Question

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

Function y = e^x + 1

Answer 1

We have,
$y=e^x+1$ ...(1)
Differentiating both sides of (1) with respect to $x$, we get
$\frac{dy}{dx}=e^x$ ...(2)
Differentiating both sides of (2) with respect to $x$, we get
$$\begin{gathered} \frac{d^{2}y}{d{x}^2}=e^x \\ \Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{dy}{dx}\left[\text{Using}\left(2\right)\right] \\ \Rightarrow \frac{d^{2}y}{d{x}^2}-\frac{dy}{dx}=0 \\ \text{It is the given differential equation.} \end{gathered}$$
$y=e^x+1$ satisfies the given differential equation; hence, it is a solution.
Also, when $x=0,y=e^0+1=1+1=2,\text{i.e.}y\left(0\right)=2.$
And, when $x=0,y\text{'}=e^0=1,\mathrm{i}.\mathrm{e}.y\text{'}\left(0\right)=1.$
Hence, $y=e^x+1$ is the solution to the given initial value problem.