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

The Correct Option is D

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)
= \(\frac{4}{e} - e\)