Question

Let \(y=y(x)\) be the solution of the differential equation \(x\frac{dy}{dx}=(2x+3)e^{x}+xy,\ x > 0\). If \(y(e)=(2e+3)e^{e}\), then \(\frac{d^2y}{dx^2}-\frac{dy}{dx}\) at \(x=1\) is equal to :

• A. \(3e\)
• B. \(e\)
• C. 0
• D. \(2e\)

Hint:

Write the equation as \(y'-y=\frac{(2x+3)e^x}{x}\) and multiply by the integrating factor \(e^{-x}\).

Answer 1

The Correct Option is D

Divide by \(x\): \(\frac{dy}{dx}-y=\frac{(2x+3)e^x}{x}=e^x\left(2+\frac3x\right)\).
Integrating factor \(=e^{-x}\), so \(\frac{d}{dx}\left(ye^{-x}\right)=2+\frac3x\).
Hence \(ye^{-x}=2x+3\ln x+C\).
Using \(y(e)=(2e+3)e^{e}\): \(2e+3=2e+3+C\), so \(C=0\) and \(y=e^x(2x+3\ln x)\).
From the equation, \(y'=y+e^x\left(2+\frac3x\right)\). Differentiate: \(y''=y'+e^x\left(2+\frac3x\right)+e^x\left(-\frac{3}{x^2}\right)\).
So \(y''-y'=e^x\left(2+\frac3x-\frac{3}{x^2}\right)\).
At \(x=1\): \(y''-y'=e(2+3-3)=2e\).
Answer: (4) \(2e\)