Question

Differential equation $\frac{d^{2}y}{d{x}^2}-3\frac{dy}{dx}+2y=0,y\left(0\right)=1,y\text{'}\left(0\right)=3$

Function y = e^x + e^2x

Answer 1

We have,
y = e^x + e^2x .....(1)
Differentiating both sides of (1) with respect to x, we get
$\frac{dy}{dx}=e^x+2e^{2x}$ .....(2)
Differentiating both sides of (2) with respect to x, we get
$$\begin{gathered} \frac{d^{2}y}{d{x}^2}=e^x+4e^{2x} \\ \Rightarrow \frac{d^{2}y}{d{x}^2}=3\left(e^x+2e^{2x}\right)-2\left(e^x+e^{2x}\right) \\ \Rightarrow \frac{d^{2}y}{d{x}^2}=3\frac{dy}{dx}-2y\left[\text{Using}\left(1\right)\text{and}\left(2\right)\right] \\ \Rightarrow \frac{d^{2}y}{d{x}^2}-3\frac{dy}{dx}+2y=0 \end{gathered}$$

$\Rightarrow \frac{d^{2}y}{d{x}^2}-3\frac{dy}{dx}+2y=0$
It is the given differential equation.
Therefore, y = e^x + e^2x satisfies the given differential equation.
Also, when $x=0;y=e^0+e^0=1+1,\text{i.e}.y\left(0\right)=2.$
And, when $x=0;y^{\text{'}}=e^0+2e^0=1+2,\text{i.e}.y^{\text{'}}\left(0\right)=3.$
Hence, y = e^x + e^2x is the solution to the given initial value problem.

Disclaimer: In the question instead of y(0) = 1, it should have been y(0) = 2.