Question

Solve the given differential equation.$\mathbf{3}{\mathit{x}}^{\mathbf{2}}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{6}\mathit{x}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathit{y}\mathbf{=}\mathbf{0}$

Short answer

The solution for the differential equation is $y=x^{-\frac{1}{2}}\left[c_1\text{cos}\left(\frac{1}{6}\sqrt{3}\text{ln}x\right)+c_2\text{sin}\left(\frac{1}{6}\sqrt{3}\text{ln}x\right)\right]$ .

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

Let,

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

So, the equation is $3x^2\left[m(m-1)x^{m-2}\right]+6x\left[m{x}^{m-1}\right]+x^m=0$

Simplify the equation as:

$$\begin{gathered} 3m{x}^m(m-1)+6m{x}^m+x^m=0 \\ 3m^2{x}^m-3m{x}^m+6m{x}^m+x^m=0 \\ 3m^2{x}^m+3m{x}^m+x^m=0 \\ {x}^m\left(3m^2+3m+1\right)=0 \end{gathered}$$

Solving for m,

$\left(3m^2+3m+1\right)=0$

Use factorize method,

$\begin{array}{rcl}{m}_{1,2}& =& \frac{-3\pm \sqrt{{\left(3\right)}^2-4\left(3\right)\left(1\right)}}{2\left(3\right)}\\ m_{1,2}& =& \frac{-3\pm \sqrt{-3}}{6}\\ & =& \frac{-3}{6}\pm \frac{\sqrt{3}i}{6}\\ & =& -\frac{1}{2}\pm \frac{1}{6}\sqrt{3}i\\ m_1& =& -\frac{1}{2}+\frac{1}{6}\sqrt{3}i;m_2=-\frac{1}{2}-\frac{1}{6}\sqrt{3}i\\ & & \end{array}$

Consider, m $=\alpha -i\beta$. By comparing it with equation, we get,

$\alpha =-\frac{1}{2}$ ; $\beta =\frac{1}{6}\sqrt{3}$

Use Case 3: Conjugate complex roots

Then the general solution becomes,

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

After substituting the values, the solution is

$y=x^{-\frac{1}{2}}\left[c_1\text{cos}\left(\frac{1}{6}\sqrt{3}\text{ln}x\right)+c_2\text{sin}\left(\frac{1}{6}\sqrt{3}\text{ln}x\right)\right]$