Question

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

Function y = xe^x + e^x

Answer 1

We have,
y = xe^x + e^x .....(1)
Differentiating both sides of (1) with respect to x, we get
$$\begin{gathered} \frac{dy}{dx}=x{e}^x+e^x+e^x \\ \Rightarrow \frac{dy}{dx}=x{e}^x+2e^x....\left(2\right) \end{gathered}$$
Differentiating both sides of (2) with respect to x, we get

$$\begin{gathered} \frac{d^{2}y}{d{x}^2}=x{e}^x+e^x+2e^x \\ \Rightarrow \frac{d^{2}y}{d{x}^2}=x{e}^x+3e^x \\ \Rightarrow \frac{d^{2}y}{d{x}^2}=2\left(x{e}^x+2e^x\right)-\left(x{e}^x+e^x\right) \\ \Rightarrow \frac{d^{2}y}{d{x}^2}=2\frac{dy}{dx}-y\left[\text{Using}\left(1\right)\text{and}\left(2\right)\right] \\ \Rightarrow \frac{d^{2}y}{d{x}^2}-2\frac{dy}{dx}+y=0 \end{gathered}$$

$\frac{d^{2}y}{d{x}^2}-2\frac{dy}{dx}+y=0$

It is the given differential equation.
Thus, y = xe^x + e^x satisfies the given differential equation.
Also, when $x=0,y=0+1=1,\text{i.e}.y\left(0\right)=1.$
And, when $x=0,y^{\text{'}}=0+2=2,\text{i.e}.y^{\text{'}}\left(0\right)=2.$
Hence, y = xe^x + e^x is the solution to the given initial value problem.