{2}}}\right) = -\frac{a\, e^{a\cos^{-1} x}}{\sqrt{1-x^{2}}} = -\frac{a}{\sqrt{1-x^{2}}}\, y. \] Next, differentiate \( \frac{dy}{dx} = -\frac{a}{\sqrt{1-x^{2}}}\, y \) to find the second derivative: \[ \frac{d^{2}y}{dx^{2}} = \frac{d}{dx}\left(-\frac{a}{\sqrt{1-x^{2}}}\, y\right) = -a\, \frac{d}{dx}\left(\frac{y}{\sqrt{1-x^{2}}}\right). \] Using the quotient rule, we have \[ \frac{d}{dx}\left(\frac{y}{\sqrt{1-x^{2}}}\right) = \frac{y’}{\sqrt{1-x^{2}}}+ y\, \frac{d}{dx}\left((1-x^{2})^{-1/2}\right). \] Note that \[ \frac{d}{dx}\left((1-x^{2})^{-1/2}\right) = -\frac{1}{2}(1-x^{2})^{-3/2}(-2x) = \frac{x}{(1-x^{2})^{3/2}}. \] Substituting \( y’ = -\frac{a}{\sqrt{1-x^{2}}}\, y \) gives \[ \frac{d}{dx}\left(\frac{y}{\sqrt{1-x^{2}}}\right) = -\frac{a\, y}{1-x^{2}} + \frac{x\, y}{(1-x^{2})^{3/2}}. \] Hence, the second derivative becomes \[ \frac{d^{2}y}{dx^{2}} = -a \left[-\frac{a\, y}{1-x^{2}} + \frac{x\, y}{(1-x^{2})^{3/2}}\right] = \frac{a^{2} y}{1-x^{2}} – \frac{a x\, y}{(1-x^{2})^{3/2}}. \] Now, multiply this expression by \((1-x^{2})\): \[ (1-x^{2})\frac{d^{2}y}{dx^{2}} = a^{2}y – \frac{a x\, y}{\sqrt{1-x^{2}}}. \] Next, we have from the first derivative: \[ \frac{dy}{dx} = -\frac{a\, y}{\sqrt{1-x^{2}}} \quad \Rightarrow \quad -x\frac{dy}{dx} = -x\left(-\frac{a\, y}{\sqrt{1-x^{2}}}\right) = \frac{a x\, y}{\sqrt{1-x^{2}}}. \] Adding the two results yields: \[ (1-x^{2})\frac{d^{2}y}{dx^{2}} – x\frac{dy}{dx} = a^{2}y – \frac{a x\, y}{\sqrt{1-x^{2}}} + \frac{a x\, y}{\sqrt{1-x^{2}}} = a^{2}y. \] Rearranging, we obtain: \[ (1-x^{2})\frac{d^{2}y}{dx^{2}} – x\frac{dy}{dx} – a^{2}y = 0. \] Thus, the given differential equation is verified.

