Question

solve the given initial-value problem. Use a graphing utility to graph the solution curve.$\mathbf{}\mathbf{25}\mathbf{.}\mathbf{}{\mathit{x}}^{\mathbf{2}}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{3}\mathit{x}\mathit{y}\mathbf{\text{'}}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{}\mathit{y}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathit{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{4}$

Short answer

The solution for the given initial–value problem is$y=2-2x^{-2}$

Step-by-step solution

Define Initial-value:

The initial value, or y-intercept, is the output value when the line function input is zero. It is the y value of the point where the line crosses the y axis.

Find a homogenous differential equation:

$x^{2}y\text{'}\text{'}+3xy\text{'}=0$

Let's consider,

$$\begin{gathered} y=x^m \\ y\text{'}=m{x}^{(m-1)} \\ y\text{'}\text{'}=m(m-1)x^{(m-2)} \end{gathered}$$

Substitute in the above equation,

$m(m-1)x^m+3m{x}^m=0$

Divide by $x^m$,

$$\begin{gathered} m(m-1)+3m=0 \\ {m}^2+2m=0 \\ m(m+2)=0 \end{gathered}$$

Make factor equal to 0,

${\text{m}}_1$=0 and ${\text{m}}_2$=-2

Use Case 1: Distinct real roots

Then the general solution becomes,

$y=c_1{x}^{m_1}+c_2{x}^{m_2}$

Substitute the values

$y=c_1+c_2{x}^{-2}$→(1)

Find y using initial value conditions:

$y\text{'}=-2c_2{x}^{-3}$

Where$y\left(1\right)=0$ and$y\text{'}\left(1\right)=4$

$\begin{array}{rcl}{c}_1+c_2& =& 0\\ -2c_2& =& 4\\ c_2& =& -2\\ c_1& =& 2\end{array}$

Therefore,

$y=2-2x^{-2}$

Graph: