The solution of the differential equation x f | vedclass

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(B) Given the differential equation: $x \frac{d^2 y}{d x^2} = 1$. Dividing by $x$ (assuming $x 
eq 0$): $\frac{d^2 y}{d x^2} = \frac{1}{x}$. Integrat

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Given the differential equation: $x \frac{d^2 y}{d x^2} = 1$. Dividing by $x$ (assuming $x 
eq 0$): $\frac{d^2 y}{d x^2} = \frac{1}{x}$. Integrating with respect to $x$: $\frac{dy}{dx} = \int \frac{1}{x} dx = \log x + C_1$. Given $\frac{dy}{dx} = 0$ at $x = 1$: $0 = \log(1) + C_1 \implies C_1 = 0$. So,$\frac{dy}{dx} = \log x$. Integrating again with respect to $x$: $y = \int \log x dx = x \log x - x + C_2$. Given $y = 1$ at $x = 1$: $1 = 1 \log(1) - 1 + C_2 \implies 1 = 0 - 1 + C_2 \implies C_2 = 2$. Thus,the solution is $y = x \log x - x + 2$.

The solution of the differential equation $x \frac{d^2 y}{d x^2} = 1$ at $x = y = 1$ with $\frac{dy}{d x} = 0$ at $x = 1$ is:

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