Question

The solution of $x\frac{dy}{dx}+y\log y=xy{e}^x$ is

• A. $x\log y=(x+1)e^x+c$
• B. $\log y=(x-1)e^x+c$
• C. $(x-1)\log y=x{e}^x+c$
• D. $x\log y=(x-1)e^x+c$

Answer 1

The correct option is D $x\log y=(x-1)e^x+c$

$x\frac{dy}{dx}+y\log y=xy{e}^x$

dividing both sides by $y$

$\frac{x}{y}\frac{dy}{dx}+\log y=xy{e}^x$

putting $\log y=t$

$\frac{1}{y}\frac{dy}{dx}=\frac{dt}{dx}$

$=\frac{xdt}{dx}+t=x{e}^x$

$=\frac{dt}{dx}+\frac{t}{x}=e^x$

$p=\frac{1}{x}Q=e^x$

$I.F=e^{{\int }^{}\frac{1}{x}dx}=e^{\log x}=x$

solution of given D.E. is $t\times I.F={\int }^{}Q.I.Fdx+c$

$t\times I.F={\int }^{}{e^{}}^{x}xdx+c$

$x\log =x{e}^x{\int }^{}i.e^{x}dx+c$

$x\log =x{e}^x-e^x+c=x\log y=e^x\left(x-1\right)+c$