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Question Number 202153 by sonukgindia last updated on 22/Dec/23

	Answered by witcher3 last updated on 22/Dec/23
	$ln(x)y′+(y/x)=(ln(x))y′+(ln(x))′y =(d/dx)(ln(x)y) ⇔ { (((d/dx)(y.(ln(x))=x^2 sin(x))),((y(π)=(π^2 /(ln(π))))) :} ⇒∫_π ^z (d/dx)(y.ln(x))=∫_π ^z x^2 sin(x)dx ⇒y(z)ln(z)−π^2 =−x^2 cos(x)−2xsin(x)−2cos(x)]_π ^z y(z)ln(z)−π^2 =−z^2 cos(z)−2zsin(z)−2cos(z)−π^2 −2 y(z)=(1/(ln(z)))(−2−2cos(z)−2zsin(z)−z^2 cos(z)]$
	$\ln (x) y^{'} + \frac{y}{x} = (\ln (x)) y^{'} + {(\ln (x))}^{'} y$ $= \frac{d}{dx} (\ln (x) y)$ $\Leftrightarrow {\begin{cases} \frac{d}{dx} (y . (\ln (x)) = x^{2} \sin (x) \\ y (π) = \frac{π^{2}}{\ln (π)} \end{cases}$ $\Rightarrow \int_{π}^{z} \frac{d}{dx} (y . \ln (x)) = \int_{π}^{z} x^{2} \sin (x) dx$ ${\Rightarrow y (z) \ln (z) - π^{2} = - x^{2} \cos (x) - 2 xsin (x) - 2 \cos (x)]}_{π}^{z}$ $y (z) \ln (z) - π^{2} = - z^{2} \cos (z) - 2 zsin (z) - 2 \cos (z) - π^{2} - 2$ $y (z) = \frac{1}{\ln (z)} (- 2 - 2 \cos (z) - 2 zsin (z) - z^{2} \cos (z)]$
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