Question

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

${\mathit{x}}^{\mathbf{2}}\mathit{y}\mathbf{"}\mathbf{-}\mathit{x}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathit{y}\mathbf{=}{\mathit{x}}^{\mathbf{3}}$

Short answer

The solution of the given Cauchy Euler differential equation$x^{2}y"-xy\text{'}+y=x^{3}as=c_{1}x+c_{2}x\ln x+\frac{1}{4}{x}^3$

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.

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

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

We have $x^{2}y"-xy\text{'}+y=x^3$

Which is in the form $$\begin{gathered} y"-\frac{1}{x}y\text{'}+\frac{1}{x^2}y=x, \\ y"+p\left(x\right)y\text{'}+q\left(x\right)y=f\left(x\right) \end{gathered}$$

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

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

When xm cannot be 0,

$$\begin{gathered} m^2-2m+1=0 \\ {\left(m-1\right)}^2=0 \end{gathered}$$

The roots are $m_{12}$which are real and distinct.

Step 3: Obtain the solution of the given Cauchy Euler differential equation

Now obtain the particular solution

When we have $$\begin{gathered} y_1=xand{y}_2=x\ln x, \\ W\left(x,x\ln x\right)=\left|\begin{array}{cc}{y}_1& y_2\\ y_1& y_2\end{array}\right| \\ =\left|\begin{array}{cc}x& x\ln x\\ 1& \ln x+1\end{array}\right| \\ =\left(x\right)\times (\ln x+1)-(x\ln x)\times \left(1\right) \\ =x\ln x+x-x\ln x \\ =x \end{gathered}$$

Now when $f\left(x\right)=x$

$$\begin{gathered} W_1=\left|\begin{array}{cc}0& y_2\\ f\left(x\right)& y_2\end{array}\right| \\ =\left|\begin{array}{cc}0& x\ln x\\ x& \ln x+1\end{array}\right| \\ =\left(0\right)\times (\ln x+1)-(x\ln x)\times \left(x\right) \\ =0-\left(x^2\ln x\right) \\ =-x^2\ln x \end{gathered}$$

$$\begin{gathered} W_2=\left|\begin{array}{cc}{y}_1& 0\\ y_1& f\left(x\right)\end{array}\right| \\ =\left|\begin{array}{cc}x& 0\\ 1& x\end{array}\right| \\ =\left(x\right)\times \left(x\right)-\left(0\right)\times \left(1\right) \\ =x^2-0 \\ =x^2 \end{gathered}$$

Step 3: we had obtained the WronskiansW1 and W2.

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

And

$$\begin{gathered} u{\text{'}}_2=\frac{W1}{W\left(y_1,y_2\right)} \\ =\frac{-x^2}{x} \\ =x \end{gathered}$$

Now integrate $$\begin{gathered} u_1=\int -x\ln xdx \\ =-x\int \ln xdx \end{gathered}$$

Set $u=\ln x,thenwehavedu=\frac{1}{x}du$

$$\begin{gathered} u_1=-\left(u\times v-\int vdu\right) \\ =-\left(\frac{1}{2}{x}^2\ln x-\int \frac{1}{2}{x}^2\times \frac{1}{x}dx\right) \\ =-\frac{1}{2}{x}^2\ln x+\frac{1}{2}\int xdx \\ =-\frac{1}{2}{x}^2\ln x+\frac{1}{2}\times \frac{1}{2}{x}^2 \\ =-\frac{1}{2}{x}^2\ln x+\frac{1}{4}{x}^2 \end{gathered}$$

Obtain the particular solution

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

The general solution$$\begin{gathered} y=y_c+y_p \\ =c_{1}x+c_{2}x\ln x+\frac{1}{4}{x}^3 \end{gathered}$$