Question

Solve the given differential equation by variation of parameters.$\mathit{x}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{4}\mathit{y}\mathbf{\text{'}}\mathbf{=}{\mathit{x}}^{\mathbf{4}}$

Short answer

The solution for the given differential equation for variation of parameters $y=c_1+c_2{x}^5-\frac{1}{25}{x}^5+\frac{1}{5}{x}^5\text{ln}x+\frac{2}{5}{x}^2$

Step-by-step solution

Define Cauchy Euler:

The Cauchy – Euler equation, or Euler equation is simply a standard equation with a variable coefficient. It is sometimes called the equidimensional equation.

Find a homogenous differential equation

$xy\text{'}\text{'}-4y\text{'}=x^4$

Divide x on both side,

$y\text{'}\text{'}-\frac{4}{x}y\text{'}=x^3$

We get,

$y\text{'}\text{'}+p\left(x\right)y\text{'}+q\left(x\right)y=f\left(x\right),$

Consider,

$xy\text{'}\text{'}-4y\text{'}=0$

Multiply x on both side,

$x^{2}y\text{'}\text{'}-4xy\text{'}=0$ →(1)

Lets consider,

$y=x^m$

$y\text{'}=m{x}^{(m-1)}$

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

Substitute in above equation,

$\begin{array}{rcl}{x}^2\left[m(m-1)x^{(m-2)}\right]-4x\left[m{x}^{(m-1)}\right]& =& 0\\ m(m-1)x^2{x}^{(m-2)}-4m{x}^1{x}^{(m-1)}& =& 0\\ m(m-1)x^m-4m{x}^m& =& 0\\ \left[m\right(m-1)-4m]x^m& =& 0\\ m(m-1-4)x^m& =& 0\\ m(m-5)x^m& =& 0\end{array}$

Where x ≠0,

$m(m-5)=0$

Taking roots equal to 0,

$m_1=0$and $m_2=5$

Hence, the homogenous differential equation is

$y=C_1{x}^{m_1}+C_2{x}^{m_2}$

Substitute the valuesf :

The solution is

$y_h=c_1+c_2{x}^5$ → (2)

Find y:

$y_p=u_1{y}_1+u_2{y}_2$

$=\left(-\frac{1}{25}{x}^5\right)\times \left(1\right)+\left(\frac{1}{5}\text{ln}x\right)\times \left(x^5\right)$

$=-\frac{1}{25}{x}^5+\frac{2}{5}{x}^2+\frac{1}{5}{x}^5\text{ln}x$

$=-\frac{1}{25}{x}^5+\frac{1}{5}{x}^5\text{ln}x+\frac{2}{5}{x}^2$ (3)

From the equation (2) and (3),

$y=y_h+y_p$

The solution is

$y=c_1+c_2{x}^5-\frac{1}{25}{x}^5+\frac{1}{5}{x}^5\text{ln}x+\frac{2}{5}{x}^2$