Question

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

Function y = e^x + e^−x

Answer 1

We have,
y = e^x + e^−x .....(1)
Differentiating both sides of (1) with respect to x, we get
$\frac{dy}{dx}=e^x-e^{-x}$ .....(2)
Differentiating both sides of (2) with respect to x, we get

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

$\Rightarrow$$\frac{d^{2}y}{d{x}^2}-y=0$
It is the given differential equation.
Therefore, y = e^x + e^−x 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 $x=0;y_1=e^0-e^0=1-1,\text{i.e}.y^{\text{'}}\left(0\right)=0.$
Hence, y = e^x + e^−x is the solution to the given initial value problem.