Question

solve the given initial-value problem. Use a graphing utility to graph the solution curve.$\mathbf{}\mathbf{26}\mathbf{.}\mathbf{}{\mathit{x}}^{\mathbf{2}}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{5}\mathit{x}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{8}\mathit{y}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{}\mathit{y}\mathbf{\left(}\mathbf{2}\mathbf{\right)}\mathbf{=}\mathbf{32}\mathbf{,}\mathit{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{2}\mathbf{\right)}\mathbf{=}\mathbf{0}$

Short answer

The solution for the given initial–value problem is$y=-2x^4+16x^2$

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:

$x^{2}y\text{'}\text{'}-5xy\text{'}+8y=0$

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,

$$\begin{gathered} x^2\left[m(m-1)x^{(m-2)}\right]-5x\left[m{x}^{(m-1)}\right]+8x^m=0 \\ m(m-1)x^2{x}^{(m-2)}-5m{x}^1{x}^{(m-1)}+8x^m=0 \\ m(m-1)x^m-5m{x}^m+8x^m=0 \\ \left[m\right(m-1)-5m+8]x^m=0 \end{gathered}$$

$$\begin{gathered} \left(m^2-m-5m+8\right)x^m=0 \\ \left(m^2-6m+8\right)x^m=0 \end{gathered}$$

So, $x^m$≠0,

$m^2-6m+8=0$

Using factorize method,

$\begin{array}{rcl}m& =& \frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ & =& \frac{-(-6)\pm \sqrt{{(-6)}^2-4\left(1\right)\left(8\right)}}{2\left(1\right)}\\ & =& \frac{6\pm \sqrt{36-32}}{2}\\ & =& \frac{6\pm \sqrt{4}}{2}\\ & =& \frac{6\pm 2}{2}\end{array}$

Therefore, $m_1=4$and$m_2=2$

Use Case 1: Distinct real roots

Then the general solution becomes,

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

Substitute the value of

$y=c_1{x}^4+c_2{x}^2$→(1)

Find y using initial value conditions:

Apply initial conditions, the points $(x,y)=(2,32)$in equation (1)

$$\begin{gathered} 32=c_1(2{)}^4+c_2(2{)}^2 \\ 16c_1+4c_2=32 \dots \left(2\right) \end{gathered}$$

Take the first derivative for the general solution,

$y\text{'}=4c_1{x}^3+2c_{2}x$

Apply the point $\left(x,y\text{'}\right)=(2,0)$

$$\begin{gathered} 0=4c_1{2}^3+2c_2\left(2\right) \\ 32c_1+4c_2=0 \dots \left(3\right) \end{gathered}$$

By solving the (2) and (3),

$c_1=-2$and $c_2=16$

Substitute values in equation (1),

The solution is$y=-2x^4+16x^2$

Graph: