Question

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

Short answer

The solution for the differential equation is$y=c_1\text{cos}\left(2\text{ln}x\right)+c_2\text{sin}\left(2\text{ln}x\right)$

Step-by-step solution

Define Differential equation

Anequation with one or more derivatives of a function. The derivative of the function is given by $\frac{\text{d}y}{\text{d}x}$is 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,

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

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:

$$\begin{gathered} x^2\frac{{\text{d}}^{2}y}{\text{d}{x}^2}+x\frac{\text{d}y}{\text{d}x}+4y=0 \\ {x}^{2}m(m-1)x^{m-2}+xm{x}^{m-1}+4x^m=0 \end{gathered}$$

Keep$x^m$commonly out,

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

Solving for m:

$m^2+4=0$

Using Factorize method,

Subtract 4 on both side:

$$\begin{gathered} m^2+4-4=0-4 \\ {m}^2=-4 \end{gathered}$$

Take square root on both side:

$m_{1,2}=\pm \sqrt{-4}$

$m_1=2i$; $m_2=-2i$

Use Case 3: Conjugate complex roots

Then the general solution becomes,

$y=x^{\alpha }\left[c_1\text{cos}\left(\beta \text{ln}x\right)+c_2\text{sin}\left(\beta \text{ln}x\right)\right]$

Where the real parts of complex solution is$\alpha =0$ and positive complex$\beta =2$ as these are the conjugate complex roots of the equation.

Substitute$\alpha \mathrm{and}\beta$ , the solution is

$y=c_1\text{cos}\left(2\text{ln}x\right)+c_2\text{sin}\left(2\text{ln}x\right)$