Question

In Problems 23 - 30 Verify that the given functions form a fundamental set of solutions of the differential equation on the indicated interval. Form the general solution.

${\mathbf{x}}^{\mathbf{3}}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{6}{\mathbf{x}}^{\mathbf{2}}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{4}\mathbf{xy}\mathbf{\text{'}}\mathbf{-}\mathbf{4}\mathbf{y}\mathbf{=}\mathbf{0}\mathbf{;}\mathbf{}\mathbf{x}\mathbf{,}{\mathbf{x}}^{\mathbf{-}\mathbf{2}}\mathbf{,}{\mathbf{x}}^{\mathbf{-}\mathbf{2}}\mathbf{ln}\mathbf{}\mathbf{x}\mathbf{,}\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{\infty }\mathbf{)}$

Short answer

The form of the general solution is $y=c_{1}x+c_2{x}^{-2}+c_3{x}^{-2}\ln x$

Step-by-step solution

Verify first order differential equation

$x^{3}y\text{'}\text{'}\text{'}+6x^{2}y\text{'}\text{'}+4xy\text{'}-4y=0······\left(1\right)$

Consider$y_1=x$,

Differentiate with respect to x

$$\begin{gathered} y_1\text{'}=1 \\ {y}_1\text{'}\text{'}=0 \\ {y}_1\text{'}\text{'}\text{'}=0 \end{gathered}$$

Substitute in equation (1),

$\begin{array}{rcl}{x}^3\left(0\right)+6x^2\left(0\right)+4x\left(1\right)-4\left(x\right)& =& 0\\ 0+0+4x\left(1\right)-4\left(x\right)& =& 0\\ 4x-4x& =& 0\\ 0& =& 0\end{array}$

From this we get to know $y_1=x$ satisfy the general equation.

Verify second order differential equation

Consider $y_2=x^{-2}$,

Differentiate with respect to x

$$\begin{gathered} y_2\text{'}=-2x^{-3} \\ {y}_2\text{'}\text{'}=6x^{-4} \\ {y}_2\text{'}\text{'}\text{'}=-24x^{-5} \end{gathered}$$

Substitute in equation (1),

$\begin{array}{rcl}{x}^3\times \left(-24x^{-5}\right)+6x^2\times \left(6x^{-4}\right)+4x\times \left(-2x^{-3}\right)-4\left(x^{-2}\right)& =& 0\\ -24x^{-2}+36x^{-2}-8x^{-2}-4x^{-2}& =& 0\\ -36x^{-2}+36x^{-2}& =& 0\\ 0& =& 0\end{array}$

From this we get to know $y_2=x^{-2}$ satisfy the general equation.

Verify third order differential equation

Consider$y_3=x^{-2}\ln x$

Differentiate with respect to x

$$\begin{gathered} y_3\text{'}=-2x^{-3}\ln x+x^{-3} \\ {y}_3\text{'}\text{'}=6x^{-4}\ln x-5x^{-4} \\ {y}_3\text{'}\text{'}\text{'}=-24x^{-5}\ln x+26x^{-5} \end{gathered}$$

Substitute in equation (1),

$\begin{array}{rcl}{x}^3\times \left(-24x^{-5}\ln x+26x^{-5}\right)+6x^2\times \left(6x^{-4}\ln x-5x^{-4}\right)+4x\times \left(-2x^{-3}\ln x+x^{-3}\right)-4\left(x^{-2}\ln x\right)& =& 0\\ -24x^{-2}\ln x+26x^{-2}+36x^{-2}\ln x-30x^{-2}-8x^{-2}\ln x+4x^{-2}-4x^{-2}\ln x& =& 0\\ -36x^{-2}\ln x+36x^{-2}\ln x+30x^{-2}-30x^{-2}& =& 0\\ 0& =& 0\end{array}$

From this we get to know $y_3=x^{-2}\ln x$ satisfy the general equation.

Use Wronskian method to find y

Substitute the values as we found before,

$\begin{array}{rcl}W\left(y_1,y_2,y_3\right)& =& \left|\begin{array}{ccc}{y}_1& y_2& y_3\\ y_1^{\text{'}}& y_2^{\text{'}}& y_3^{\text{'}}\\ y_1^{\text{'}\text{'}}& y_2^{\text{'}\text{'}}& y_3^{\text{'}\text{'}}\end{array}\right|\\ & =& \left|\begin{array}{ccc}x& x^2& x^{-2}\ln x\\ 1& -2x^{-3}& -2x^{-3}\ln x+x^{-3}\\ 0& 6x^{-4}& 6x^{-4}\ln x-5x^{-4}\end{array}\right|\end{array}$

Using determinant method and solve matrix,

$\begin{array}{rcl}W\left(f_1,f_2,f_3\right)& =& x\times \left|\begin{array}{cc}-2x^{-3}& -2x^{-3}\ln x+x^{-3}\\ 6x^{-4}& 6x^{-4}\ln x-5x^{-4}\end{array}\right|-x^{-2}\times \left|\begin{array}{cc}1& -2x^{-3}\ln x+x^{-3}\\ 0& 6x^{-4}\ln x-5x^{-4}\end{array}\right|+x^{-2}\ln x\times \left|\begin{array}{cc}1& -2x^{-3}\\ 0& 6x^{-4}\end{array}\right|\\ W\left(f_1,f_2,f_3\right)& =& x\left[\left(-2x^{-3}\right)\times \left(6x^{-4}\ln x-5x^{-4}\right)-\left(-2x^{-3}\ln x+x^{-3}\right)\times \left(6x^{-4}\right)\right]\\ & & -x^{-2}\left[\left(1\right)\times \left(6x^{-4}\ln x-5x^{-4}\right)-0\right]+\left(x^{-2}\ln x\right)\left[\left(1\right)\times \left(6x^{-4}\right)-0\right]\\ W\left(f_1,f_2,f_3\right)& =& \left[-12x^{-6}\ln x+10x^{-6}+12x^{-6}\ln x-6x^{-6}\right]+\left[-6x^{-6}\ln x+5x^{-6}\right]+\left[\left(6x^{-6}\right)\ln x\right]\\ & =& 4x^{-6}+5x^{-6}\\ & =& 9x^{-6}\end{array}$

Substitute $y_1=x;y_2=x^{-2};y_3=x^{-2}\ln x$in general equation

We get,

Hence, the solution is $y=c_{1}x+c_2{x}^{-2}+c_3{x}^{-2}\ln x$