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

The Correct Option is C

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
\((x² - 2x)e^x = t\)
\(∴ (x² - 2)e^x dx = dt\)
\(y.e(x² - 2x)e^x = ∫ t.e^tdt\)
\(y.e(x² - 2x)e^x = (x² - 2x - 1)e^{(x² - 2x)e^x} + c\)
\(∴ y(0) = 0\)
\(∴ c = 1\)
\(∴ y = (x² - 2x - 1) + e(2x - x²)e^x\)
\(∴ y(2) = -1 + 1 = 0\)