Question

Show that the function y = A cos 2x − B sin 2x is a solution of the differential equation \[\frac{d^2 y}{d x^2} + 4y = 0\].

Answer 1

We have,
\[y = A \cos 2x - B \sin 2x............(1)\]
Differentiating both sides of (1) with respect to x, we get
\[\frac{dy}{dx} = - 2A \sin 2x - 2B \cos 2x........(2)\]
Differentiating both sides of (2) with respect to x, we get
\[\frac{d^2 y}{d x^2} = - 4A \cos 2x + 4B \sin 2x\]
\[ \Rightarrow \frac{d^2 y}{d x^2} = - 4\left( A \cos 2x - B \sin 2x \right)\]
\[ \Rightarrow \frac{d^2 y}{d x^2} = - 4y ........\left[\text{Using }\left( 1 \right) \right]\]
\[\Rightarrow\] \[\frac{d^2 y}{d x^2} + 4y = 0\]
Hence, the given function is the solution to the given differential equation.