Question

Solve the given initial-value problem. Use a graphing utility to graph the solution curve.${\mathit{x}}^{\mathbf{2}}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{5}\mathit{x}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{8}\mathit{y}\mathbf{=}\mathbf{8}{\mathit{x}}^{\mathbf{6}}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}\mathit{y}\left(\frac{1}{2}\right)\mathbf{=}\mathbf{0}\mathbf{,}\mathit{y}\mathbf{\text{'}}\left(\frac{1}{2}\right)\mathbf{=}\mathbf{0}$

Short answer

The solution of initial value problem using graphing utility is $y=x^6-\frac{1}{2}{x}^4+\frac{1}{16}{x}^2$

Step-by-step solution

Deriving the given solution by differential equation

We have the Cauchy Euler equation and we get,

$$\begin{gathered} x^{2}y\text{'}\text{'}-5xy\text{'}+8y=8x^6 \\ y\text{'}\text{'}+p\left(x\right)y\text{'}+q\left(x\right)y=f\left(x\right) \end{gathered}$$

We have to get its general solution by the following technique:

First, we have to find the solution of the corresponding homogeneous differential equation as

$y\left(\frac{1}{2}\right)=0andy\text{'}\left(\frac{1}{2}\right)=0$

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

By assuming that $y=x^m$is a solution for the homogeneous differential equation, then follow the following technique:

Differentiate this assumption with respect to then we have

$$\begin{gathered} y\text{'}=m{x}^{(m-1)} \\ y\text{'}\text{'}=m(m-1)x^{(m-2)} \\ {x}^2\left[m(m-1)x^{(m-2)}\right]-5x\left[m{x}^{(m-1)}\right]+8x^m=0 \\ m(m-1)x^2{x}^{(m-2)}-5m{x}^1{x}^{(m-1)}+8x^m=0 \\ m(m-1)x^m-5m{x}^m+8x^m=0 \\ \left[m(m-1)-5m+8\right]x^m=0 \\ \left(m^2-m-5m+8\right)x^m=0 \\ \left(m^2-6m+8\right)x^m=0 \end{gathered}$$

Solving the coefficient and find wronskian

Since$x^m$cannot be equal 0 , the we have

$$\begin{gathered} m^2-6m+8=0 \\ (m-4)(m-2)=0 \end{gathered}$$

The roots are,

$m_1=4and{m}_2=2$

Then we can obtain the general solution of this homogeneous differential equation as

$y=c_1{x}^4+c_2{x}^2$

Second, now we have to find the particular solution $y_p$ of the given differential equation as the following technique :

Since we have $y_1=xand{y}_2=x\ln x$ then we can obtain the wronskian as

Finding W1  and  W2 using wronskian

After that, since we have the right side of the given differential equation $f\left(x\right)=8x^4$ then we can obtain the wronskian W₁ andW₂as

After we had obtained the wronskian W₁ and W₂as we have to obtain $u_1^{\text{'}}and{u}_2^{\text{'}}$as the following

$\begin{array}{rcl}{u}_1^{\text{'}}& =& \frac{W}{W\left(y_1,y_2\right)}\\ & =& \frac{-8x^6}{-2x^5}\\ & =& 4x\\ & & \end{array}$

$\begin{array}{rcl}{u}_2^{\text{'}}& =& \frac{W_2}{W\left(y_1,y_2\right)}\\ & =& \frac{8x^8}{-2x^5}\\ & =& -4x^3\\ & & \end{array}$

Solve the integrals and find partial solution

After that, we have to do integration for both$u_1^{\text{'}}and{u}_2^{\text{'}}$neglecting integration constants to obtain$u_{1}and{u}_2$as the following

$\begin{array}{rcl}{u}_1& =& \int u_1^{\text{'}}dx\\ & =& \int 4xdx\\ & =& 4\times \frac{1}{2}{x}^2\\ & =& 2x^2\end{array}$

$\begin{array}{rcl}{u}_2& =& \int u_1^{\text{'}}dx\\ & =& \int -4x^{3}dx\\ & =& -4\times \frac{1}{4}{x}^4\\ & =& -x^4\end{array}$

After that, using the values of $u_{1}and{u}_2$and the functions $y_{1}and{y}_2$. We can obtain the particular solution as

$\begin{array}{rcl}{y}_p& =& u_1{y}_1+u_2{y}_2\\ & =& \left(2x^2\right)\times \left(x^4\right)+\left(-x^4\right)\times \left(x^2\right)\\ & =& 2x^6-x^6\\ & =& x^6\\ & & \end{array}$

Then from equations (a) and (b) we can obtain the general solution of the differential equation

$x^{2}y\text{'}\text{'}-5xy\text{'}+y=8x^{6}as$

$\begin{array}{rcl}y& =& y_h+y_p\\ & =& x^6+c_1{x}^4+c_2{x}^2\\ & & \end{array}$

Deriving the given solution using initial value

After that, we have to apply the initial conditions into the solution of the given D.E as the following:

We have to apply the point $(x,y)=\left(\frac{1}{2},0\right)$ into equation (c), then we have

$$\begin{gathered} 0={\left(\frac{1}{2}\right)}^6+c_1\times {\left(\frac{1}{2}\right)}^4+c_2\times {\left(\frac{1}{2}\right)}^2 \\ \frac{1}{16}{c}_1+\frac{1}{4}{c}_2=-\frac{1}{64} \\ 4c_1+16c_2=-1 \end{gathered}$$

After that, we have to take the first derivative for the general solution of the given D.E as

$y\text{'}=6x^5+4c_1{x}^3+2c_{2}x$

Then apply the point $\left(x,y\text{'}\right)=\left(\frac{1}{2},0\right)$ into this derivative, then we have

$$\begin{gathered} 0=6\times {\left(\frac{1}{2}\right)}^5+4c_1\times {\left(\frac{1}{2}\right)}^3+2c_2\times \left(\frac{1}{2}\right) \\ \frac{1}{2}{c}_1+c_2=-\frac{3}{16} \\ 8c_1+16c_2=-3 \end{gathered}$$

Then by solving these two equations, we find that

$c_1=-\frac{1}{2}and{c}_2=\frac{1}{16}$

Finally, substitute with these values into the general solution (c) of the given Cauchy Euler equation, then we obtain

$y=x^6-\frac{1}{2}{x}^4+\frac{1}{16}{x}^2$

Draw the graph using graphing utility

Using a graphing utility, we can graph the solution curve as the following: