Question

Solve the given differential equation.${\mathit{x}}^{\mathbf{2}}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{3}\mathit{x}\mathit{y}\mathbf{\text{'}}\mathbf{-}\mathbf{4}\mathit{y}\mathbf{=}\mathbf{0}$

Short answer

The solution for the differential equation is$y=c_1{x}^{-1+\sqrt{5}}+c_2{x}^{-1-\sqrt{5}}$

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 the differential equation

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

Consider$y=x^r$ , this implies that:

$x^2{\left(\left(x^r\right)\right)}^{\text{'}\text{'}}+3x{\left(\left(x^r\right)\right)}^{\text{'}}-4x^r=0$

$x^2{\left(\left(x^r\right)\right)}^{\text{'}\text{'}}+3x{\left(\left(x^r\right)\right)}^{\text{'}}-4x^r=0$

Simplify forrby Roots method,

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

$r_1$$=-1+\sqrt{5}$;$r_2$$=-1-\sqrt{5}$

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 $r_2$.

Therefore, the solution is$y=c_1{x}^{-1+\sqrt{5}}+c_2{x}^{-1-\sqrt{5}}$ .

Graph

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

We can plot,