Question

Solve the given differential equation.${\mathit{x}}^{\mathbf{2}}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{8}\mathit{x}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{6}\mathit{y}\mathbf{=}\mathbf{0}$

Short answer

The solution for the differential equation is $y=\frac{c_1}{x}+\frac{c_2}{x^6}$.

Step-by-step solution

Define Euler homogenous

The second-order equation is a differential equation because it includes the second derivative of y. It's homogeneous because the right side is 0. If the right side of the equation is non-zero, the differential equation is called nonhomogeneous.

Solve differential equation

$x^{2}y\text{'}\text{'}+8xy\text{'}+6y=0$

Use the form $a{x}^{2}y\text{'}\text{'}+bxy\text{'}+cy=0$ from a second order Euler homogenous ODE

Consider $y=x^r$,

$$\begin{gathered} x^2{\left(\left(x^r\right)\right)}^{\text{'}\text{'}}+8x{\left(\left(x^r\right)\right)}^{\text{'}}+6x^r=0 \\ {x}^2{\left(\left(x^r\right)\right)}^{\text{'}\text{'}}+8x{\left(\left(x^r\right)\right)}^{\text{'}}+6x^r=0 \end{gathered}$$

Simplify for $x^r$:

$x^r\left(r^2+7r+6\right)=0$
Use Factorize method, where $a=1,b=7,andc=6$:

$\begin{array}{rcl}r& =& \frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ & =& \frac{-7\pm \sqrt{7^2-4\left(1\right)\left(6\right)}}{2\left(1\right)}\\ & =& \frac{-7\pm \sqrt{49-24}}{2}\\ & =& \frac{-7\pm \sqrt{25}}{2}\\ & =& \frac{-7\pm 5}{2}\\ r_1& =& -1;r_2=-6\\ & & \end{array}$

Two real roots are not equal $r_1\ne r_2$,

Use Case 1: Distinct real roots

Then the general solution becomes,

$y=C_1{x}^{r_1}+C_2{x}^{r2}$

Substitute the values :

$y=c_1{x}^{-1}+c_2{x}^{-6}$

Therefore, the solution is $y=\frac{c_1}{x}+\frac{c_2}{x^6}$.