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
$$\begin{gathered} \frac{d^{2}y}{d{x}^2}=e^x\left(A\cos x+B\sin x\right)+e^x\left(-A\sin x+B\cos x\right)+e^x\left(-A\sin x+B\cos x\right)+e^x\left(-A\cos x-B\sin x\right) \\ \Rightarrow \frac{d^{2}y}{d{x}^2}=2e^x\left(-A\sin x+B\cos x\right) \\ \Rightarrow \frac{d^{2}y}{d{x}^2}=2e^x\left(-A\sin x+B\cos x\right)+2e^x\left(A\cos x+B\sin x\right)-2e^x\left(A\cos x+B\sin x\right) \\ \Rightarrow \frac{d^{2}y}{d{x}^2}=2\frac{dy}{dx}-2y\left[\text{Using}\left(1\right)\text{and}\left(2\right)\right] \\ \Rightarrow \frac{d^{2}y}{d{x}^2}-2\frac{dy}{dx}+2y=0 \end{gathered}$$
Hence, the given function is the solution to the given differential equation.