Solution: First, observe that \[ y=\log\left(\frac{x}{a+bx}\right)^{x}= x\log\left(\frac{x}{a+bx}\right). \] Thus, we have \[ \frac{y}{x}=\log x-\log(a+bx). \] Let \[ f(x)=\log x-\log(a+bx). \] Then, \[ y=x\,f(x). \] **Step 1. Compute \( \frac{dy}{dx}=y_{1} \):** By the product rule: \[ y_{1}=\frac{dy}{dx} = f(x) + x\,f'(x). \] We know: \[ f(x)=\log x-\log(a+bx), \] hence, \[ f'(x)=\frac{1}{x} – \frac{b}{a+bx}. \] Therefore, \[ y_{1}=\log x-\log(a+bx) + x\left(\frac{1}{x}-\frac{b}{a+bx}\right) = \log x-\log(a+bx) + 1 – \frac{b\,x}{a+bx}. \] **Step 2. Compute \( x\frac{dy}{dx}-y \):** Notice that \[ x\frac{dy}{dx} = x\,y_{1} = x\left[\log x-\log(a+bx)+1-\frac{b\,x}{a+bx}\right], \] and since \[ y=x\left[\log x-\log(a+bx)\right], \] we have \[ x\frac{dy}{dx}-y = x\left[\log x-\log(a+bx)+1-\frac{b\,x}{a+bx}\right] – x\left[\log x-\log(a+bx)\right]. \] This simplifies to: \[ x\frac{dy}{dx}-y = x\left[1-\frac{b\,x}{a+bx}\right]. \] Write the term inside the bracket as: \[ 1-\frac{b\,x}{a+bx}=\frac{a+bx-bx}{a+bx}=\frac{a}{a+bx}. \] Thus, \[ x\frac{dy}{dx}-y=\frac{a\,x}{a+bx}. \] **Step 3. Compute \( \frac{d^{2} y}{dx^{2}}=y_{2} \):** Differentiate \( y_{1} \) with respect to \( x \): \[ y_{1}=\log x-\log(a+bx)+1-\frac{b\,x}{a+bx}. \] Differentiating term by term: – \(\frac{d}{dx}(\log x)=\frac{1}{x},\) – \(\frac{d}{dx}[-\log(a+bx)] = -\frac{b}{a+bx},\) – \(\frac{d}{dx}(1)=0,\) – For \(-\frac{b\,x}{a+bx}\), apply the quotient rule: \[ \frac{d}{dx}\left(\frac{b\,x}{a+bx}\right)=\frac{b(a+bx)-b\,x\cdot b}{(a+bx)^2}=\frac{ab}{(a+bx)^2}. \] Therefore, \[ \frac{d}{dx}\left(-\frac{b\,x}{a+bx}\right)=-\frac{ab}{(a+bx)^2}. \] Hence, \[ y_{2}=\frac{1}{x}-\frac{b}{a+bx}-\frac{ab}{(a+bx)^2}. \] **Step 4. Form the expression \( x^{3}y_{2} \):** Multiply \( y_{2} \) by \( x^{3} \): \[ x^{3}y_{2} = x^{3}\left[\frac{1}{x}-\frac{b}{a+bx}-\frac{ab}{(a+bx)^2}\right] = x^{2} – \frac{b\,x^{3}}{a+bx} – \frac{ab\,x^{3}}{(a+bx)^2}. \] **Step 5. Compute \( \left(x\frac{dy}{dx}-y\right)^{2} \):** We found earlier that \[ x\frac{dy}{dx}-y=\frac{a\,x}{a+bx}. \] Squaring both sides: \[ \left(x\frac{dy}{dx}-y\right)^{2} = \frac{a^{2}x^{2}}{(a+bx)^{2}}. \] **Step 6. Show that \( x^{3}y_{2}=\left(x\frac{dy}{dx}-y\right)^{2} \):** Express \( x^{3}y_{2} \) over the common denominator \( (a+bx)^{2} \): \[ x^{3}y_{2} = \frac{x^{2}(a+bx)^{2} – b\,x^{3}(a+bx) – ab\,x^{3}}{(a+bx)^{2}}. \] Expand \( (a+bx)^{2}=a^{2}+2abx+b^{2}x^{2} \): \[ x^{2}(a^{2}+2abx+b^{2}x^{2})- b\,x^{3}(a+bx)-ab\,x^{3}. \] Compute: \[ x^{2}(a^{2}+2abx+b^{2}x^{2}) = a^{2}x^{2}+2ab\,x^{3}+b^{2}x^{4}, \] and \[ b\,x^{3}(a+bx)= ab\,x^{3}+b^{2}x^{4}. \] Thus, the numerator becomes: \[ a^{2}x^{2}+2ab\,x^{3}+b^{2}x^{4} – ab\,x^{3} – b^{2}x^{4} – ab\,x^{3} = a^{2}x^{2}. \] Therefore, \[ x^{3}y_{2} = \frac{a^{2}x^{2}}{(a+bx)^{2}}, \] which matches exactly with \[ \left(x\frac{dy}{dx}-y\right)^{2}. \] Thus, we have shown that \[ x^{3}\frac{d^{2}y}{dx^{2}}=\left(x\frac{dy}{dx}-y\right)^{2}. \]

Solution: We begin with \[ \sqrt{x}+\sqrt{y}=\sqrt{a}. \] Isolate \(\sqrt{y}\) to express \(y\) explicitly: \[ \sqrt{y}=\sqrt{a}-\sqrt{x}. \] Squaring both sides, we obtain \[ y=\left(\sqrt{a}-\sqrt{x}\right)^{2}=a-2\sqrt{a}\sqrt{x}+x. \] **Step 1. Compute the first derivative \( \frac{dy}{dx} \):** Differentiating term by term: \[ \frac{d}{dx}(a)=0,\quad \frac{d}{dx}\left(-2\sqrt{a}\sqrt{x}\right)=-2\sqrt{a}\cdot\frac{1}{2\sqrt{x}}=-\frac{\sqrt{a}}{\sqrt{x}},\quad \frac{d}{dx}(x)=1. \] Therefore, \[ y’ = \frac{dy}{dx} = -\frac{\sqrt{a}}{\sqrt{x}}+1. \] **Step 2. Compute the second derivative \( \frac{d^{2}y}{dx^{2}} \):** Differentiate \( y’ \) with respect to \( x \): \[ \frac{d}{dx}\left(-\frac{\sqrt{a}}{\sqrt{x}}\right) = -\sqrt{a}\cdot \frac{d}{dx}\left(x^{-1/2}\right) = -\sqrt{a}\left(-\frac{1}{2}x^{-3/2}\right) = \frac{\sqrt{a}}{2x^{3/2}}. \] The derivative of the constant 1 is 0, so: \[ y” = \frac{d^{2}y}{dx^{2}} = \frac{\sqrt{a}}{2x^{3/2}}. \] **Step 3. Evaluate at \( x=a \):** Substitute \( x=a \) into the expression for \( y” \): \[ y”\Big|_{x=a} = \frac{\sqrt{a}}{2a^{3/2}} = \frac{1}{2a}. \]

Solution: Taking natural logarithms on both sides of \[ x^{m}y^{n}=(x+y)^{m+n}, \] we obtain \[ m\ln x+n\ln y=(m+n)\ln(x+y). \] Differentiate both sides with respect to \( x \): \[ \frac{m}{x}+\frac{n}{y}\frac{dy}{dx}=\frac{m+n}{x+y}\left(1+\frac{dy}{dx}\right). \] Let \( y_{1}=\frac{dy}{dx} \). Rearranging the equation: \[ \frac{m}{x}+\frac{n}{y}y_{1}=\frac{m+n}{x+y}+\frac{m+n}{x+y}y_{1}. \] We now make an ansatz that \[ y_{1}=\frac{y}{x}. \] Substitute \( y_{1}=\frac{y}{x} \) into the left-hand side: \[ \frac{m}{x}+\frac{n}{y}\cdot\frac{y}{x}=\frac{m+n}{x}. \] And into the right-hand side: \[ \frac{m+n}{x+y}\left(1+\frac{y}{x}\right) =\frac{m+n}{x+y}\cdot\frac{x+y}{x}=\frac{m+n}{x}. \] Since both sides are equal, we conclude that \[ \frac{dy}{dx}=\frac{y}{x}. \] For part (ii), differentiate \[ \frac{dy}{dx}=\frac{y}{x} \] with respect to \( x \). Writing \( y_{1}=\frac{dy}{dx} \), we have: \[ y_{1}=\frac{y}{x}. \] Differentiating by the quotient rule: \[ \frac{d^{2}y}{dx^{2}}=y_{2}=\frac{x\frac{dy}{dx}-y}{x^{2}}. \] Substitute \(\frac{dy}{dx}=\frac{y}{x}\) into the expression: \[ y_{2}=\frac{x\left(\frac{y}{x}\right)-y}{x^{2}} =\frac{y-y}{x^{2}}=0. \]

