Question

Show that y = e^x (A cos x + B sin x) is the solution of the differential equation \[\frac{d^2 y}{d x^2} - 2\frac{dy}{dx} + 2y = 0\]

Answer 1

We have,
\[y = e^x \left( A \cos x + B \sin x \right).........(1)\]

Differentiating both sides of (1) with respect to x, we get

\[\frac{dy}{dx} = e^x \left( A \cos x + B \sin x \right) + e^x \left( - A \sin x + B \cos x \right)..........(2)\]
Differentiating both sides of (2) with respect to x, we get
\[\frac{d^2 y}{d x^2} = e^x \left( A \cos x + B \sin x \right) + e^x \left( - A \sin x + B \cos x \right) + e^x \left( - A \sin x + B \cos x \right) + e^x \left( - A \cos x - B \sin x \right)\]
\[ \Rightarrow \frac{d^2 y}{d x^2} = 2 e^x \left( - A \sin x + B \cos x \right)\]
\[ \Rightarrow \frac{d^2 y}{d x^2} = 2 e^x \left( - A \sin x + B \cos x \right) + 2 e^x \left( A \cos x + B \sin x \right) - 2 e^x \left( A \cos x + B \sin x \right)\]
\[ \Rightarrow \frac{d^2 y}{d x^2} = 2\frac{dy}{dx} - 2y ........\left[  \text{Using (1) and (2)}\right]\]
\[ \Rightarrow \frac{d^2 y}{d x^2} - 2\frac{dy}{dx} + 2y = 0\]

Hence, the given function is the solution to the given differential equation.