Question

use the procedures developed in this chapter to find the general solution of each differential equation.

${\mathit{x}}^{\mathbf{2}}\mathit{y}\mathbf{"}\mathbf{-}\mathbf{4}\mathit{x}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{6}\mathit{y}\mathbf{}\mathbf{=}\mathbf{2}{\mathit{x}}^{\mathbf{4}}\mathbf{}\mathbf{+}{\mathit{x}}^{\mathbf{2}}$

Short answer

The solution of the given Cauchy Euler differential equation

$x^{2}y"-4xy\text{'}+6y=2x^4+x^{2}asy=x^4+c_1{x}^3+c_3{x}^2-x^2\ln x$

Step-by-step solution

Define the second order Cauchy Euler differential equation

The Cauchy-Euler differential equation is a type of linear ordinary differential equation with variable coefficients. In time-harmonic vibrations of a thin elastic rod, issues on yearly and solid discs, wave mechanics, and other domains of science and engineering, the second order Cauchy–Euler equations are applied.

${\mathit{a}}_{\mathbf{n}}{\mathit{x}}^{\mathbf{n}}{\mathit{y}}^{\left(n\right)}\mathbf{+}\mathbf{}{\mathit{a}}_{\mathbf{n}\mathbf{-}\mathbf{1}}{\mathit{x}}^{\mathbf{n}\mathbf{-}\mathbf{1}}{\mathit{y}}^{\left(n-1\right)}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{+}{\mathit{a}}_{\mathbf{0}}\mathit{y}\mathbf{=}\mathbf{0}$

Step 2: Obtain the solution of the given Cauchy Euler differential equation

Here

$$\begin{gathered} x^{2}y"-4xy\text{'}+6y=2x^4+x^2 \\ y"-4x^{-1}y\text{'}+6x^{-2}y=2x^2+1 \\ y"+p\left(x\right)y\text{'}+q\left(x\right)y=f\left(x\right) \end{gathered}$$

Differentiate this assumption with respect to x then we have

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

Now substitute $y\text{'}andy"$

$$\begin{gathered} x2\left(m\left(m-1\right)x^{\left(m-2\right)}\right)-4x\left(m{x}^{\left(m-1\right)}+6x^m\right)=0 \\ m\left(m-1\right)x^2.x^{\left(m-2\right)}-4mx.x^{\left(m-1\right)}+6x^m=0 \\ m\left(m-1\right)x^m-4m{x}^m+6x^m=0 \\ \left(m^2-5m+6\right)x^m=0 \end{gathered}$$

Since $x^m$cannot be negative $$\begin{gathered} m^2-5m+6=0 \\ \left(m-3\right)\left(m-2\right)=0 \end{gathered}$$

$m_1=3and{m}_2=3$

The roots are real and distinct.

Now obtain the homogeneous solution

Here $$\begin{gathered} x^{2}y"-4xy\text{'}+6y=2x^4+x^4 \\ y=c_1{x}^3+c_2{x}^2 \end{gathered}$$

We have $y_1=x^{3}and{y}_2=x^2$

$$\begin{gathered} W\left(x^3,x^2\right)=\left|\begin{array}{cc}{y}_1& y_2\\ y_1& y_2\end{array}\right| \\ =\left|\begin{array}{cc}{x}^3& x^2\\ 3x^2& 2x\end{array}\right| \\ =\left(x^3\right)\times \left(2x\right)-\left(x^2\right)\times \left(3x^2\right) \\ =2x^4-3x^4 \\ =-x^4 \end{gathered}$$

Since in the right side $f\left(x\right)=2x^2+1$

$$\begin{gathered} W_1=\left|\begin{array}{cc}0& y_2\\ f\left(x\right)& y_2\end{array}\right| \\ =\left|\begin{array}{cc}0& x^2\\ 2x^2+1& 2x\end{array}\right| \\ =\left(0\right)\times \left(2x\right)-\left(x^2\right)\times \left(2x^2+1\right) \\ =-2x^4-x^2 \\ Andfor{W}_2 \\ {W}_2=\left|\begin{array}{cc}{y}_1& 0\\ y_1& f\left(x\right)\end{array}\right| \\ =\left|\begin{array}{cc}x3& 0\\ 3x^2& 2x^2+1\end{array}\right| \\ =\left(x^3\right)\times \left(2x^2+1\right)-\left(0\right)\times \left(3x^2\right) \\ =2x^5+x^3-0 \\ =2x^5+x^3 \end{gathered}$$

The Wronskians W1 and W2

Now obtain $u_1^{\text{'}}and{u}_2^{\text{'}}$

$$\begin{gathered} u_1^{\text{'}}=\frac{W_1}{W\left(y_1,y_2\right)} \\ =\frac{-2x^4-x^2}{-x^4} \\ =2+x^{-2} \\ {u}_2^{\text{'}}=\frac{W}{W} \\ =\frac{2x5+x_1}{-y_2} \\ =-2x^4 \end{gathered}$$

Now Integrate the values u1' and u2'

$$\begin{gathered} u_1=\int \left(2+x^{-2}\right)dx \\ =\int 2dx+\int x^{-2}dx \\ =2x-x^{-1} \\ {u}_2=\int \left(-2x-x^{-1}\right)dx \\ =\int -2xdx-\int \frac{1}{x}dx \\ =-2\times \frac{1}{2}{x}^2-\ln x \\ =-x^2-\ln x \end{gathered}$$

Now find the particular solution

$$\begin{gathered} y_p=u_1{y}_1+u_1{y}_2 \\ =\left(2x-x^{-1}\right)\times \left(x^3\right)+\left(-x^2-\ln x\right)\times \left(x^2\right) \\ =\left(2x^4-x^2\right)+\left(-x^4-x^2\ln x\right) \\ =x^4-x^2-x^2\ln x \end{gathered}$$

Now combine for general solution $y=y_c+y_p$

$$\begin{gathered} =c_1{x}^3+c_2{x}^2+x^4-x^2-x^2\ln x \\ =c_1{x}^3+\left(c_2-1\right)x^2+x^4-x^2\ln x \\ =x^4+c_1{x}^3+c_3{x}^2-x^2\ln x \\ Where{c}_3=c_2-1 \\ y=x^4+c_1{x}^3+c_3{x}^2-x^2\ln x \end{gathered}$$