Solution: We begin with the given equation \[ x\cos(a+y)=\cos y. \] **Step 1. Differentiate once to obtain an expression for \(\frac{dy}{dx}\):** Differentiate both sides with respect to \( x \). Using the product rule on the left-hand side and the chain rule, we have: \[ \frac{d}{dx}\left[x\cos(a+y)\right] = \cos(a+y) – x\sin(a+y)\frac{dy}{dx}, \] and for the right-hand side: \[ \frac{d}{dx}[\cos y] = -\sin y\,\frac{dy}{dx}. \] Hence, the differentiated equation is: \[ \cos(a+y) – x\sin(a+y)\frac{dy}{dx} = -\sin y\,\frac{dy}{dx}. \] Rearranging, \[ \cos(a+y) = \frac{dy}{dx}\Bigl[x\sin(a+y)-\sin y\Bigr]. \] We denote \[ y_{1}=\frac{dy}{dx}. \] Thus, \[ y_{1}=\frac{\cos(a+y)}{x\sin(a+y)-\sin y}. \] **Step 2. Differentiate a second time to obtain \(\frac{d^{2} y}{dx^{2}}\):** Differentiate the relation \[ \cos(a+y)=y_{1}\Bigl[x\sin(a+y)-\sin y\Bigr] \] with respect to \( x \). The derivative of the left-hand side is obtained using the chain rule: \[ \frac{d}{dx}\left[\cos(a+y)\right] = -\sin(a+y)\frac{d}{dx}(a+y) = -\sin(a+y)y_{1}, \] since \( a \) is constant. For the right-hand side, apply the product rule: \[ \frac{d}{dx}\left[y_{1}\Bigl(x\sin(a+y)-\sin y\Bigr)\right] = y_{2}\Bigl[x\sin(a+y)-\sin y\Bigr] + y_{1}\frac{d}{dx}\Bigl[x\sin(a+y)-\sin y\Bigr], \] where \[ y_{2}=\frac{d^{2}y}{dx^{2}}. \] Now, compute the derivative of the bracketed term: \[ \frac{d}{dx}\left[x\sin(a+y)-\sin y\right] = \sin(a+y) + x\cos(a+y)y_{1} -\cos y\,y_{1}. \] Hence, the second derivative equation becomes: \[ -\sin(a+y)y_{1} = y_{2}\Bigl[x\sin(a+y)-\sin y\Bigr] + y_{1}\left[\sin(a+y) + x\cos(a+y)y_{1} -\cos y\,y_{1}\right]. \] **Step 3. Simplify using the given relationship:** Notice from the original equation that \[ x\cos(a+y)=\cos y, \] which implies \[ x\cos(a+y)-\cos y=0. \] This observation will cancel the terms involving \( y_{1}^{2} \) in the differentiated equation. In fact, we can regroup the equation as: \[ y_{2}\Bigl[x\sin(a+y)-\sin y\Bigr] = -\sin(a+y)y_{1} – y_{1}\sin(a+y) = -2\sin(a+y)y_{1}. \] Next, express the bracket \( x\sin(a+y)-\sin y \) in a more convenient form. Using the sine subtraction formula, write: \[ x\sin(a+y)-\sin y = \frac{\cos y}{\cos(a+y)}\sin(a+y)-\sin y = \frac{\sin(a+y)\cos y -\sin y\cos(a+y)}{\cos(a+y)} = \frac{\sin\bigl((a+y)-y\bigr)}{\cos(a+y)} = \frac{\sin a}{\cos(a+y)}. \] Thus, the equation for \( y_{2} \) becomes: \[ y_{2}\left(\frac{\sin a}{\cos(a+y)}\right) = -2\sin(a+y)y_{1}. \] Multiply both sides by \(\cos(a+y)\): \[ \sin a\,y_{2} = -2\sin(a+y)\cos(a+y)y_{1}. \] Recognize that \[ 2\sin(a+y)\cos(a+y)=\sin2(a+y). \] Therefore, we obtain: \[ \sin a\,y_{2} = -\sin2(a+y)y_{1}. \] Rearranging, we finally have: \[ \sin a\,y_{2}+\sin2(a+y)y_{1}=0. \]

