Question

Determine by inspection at least two solutions of the given first-order IVP.

$\mathit{x}\mathit{y}\mathbf{\text{'}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{2}\mathit{y}\mathbf{,}\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}$

Short answer

The two solutions of the initial value problem are $y\left(x\right)=0$and $y\left(x\right)=x^2$.

Step-by-step solution

Note the given data 

$y\left(0\right)=0$and $xy\text{'}=2y$

Finding the first solution of IVP

Let $y\left(x\right)=0$be a solution, then put it in given equation, we get

$$\begin{gathered} x\times 0=2\times 0 \\ \Rightarrow 0=0 \end{gathered}$$

Now, we check the initial condition and we get

Given initial value is $y\left(0\right)=0$

We substitute the value of $x=0,y=0$ in $y\left(x\right)=0$.

So,

$$\begin{gathered} y\left(0\right)=0 \\ \Rightarrow 0=0 \end{gathered}$$

Therefore,$y\left(x\right)=0$ satisfies all the conditions and initial condition $y\left(0\right)=0$.

Finding the second solution of IVP

Let $y\left(x\right)=x^2$be a solution and put it in given equation

Differentiating both sides of the above equation and we get

$y\text{'}=2x$

$$\begin{gathered} xy\text{'}=2y \\ \Rightarrow x\times 2x=2x^2 \\ \Rightarrow 2x^2=2x^2 \end{gathered}$$

This shows that It satisfied the condition

Now we check initial condition $y\left(0\right)=0$

Put the values of $x=0,y=0$, we get

$$\begin{gathered} y\left(x\right)=x^2 \\ \Rightarrow y\left(0\right)=0 \\ \Rightarrow 0=0 \end{gathered}$$

So, $y\left(x\right)=x^2$satisfies all the conditions.

Therefore, the solutions of the initial value problem are $y\left(x\right)=0$ and $y\left(x\right)=x^2$.