Question

Show that y = be^x + ce^2x is a solution of the differential equation, $\frac{d^{2}y}{d{x}^2}-3\frac{dy}{dx}+2y=0$.

Answer 1

We have,
$y=b{e}^x+c{e}^{2x}$ ...(1)
Differentiating both sides of equation (1) with respect to $x$, we get
$\frac{dy}{dx}=b{e}^x+2c{e}^{2x}$ ...(2)
Differentiating both sides of equation (2) with respect to $x$, we get
$$\begin{gathered} \frac{d^{2}y}{d{x}^2}=b{e}^x+4c{e}^{2x} \\ =3b{e}^x+6c{e}^{2x}-2b{e}^x-2c{e}^{2x} \\ =3\left(b{e}^x+2c{e}^{2x}\right)-2\left(b{e}^x+c{e}^{2x}\right) \\ =3\frac{dy}{dx}-2y\left[\text{Using equations}\left(1\right)\text{and}\left(2\right)\right] \end{gathered}$$
$\Rightarrow \frac{d^{2}y}{d{x}^2}-3\frac{dy}{dx}+2y=0$
Hence, the given function is the solution to the given differential equation.