Question

If y = y(x) is the solution of the differential equation
$x$ $\frac{dy}{dx}$ $+ 2y =$ $xe^x , y(1) = 0$
then the local maximum value of the function
$z(x) = x²y(x) - e^x , x ∈ R$
is

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

Answer 1

Answer: (D)

$\frac{4}{e} - e$

Answer 2

The correct answer is (D) : $\frac{4}{e} - e$
$x \frac{dy}{dx} + 2y = xe^x , y(1) = 0$
\frac{dy}{dx} + \frac{2}{x} y = e^x , then$$e^{\int\frac{2}{x} dx} dx = x²
$y.x² = ∫ x²exdx$
$yx² = x²e^x - ∫ 2xe^xdx$
$= x²e^x - 2(xe^x - e^x ) + c$
$yx² = x²e^x - 2xe^x + 2e^x + c$
$yx² = (x² - 2x + 2)e^x + c$
$0 = e + c ⇒ c = -e$
$y(x).x² - e^x = (x - 1)²e^x - e$
$z(x) = (x - 1)^2e^x - e$

For local maximum z′(x) = 0
∴ $2(x - 1)e^x + (x - 1)^2e^x = 0$
∴$x = -1$
And local maximum value = z(-1)
= $\frac{4}{e} - e$