Solution: We are given the parametric equations: \[ x = a\left(\theta-\sin \theta\right), \quad y = a\left(1+\cos \theta\right). \] **Step 1. Compute the first derivative \(\frac{dy}{dx}\):** Differentiate with respect to \(\theta\): \[ \frac{dx}{d\theta} = a\left(1-\cos \theta\right),\quad \frac{dy}{d\theta} = -a\sin \theta. \] Thus, \[ \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{-a\sin \theta}{a\left(1-\cos \theta\right)} = -\frac{\sin \theta}{1-\cos \theta}. \] Using the half-angle identities: \[ 1-\cos \theta=2\sin^2\frac{\theta}{2},\quad \sin \theta=2\sin\frac{\theta}{2}\cos\frac{\theta}{2}, \] we simplify: \[ \frac{dy}{dx} = -\frac{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2\sin^2\frac{\theta}{2}} = -\cot\frac{\theta}{2}. \] **Step 2. Compute the second derivative \(\frac{d^{2}y}{dx^{2}}\):** Differentiate \(\frac{dy}{dx}=-\cot\frac{\theta}{2}\) with respect to \(\theta\): \[ \frac{d}{d\theta}\left(-\cot\frac{\theta}{2}\right) = -\left(-\csc^2\frac{\theta}{2}\cdot \frac{1}{2}\right) = \frac{1}{2}\csc^2\frac{\theta}{2}. \] Then, by the chain rule: \[ \frac{d^{2}y}{dx^{2}}=\frac{\frac{d}{d\theta}\left(\frac{dy}{dx}\right)}{\frac{dx}{d\theta}} =\frac{\frac{1}{2}\csc^2\frac{\theta}{2}}{a\left(1-\cos \theta\right)}. \] Substitute \( 1-\cos \theta = 2\sin^2\frac{\theta}{2} \): \[ \frac{d^{2}y}{dx^{2}}=\frac{\frac{1}{2}\csc^2\frac{\theta}{2}}{a\cdot 2\sin^2\frac{\theta}{2}} =\frac{1}{4a}\cdot\frac{1}{\sin^2\frac{\theta}{2}\,\sin^2\frac{\theta}{2}} =\frac{1}{4a}\,\mathrm{cosec}^{4}\left(\frac{\theta}{2}\right). \]

Solution: Since \[ x=\log t, \] we have \[ t=e^{x}. \] Given \[ y=\frac{1}{t}, \] then in terms of \( x \), the function becomes: \[ y=\frac{1}{e^{x}}=e^{-x}. \] Now, differentiate \( y \) with respect to \( x \): \[ \frac{dy}{dx}=\frac{d}{dx}\left(e^{-x}\right)=-e^{-x}. \] Next, differentiate the first derivative to find the second derivative: \[ \frac{d^{2}y}{dx^{2}}=\frac{d}{dx}\left(-e^{-x}\right)=e^{-x}. \] Adding the two derivatives gives: \[ \frac{d^{2}y}{dx^{2}}+\frac{dy}{dx}=e^{-x}-e^{-x}=0. \]

