Question

If the solution of the differential equation
$\frac{dy}{dx}$ $+ e^x(x² - 2)y = (x^2 - 2x)(x^2 - 2)e^{2x}$
satisfies y(0) = 0, then the value of y(2) is ______.

• A. -1
• B. 1
• C. 0
• D. $e$

Answer 1

Answer: (C)

0

Answer 2

The correct answer is (C) : 0
$\frac{dy}{dx} + e^x(x^2-2)y=(x^2-2x)(x^2-2)e^{2x }$
Here, I.F.
$=$ $e^{\int{e^x(x² - 2)dx}}$
$=$ $e^{(x² - 2x)e^x}$
∴ Solution of the differential equation is
$y.e^{(x² - 2x)e^x} = \int{(x² - 2x)(x² - 2)e^{2x}.e^{(x² - 2x)e^x }dx}$
$= \int{ (x² - 2x)e^x.(x² - 2)e^x.e(x² - 2x)e^x dx}$
Let
$(x² - 2x)e^x = t$
$∴ (x² - 2)e^x dx = dt$
$y.e(x² - 2x)e^x = ∫ t.e^tdt$
$y.e(x² - 2x)e^x = (x² - 2x - 1)e^{(x² - 2x)e^x} + c$
$∴ y(0) = 0$
$∴ c = 1$
$∴ y = (x² - 2x - 1) + e(2x - x²)e^x$
$∴ y(2) = -1 + 1 = 0$