Question

\(y=\mathrm{e}^x(\mathrm{~A} \cos x+\mathrm{B} \sin x)\) is the solution of the differential equation

• A. \(x^{2}\frac{\text{ }\mathrm{d}^{2}y}{\text{ }\mathrm{d}x^{2}}+(1+y^{2})=0\)
• B. \(\frac{\mathrm{d}^{2}y}{\text{ }\mathrm{d}x^{2}}-\frac{\mathrm{d}y}{\text{ }\mathrm{d}x}+y=0\)
• C. \(\frac{\mathrm{d}^{2}y}{\text{ }\mathrm{d}x^{2}}-2\frac{\text{ }\mathrm{d}y}{\text{ }\mathrm{d}x}+2y=0\)
• D. \(x\frac{\text{ }\mathrm{d}^{2}y}{\text{ }\mathrm{d}x^{2}}-2\frac{\text{ }\mathrm{d}y}{\text{ }\mathrm{d}x}+2y=0\)

Hint:

Differentiate twice and eliminate the two arbitrary constants A and B.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
With two arbitrary constants, the differential equation is of second order.

Step 2: First derivative
\(y'=e^x(A\cos x+B\sin x)+e^x(-A\sin x+B\cos x)=y+e^x(-A\sin x+B\cos x)\).

Step 3: Second derivative
Let \(w=e^x(-A\sin x+B\cos x)\), so \(y'=y+w\) and \(w'=w-y\). Then \(y''=y'+w'=y'+(y'-y)-y=2y'-2y\).

Step 4: Result
So \(y''=2y'-2y\), i.e. \(\frac{d^2y}{dx^2}-2\frac{dy}{dx}+2y=0\), option (C).

Final Answer:
The differential equation is \(y''-2y'+2y=0\). \[ \boxed{\text{(C) }\dfrac{d^{2}y}{dx^{2}}-2\dfrac{dy}{dx}+2y=0} \]