Question

Solve the differential equation: $x\frac{dy}{dx}+y\log y=xy{e}^x.$

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

Answer 1

Answer: (D)

$x\log y=e^x(x-1)+c.$

$x\frac{dy}{dx}+y\log y=xy{e}^x\Rightarrow \frac{1}{y}\frac{dy}{dx}+\frac{\log y}{x}=e^x$
Put $\log y=v\Rightarrow \frac{dy}{y}=dv$
$\therefore \frac{dv}{dx}+\frac{v}{x}=e^x$ ...(1)
Here $P=\frac{1}{x}\Rightarrow \int Pdx=\int \frac{1}{x}dx=\log x$
$\therefore I.F.=e^{\log x}=x$
Multiplying (1) by $I.F.$ we get
$x\frac{dv}{dx}+v={xe}^x$
Integrating both sides we get
$xv=\int {xe}^{x}dx+c={xe}^x-e^x+c$

$\Rightarrow x\log y=e^x(x-1)+c$