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)]](https://www.tinkutara.com/question/Q202174.png)
$$\mathrm{ln}\left(\mathrm{x}\right)\mathrm{y}'+\frac{\mathrm{y}}{\mathrm{x}}=\left(\mathrm{ln}\left(\mathrm{x}\right)\right)\mathrm{y}'+\left(\mathrm{ln}\left(\mathrm{x}\right)\right)'\mathrm{y} \\ $$$$=\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{ln}\left(\mathrm{x}\right)\mathrm{y}\right) \\ $$$$\Leftrightarrow\begin{cases}{\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{y}.\left(\mathrm{ln}\left(\mathrm{x}\right)\right)=\mathrm{x}^{\mathrm{2}} \mathrm{sin}\left(\mathrm{x}\right)\right.}\\{\mathrm{y}\left(\pi\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{ln}\left(\pi\right)}}\end{cases} \\ $$$$\Rightarrow\int_{\pi} ^{\mathrm{z}} \frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{y}.\mathrm{ln}\left(\mathrm{x}\right)\right)=\int_{\pi} ^{\mathrm{z}} \mathrm{x}^{\mathrm{2}} \mathrm{sin}\left(\mathrm{x}\right)\mathrm{dx} \\ $$$$\left.\Rightarrow\mathrm{y}\left(\mathrm{z}\right)\mathrm{ln}\left(\mathrm{z}\right)−\pi^{\mathrm{2}} =−\mathrm{x}^{\mathrm{2}} \mathrm{cos}\left(\mathrm{x}\right)−\mathrm{2xsin}\left(\mathrm{x}\right)−\mathrm{2cos}\left(\mathrm{x}\right)\right]_{\pi} ^{\mathrm{z}} \\ $$$$\mathrm{y}\left(\mathrm{z}\right)\mathrm{ln}\left(\mathrm{z}\right)−\pi^{\mathrm{2}} =−\mathrm{z}^{\mathrm{2}} \mathrm{cos}\left(\mathrm{z}\right)−\mathrm{2zsin}\left(\mathrm{z}\right)−\mathrm{2cos}\left(\mathrm{z}\right)−\pi^{\mathrm{2}} −\mathrm{2} \\ $$$$\mathrm{y}\left(\mathrm{z}\right)=\frac{\mathrm{1}}{\mathrm{ln}\left(\mathrm{z}\right)}\left(−\mathrm{2}−\mathrm{2cos}\left(\mathrm{z}\right)−\mathrm{2zsin}\left(\mathrm{z}\right)−\mathrm{z}^{\mathrm{2}} \mathrm{cos}\left(\mathrm{z}\right)\right] \\ $$