Question

If $y=y(x)$ is the solution of the differential equation $x\frac{dy}{dx}+2y=x{e}^x,y(1)=0$ then the local maximum value of the function $z(x)=x^{2}y(x)-e^x,x\in R$ is :

• A. $1-e$
• B. $0$
• C. $\frac{1}{2}$
• D. $\frac{4}{e}-e$

Answer 1

Answer: (D)

$x\frac{dy}{dx}+2y=x{e}^x$
$\frac{dy}{dx}+\frac{2y}{x}=e^x$
$I.F.=x^2$
$y.x{x}^2=\int x^2{e}^{x}dx$
$=\int e^x\left(x^2+2x-2x-2+2\right)dx$
$y{x}^2=e^x\left(x^2-2x+2\right)+c$
$y(1)=0$
$0=e(1+0)+c$
$c=-e$
$z(x)=x^{2}y(x)-e^x$
$=e^x\left(x^2-2x+2\right)-e-e^x$
$=e^x(x-1{)}^2-e$
$\frac{dz}{dx}=e^x\cdot 2(x-1)+e^x(x-1{)}^2=0$
$x^x(x-1)(2+x-1)=0$
$e^x(x-1)(x+1)=0$
$x=-1,1$

$x=-1$ local maxima. Then maximum value is
$z(-1)=\frac{4}{e}-e$