The solution of the equation frac{x^2 d^2y}{d | vedclass

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Question

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

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$-\frac{1}{2}(\ln x)^2 - \ln x$

Vedclass · Answer

(D) Given the differential equation: $\frac{d^2y}{dx^2} = \frac{\ln x}{x^2}$.Integrating both sides with respect to $x$:$\frac{dy}{dx} = \int \frac{\ln x}{x^2} dx$.Using integration by parts,let $u = \ln x$ and $dv = x^{-2} dx$. Then $du = \frac{1}{x} dx$ and $v = -\frac{1}{x}$.$\frac{dy}{dx} = -\frac{\ln x}{x} - \int (-\frac{1}{x}) \cdot \frac{1}{x} dx = -\frac{\ln x}{x} + \int x^{-2} dx = -\frac{\ln x}{x} - \frac{1}{x} + C_1 = -\frac{\ln x + 1}{x} + C_1$.Given $\frac{dy}{dx} = -1$ at $x = 1$:$-1 = -\frac{\ln(1) + 1}{1} + C_1 \Rightarrow -1 = -1 + C_1 \Rightarrow C_1 = 0$.So,$\frac{dy}{dx} = -\frac{\ln x + 1}{x}$.Integrating again:$y = -\int \frac{\ln x + 1}{x} dx$.Let $t = \ln x$,then $dt = \frac{1}{x} dx$:$y = -\int (t + 1) dt = -(\frac{t^2}{2} + t) + C_2 = -\frac{(\ln x)^2}{2} - \ln x + C_2$.Given $y = 0$ at $x = 1$:$0 = -\frac{(\ln 1)^2}{2} - \ln(1) + C_2 \Rightarrow 0 = 0 - 0 + C_2 \Rightarrow C_2 = 0$.Thus,the solution is $y = -\frac{1}{2}(\ln x)^2 - \ln x$.

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

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