Question

$x\frac{dy}{dx}+2y=lnx$ then $e^2.y\left(e\right)-y\left(1\right)=$?

Answer 1

$x\frac{dy}{dx}+2y=lnx$

$\frac{dy}{dx}+\frac{2y}{x}=\frac{\ln x}{x}$

It is first order linear differential equation

$P=\frac{2}{x},Q=\frac{\ln x}{x}$

$I.F.=e^{\int P.dx}=e^{\int \frac{2}{x}dx}=e^{2\ln x}=x^2$

solution of D.E. is

$y\times I.F.=\int Q\times I.F.dx+C$

$y\times x^2=\int \frac{\ln x}{x}\times x^{2}dx+C$

Putting $\ln x==t,x=e^t,\frac{1}{x}dx=dt$

$y{x}^2=\int t.e^{2t}dt+C$

Integration by parts taking $t$ as first function

$x^{2}y=t\frac{e^{2t}}{2}-\int 1.\frac{e^{2t}}{2}dt+C$

$x^{2}y=\frac{t{e}^{2t}}{2}-\frac{e^{2t}}{4}+C$

$x^{2}y=\frac{\ln x\times x^2}{2}-\frac{x^2}{4}+C$

Putting $x=e$, we get

$e^{2}y\left(e\right)=\frac{e^2}{2}-\frac{e^2}{4}+C$ ---- (i)

Putting $x=1$, we get

$y\left(1\right)=-\frac{1}{4}+C$ ----- (ii)

Subtracting (i) from (ii)

$e^{2}y\left(e\right)-y\left(1\right)=\frac{e^2+1}{4}$