Question

Use the substitution t= -x to solve the given initial-value problem on the interval$\mathbf{(}\mathbf{-}\mathbf{\infty }\mathbf{,}\mathbf{0}\mathbf{)}$.${\mathit{x}}^{\mathbf{2}}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{4}\mathit{x}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{6}\mathit{y}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{}\mathit{y}\mathbf{(}\mathbf{-}\mathbf{2}\mathbf{)}\mathbf{=}\mathbf{8}\mathbf{,}\mathit{y}\mathbf{\text{'}}\mathbf{(}\mathbf{-}\mathbf{2}\mathbf{)}\mathbf{=}\mathbf{0}$

Short answer

The solution of the given initial value by substituting the given equation is $y=2x^3+6x^2$.

Step-by-step solution

Deriving the given solution by the Cauchy Euler equation

We have the Cauchy Euler equation

$x^{2}y\text{'}\text{'}-4xy\text{'}+6y=0$

On the interval $(-\infty ,0)$with the initial conditions

$y(-2)=8andy\text{'}(-2)=0$

And we have to obtain its solution as the following technique:

Assume that$y=x^m$ and then differentiate with respect to x, then we have

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

Then, substitute with$y=t^m$ and equations (2) and (3) into equation (1), then we have

$$\begin{gathered} x^2\left(m(m-1)x^{(m-2)}\right)-4x\left(m{x}^{(m-1)}\right)+6x^m=0 \\ m(m-1)x^2{x}^{(m-2)}-4mx{x}^{(m-1)}+6x^m=0 \\ m(m-1)x^m-4m{x}^m+6x^m=0 \\ \left(m\right(m-1)-4m+6)x^m=0 \\ \left(m^2-m-4m+6\right)x^m=0 \\ \left(m^2-5m+6\right)x^m=0 \end{gathered}$$

Since $x^m$cannot be equal to 0, then we obtain

$$\begin{gathered} m^2-5m+6=0 \\ (m-3)(m-2)=0 \end{gathered}$$

Then the roots are$m_1=3and{m}_2=2$

Then we can obtain the general solution of this Cauchy Euler equation as

$y=c_1{x}^3+c_2{x}^2$

Appling the initial conditions to find the value of constants

After that, use the substitution x = - t into equation (1), then we have

$\begin{array}{rcl}y& =& c_1{(-t)}^3+c_2{\left(t\right)}^2\\ & =& -c_1{t}^3+c_2{t}^2\end{array}$

After that, we have to apply the initial conditions to find the value of constants as the following :

We have to apply the point$(t,y)=(2,8)$ to the general solution shown in equation (2), then we have

$$\begin{gathered} 8=-c_1\times {\left(2\right)}^3+c_2\times {\left(2\right)}^2 \\ -8c_1+4c_2=8 \end{gathered}$$

After that, take the first derivative for the general solution into equation (a) as

$y\text{'}=-3c_1{t}^2+2c_{2}t$

Then apply the point $\left(t,y\text{'}\right)=(2,0)$,then we have

$$\begin{gathered} 0=-3c_1\times {\left(2\right)}^2+2c_2\times \left(2\right) \\ -12c_1+4c_2=0 \end{gathered}$$

Then by solving the two equations (a) and (b), we find that

$c_1=2and{c}_2=6$

Then, substitute with the constantsinto equation (2), then we obtain

$y=-2t^3+6t^2$

Finally, back substitute t = - x withinto equation (3), then we have

$\begin{array}{rcl}y& =& -2{(-x)}^3+6{\left(x\right)}^2\\ & =& 2x^3+6x^2\end{array}$

The solution of the given differential equation $x^{2}y\text{'}\text{'}-4xy\text{'}+6y=0$