Question

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

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

Short answer

The solution of the given Cauchy Euler differential equation

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

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.

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

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

Here

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

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

Differentiate this assumption with respect to then we have

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

Now substitute $y^{\text{'}}\text{and}{y}^{\text{'}\text{'}}$

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

Since $x^m$cannot be negative

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

$m_1=3\text{and}{m}_2=2$

The roots are real and distinct.

Now obtain the homogeneous solution 

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

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

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

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

$=\left|\begin{array}{cc}{x}^1& x^2\\ 3x^2& 2x\end{array}\right|$

$=\left(x^3\right)\times \left(2x\right)-\left(x^2\right)\times \left(3x^2\right)$

$=2x^4-3x^4$

$=-x^4$,

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

$W_1=\left|\begin{array}{cc}0& y_2\\ f\left(x\right)& y_2^{\text{'}}\end{array}\right|$

$=\left|\begin{array}{cc}0& x^2\\ 2x^2+1& 2x\end{array}\right|$

$=\left(0\right)\times \left(2x\right)-\left(x^2\right)\times \left(2x^2+1\right)$

$=-2x^4-x^2$

And for w2

$W_2=\left|\begin{array}{cc}{y}_1& 0\\ y_1^{\text{'}}& f\left(x\right)\end{array}\right|$

$=\left|\begin{array}{cc}{x}^3& 0\\ 3x^2& 2x^2+1\end{array}\right|$

$=\left(x^3\right)\times \left(2x^2+1\right)-\left(0\right)\times \left(3x^2\right)$

$=2x^5+x^3-0$

$=2x^5+x^3$

The Wronskians W1 and W2

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

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

$$\begin{gathered} u_2^{\text{'}}=\frac{W}{W} \\ =\frac{2x^5+x_1}{\left.-y_2\right)} \\ =-2x^4 \end{gathered}$$

Now Integrate the values u1' and u2'

$$\begin{gathered} u_1=\int \left(2+x^{-2}\right)dx \\ =\int 2dx+\int x^{-2}dx \\ =2x-x^{-1} \end{gathered}$$

And

$$\begin{gathered} u_2=\int \left(-2x-x^{-1}\right)dx \\ =\int -2xdx-\int x^{-1}dx \\ =-2\int xdx-\int \frac{1}{x}dx \\ =-2\times \frac{1}{2}{x}^2-\ln x \\ =-x^2-\ln x \end{gathered}$$

Now find the particular solution

$$\begin{gathered} y_p=u_1{y}_1+u_2{y}_2 \\ =\left(2x-x^{-1}\right)\times \left(x^3\right)+\left(-x^2-\ln x\right)\times \left(x^2\right) \\ =\left(2x^4-x^2\right)+\left(-x^4-x^2\ln x\right) \\ =x^4-x^2-x^2\ln x \end{gathered}$$

Now combine for general solution ${\text{y =y}}_c{\text{+y}}_p$

$=c_1{x}^3+c_2{x}^2+x^4-x^2-x^2\ln x$

$=c_1{x}^3+\left(c_2-1\right)x^2+x^4-x^2\ln x$

$=x^4+c_1{x}^3+c_3{x}^2-x^2\ln x$

When$\text{c}{}_3\text{=c}_2\text{- 1}$

$y=x^4+c_1{x}^3+c_3{x}^2-x^2\ln x$