Mathematics - Definite Integration Question with Solution | TestHub

$x\underset{1}{\overset{x}{\int }}y\left(t\right)dt=x\underset{1}{\overset{x}{\int }}ty\left(t\right)dt+ \underset{1}{\overset{x}{\int }}ty\left(t\right)dt$

Differentiate w.r.t.$x$

$\underset{1}{\overset{x}{\int }}y\left(t\right)dt+xy\left(x\right)=\underset{1}{\overset{x}{\int }}ty\left(t\right)dt+x\left[xy\right(x\left)\right]+xy\left(x\right)$

$\underset{1}{\overset{x}{\int }}y\left(t\right)dt=\underset{1}{\overset{x}{\int }}ty\left(t\right)dt+x^{2}y\left(x\right)$

Differentiate again w.r.t.$x$

$y\left(x\right)=xy\left(x\right)+2x y\left(x\right)+x^2{y}^{\text{'}}\left(x\right)$

$\left(1-3x\right)y\left(x\right)=x^2{y}^{\text{'}}\left(x\right)$

$\frac{y^{\text{'}}\left(x\right)}{y\left(x\right)}=\frac{1-3x}{x^2}$

$\frac{1}{y}\frac{dy}{dx}=\frac{1-3x}{x^2}$

Integrating on both sides

$\Rightarrow \ln y=-\frac{1}{x}-3\ln x+C$

$\Rightarrow \ln \left(y{x}^3\right)=-\frac{1}{x}+C$

$\Rightarrow y{x}^3= e^{-\frac{1}{x}+C}$

$\Rightarrow y=\frac{e^{-\frac{1}{x}+C}}{x^3}$

$\Rightarrow y=\frac{C}{x^3}{e}^{-\frac{1}{x}}$