The solution of the differential equation x f | vedclass

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Question

(B) Given the differential equation $x \frac{d^2y}{dx^2} = 1$. Dividing by $x$,we get $\frac{d^2y}{dx^2} = \frac{1}{x}$. Integrating with respect to $

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$y = x \log x - x + 2$

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(B) Given the differential equation $x \frac{d^2y}{dx^2} = 1$. Dividing by $x$,we get $\frac{d^2y}{dx^2} = \frac{1}{x}$. Integrating with respect to $x$,we get $\frac{dy}{dx} = \int \frac{1}{x} dx = \log x + c_1$. Given that $\frac{dy}{dx} = 0$ when $x = 1$,we have $0 = \log(1) + c_1$,which implies $c_1 = 0$. So,$\frac{dy}{dx} = \log x$. Integrating again with respect to $x$,we get $y = \int \log x dx = x \log x - x + c_2$. Given that $y = 1$ when $x = 1$,we have $1 = 1 \log(1) - 1 + c_2$,which implies $1 = 0 - 1 + c_2$,so $c_2 = 2$. Substituting the values of $c_1$ and $c_2$,the required solution is $y = x \log x - x + 2$.

The solution of the differential equation $x \frac{d^2y}{dx^2} = 1$,given that $y = 1$ and $\frac{dy}{dx} = 0$ when $x = 1$,is

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