Question

solve the given initial-value problem. Use a graphing utility to graph the solution curve.$\mathbf{}\mathbf{28}\mathbf{.}\mathbf{}{\mathit{x}}^{\mathbf{2}}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{3}\mathit{x}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{4}\mathit{y}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{}\mathit{y}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{5}\mathbf{,}\mathit{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{3}$

Short answer

The solution for the given initial–value problem is $y=5x^2-7x^2\ln x$

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:

$x^{2}y\text{'}\text{'}-3xy\text{'}+4y=0$

Let's consider,

$y=x^m$

$y\text{'}=m{x}^{(m-1)}$

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

Substitute in the above equation,

$m(m-1)x^m-3m{x}^m+4x^m=0$

Divide $x^m$ into both sides,

$$\begin{gathered} m(m-1)-3m+4=0 \\ {m}^2-4m+4=0 \\ (m-2{)}^2=0 \\ m=2,2 \end{gathered}$$

So the equation is $y=c_1{x}^2+c_2{x}^2\ln x$,

Find y using initial value conditions:

$y=2c_{1}x+2c_{2}x\ln x+c_{2}x$

Where$y\left(1\right)=5$ and $y\text{'}\left(1\right)=3$

role="math" localid="1664234138152" $$\begin{gathered} c_1+0=5 \\ {c}_1=5 \end{gathered}$$

$$\begin{gathered} 2c_1+c_2=3 \\ {c}_2=-7 \end{gathered}$$

Substitute in the above equation,

The solution is $y=5x^2-7x^2\ln x$

Graph: