Question

solve the given differential equation.$\mathit{x}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{3}\mathit{y}\mathbf{\text{'}}\mathbf{=}\mathbf{0}$

Short answer

The solution for the differential equation is $y=c_1+c_2{x}^4$.

Step-by-step solution

Define Differential equation:

Anequation with one or more derivatives of a function. The derivative of the function is given by$\frac{dy}{dx}$ is known as a Differential equation.

Solve differential equation:

Consider,

$y\text{'}=\frac{dy}{dx}$

$y\text{'}\text{'}=\frac{{\text{d}}^{2}y}{\text{d}{x}^2}$

Substitute in the above equation,

$x\frac{{\text{d}}^{2}y}{\text{d}{x}^2}-3\frac{\text{d}y}{\text{d}x}=0$

Multiply both sides with x,

$x^2\frac{{\text{d}}^{2}y}{\text{d}{x}^2}-3x\frac{\text{d}y}{\text{d}x}=0$

Let,

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

Then the derivatives become,

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

So, the equation is:

$x^2\frac{{\text{d}}^{2}y}{\text{d}{x}^2}-3x\frac{\text{d}y}{\text{d}x}=0$

$x^{2}m(m-1)x^{m-2}-3xm{x}^{m-1}=0$

Keep $x^m$ commonly out,

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

Solving for m,

$m^2-4m=0$

Using Factorize method,

$m\left(m-4\right)=0$

Set factors equal to 0:

$m_1=0;m_2=4$

Use Case 1: Distinct real roots

Then the general solution becomes,

$y=c_1{x}^{m_1}+c_2{x}^{m_2}$

Substitute the value of $m_{1}and{m}_2$:

The solution is $y=c_1+c_2{x}^4$.