The solution of frac{d^2y}{dx^2} = sec^2 x + | vedclass

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Question

(A) Given the differential equation: $\frac{d^2y}{dx^2} = \sec^2 x + x e^x$. Integrating both sides with respect to $x$: $\frac{dy}{dx} = \int \sec^2

Vedclass · Accepted Answer

$y = \log(\sec x) + (x - 2)e^x + c_1 x + c_2$

Vedclass · Answer

(A) Given the differential equation: $\frac{d^2y}{dx^2} = \sec^2 x + x e^x$. Integrating both sides with respect to $x$: $\frac{dy}{dx} = \int \sec^2 x \, dx + \int x e^x \, dx + c_1$. Using the integral formula $\int x e^x \, dx = x e^x - e^x$: $\frac{dy}{dx} = 	an x + (x e^x - e^x) + c_1$. Integrating again with respect to $x$: $y = \int 	an x \, dx + \int x e^x \, dx - \int e^x \, dx + \int c_1 \, dx + c_2$. Since $\int 	an x \, dx = \log(\sec x)$: $y = \log(\sec x) + (x e^x - e^x) - e^x + c_1 x + c_2$. Simplifying the expression: $y = \log(\sec x) + x e^x - 2e^x + c_1 x + c_2$. $y = \log(\sec x) + (x - 2)e^x + c_1 x + c_2$.

The solution of $\frac{d^2y}{dx^2} = \sec^2 x + x e^x$ is

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