Solution: We first compute the derivatives with respect to the parameter \( t \). Since \[ x=a\sin(pt), \quad \text{we have} \quad \frac{dx}{dt}=a p\cos(pt), \] and \[ y=b\cos(pt), \quad \text{so} \quad \frac{dy}{dt}=-b p\sin(pt). \] Then, the first derivative is given by: \[ \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{-b p\sin(pt)}{a p\cos(pt)}=-\frac{b}{a}\tan(pt). \] To find the second derivative, differentiate \(\frac{dy}{dx}\) with respect to \(t\) and multiply by \(\frac{dt}{dx}\): \[ \frac{d^2y}{dx^2}=\frac{d}{dt}\left(-\frac{b}{a}\tan(pt)\right)\cdot\frac{dt}{dx}. \] Differentiate the first derivative with respect to \(t\): \[ \frac{d}{dt}\left(-\frac{b}{a}\tan(pt)\right)=-\frac{b}{a} \cdot p\sec^2(pt). \] Since \[ \frac{dx}{dt}=a p\cos(pt), \quad \text{we have} \quad \frac{dt}{dx}=\frac{1}{a p\cos(pt)}. \] Therefore, \[ \frac{d^2y}{dx^2}=-\frac{b}{a}\, p\,\sec^2(pt)\cdot\frac{1}{a p\cos(pt)} =-\frac{b}{a^2}\cdot\frac{\sec^2(pt)}{\cos(pt)} =-\frac{b}{a^2}\cdot\frac{1}{\cos^3(pt)}. \] Evaluating at \( t=0 \): \[ \cos(p\cdot0)=1 \quad \text{and thus} \quad \frac{d^2y}{dx^2}\Big|_{t=0}=-\frac{b}{a^2}. \]

Solution: We first compute the derivatives with respect to \( t \). Since \[ x=a(1+\cos t), \quad \frac{dx}{dt}=a(-\sin t), \] and \[ y=a(t+\sin t), \quad \frac{dy}{dt}=a\left(1+\cos t\right). \] Hence, the first derivative is \[ \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{a(1+\cos t)}{a(-\sin t)} =-\frac{1+\cos t}{\sin t}. \] Next, we differentiate \(\frac{dy}{dx}\) with respect to \( t \): \[ \frac{d}{dt}\left(-\frac{1+\cos t}{\sin t}\right) = -\frac{d}{dt}\left(\frac{1+\cos t}{\sin t}\right). \] Using the quotient rule with \( f(t)=1+\cos t \) and \( g(t)=\sin t \): \[ f'(t)=-\sin t, \quad g'(t)=\cos t, \] we have \[ \frac{d}{dt}\left(\frac{1+\cos t}{\sin t}\right) = \frac{f'(t)g(t)-f(t)g'(t)}{g^2(t)} = \frac{(-\sin t)(\sin t)-(1+\cos t)(\cos t)}{\sin^2 t}. \] Simplifying the numerator: \[ -\sin^2 t – (1+\cos t)\cos t = -\sin^2 t -\cos t -\cos^2 t. \] Since \(\sin^2 t + \cos^2 t =1\), it follows that \[ -\sin^2 t -\cos^2 t = -1. \] Thus, the numerator becomes: \[ -1-\cos t. \] Therefore, \[ \frac{d}{dt}\left(\frac{1+\cos t}{\sin t}\right) = \frac{-1-\cos t}{\sin^2 t}. \] Including the negative sign from earlier, we get: \[ \frac{d}{dt}\left(\frac{dy}{dx}\right) = -\left(\frac{-1-\cos t}{\sin^2 t}\right) = \frac{1+\cos t}{\sin^2 t}. \] Now, the second derivative is given by: \[ \frac{d^2y}{dx^2}=\frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}} = \frac{\frac{1+\cos t}{\sin^2 t}}{a(-\sin t)} = -\frac{1+\cos t}{a\sin^3 t}. \] Evaluating at \( t=\frac{\pi}{2} \): \[ \cos \frac{\pi}{2}=0, \quad \sin \frac{\pi}{2}=1. \] Therefore, \[ \frac{d^2y}{dx^2}\Bigg|_{t=\frac{\pi}{2}} = -\frac{1+0}{a(1)^3} = -\frac{1}{a}. \]