Question

solve the given differential equation.${\mathit{x}}^{\mathbf{2}}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{5}\mathit{x}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{4}\mathit{y}\mathbf{=}\mathbf{0}$

Short answer

The solution for the differential equation is$y=c_1{x}^{-2}+c_2{x}^{-2}\text{ln}x$

Step-by-step solution

Define Differential equation

An equation with one or more derivatives of a function. The derivative of the function is given by dy/dxis known as Differential equation.

Solve differential equation

Consider,

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

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

Substitute in the above equation,

$x^2\frac{{\text{d}}^{2}y}{\text{d}{x}^2}+5x\frac{\text{d}y}{\text{d}x}+4y=0$

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$x^{2}m(m-1)x^{m-2}+5xm{x}^{m-1}+4x^m=0$

Keep$x^m$commonly out,

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

Solving for m,

$m^2+4m+4=0$

Use factorize method, square on both sides,

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

Where$m_1=-2$ .

Use case 2: Repeated real roots

Then the general solution becomes,

$y=c_1{x}^{m_1}+c_2{x}^{m_1}\text{ln}x$

After substituting $m_1$value in above equation, the solution is$y=c_1{x}^{-2}+c_2{x}^{-2}\text{ln}x$ .