Question

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

Short answer

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

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

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

Consider$y=x^r$,

$4x^2{\left(\left(x^r\right)\right)}^{\text{'}\text{'}}+4x{\left(\left(x^r\right)\right)}^{\text{'}}-x^r=0$

$4x^2{\left(\left(x^r\right)\right)}^{\text{'}\text{'}}+4x{\left(\left(x^r\right)\right)}^{\text{'}}-x^r=0$

Simplify for$x^r$:

$x^r\left(4r^2-1\right)=0$

Use Factorize method, $4r^2-1=0$, we have $a=4,b=0,andc=-1$. Therefore:

$\begin{array}{rcl}r& =& \frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ & =& \frac{-0\pm \sqrt{0^2-4\left(4\right)(-1)}}{2\left(4\right)}\\ & =& \frac{-0\pm \sqrt{0+16}}{8}\\ & =& \frac{-0\pm \sqrt{16}}{8}\\ & =& \frac{-0\pm 4}{8}\\ r& =& \frac{1}{2};r=-\frac{1}{2}\\ & & \end{array}$

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 value of$r_1$ and$r_2$ :

$C_1\sqrt{x}+C_2{x}^{-\frac{1}{2}}$

Therefore, the solution is $C_1\sqrt{x}+C_2{x}^{-\frac{1}{2}}$.

Graph

Assume$c_1=1$and$c_2=1$

We can plot,