Question

Solve the following equations using method (d) above. $x^2{y}^{\text{'}\text{'}}+x{y}^{\text{'}}-16y=8x^4$

Short answer

The general solution of the equation is$y=c_1{x}^{-4}+c_2{x}^4+(lnx)x^4$

Step-by-step solution

Given information

The given differential equation is $x^2{y}^{\text{'}\text{'}}+3x{y}^{\text{'}}-3y=8x^4$.

Auxiliary equation

Auxiliary equation is an algebraic equation of degree $n$upon which depends the solution of a given $n^{th}$-order differential equation or difference equation.

Solve for double differential

Consider the given equation.

$$\begin{gathered} x^2{y}^{\text{'}\text{'}}+3x{y}^{\text{'}}-3y=0 \\ \text{Let}x=e^z. \\ \text{So,}x=e^z \\ z=lnx \end{gathered}$$

$\frac{dz}{dx}=\frac{1}{x}$

Now,

And,

$x\frac{dy}{dx}=\frac{dy}{dz}$

Roots of equation

The given differential equation can be written as,

$\left(D\right(D-1)+D-16)y=8e^{4z}$

The auxiliary equation of the above equation is,

$m(m-1)+m-16=0$

The solution of the auxiliary equation is,

$m=\pm 4$

Let the roots be represented as,

$$\begin{gathered} a=-4 \\ b=4 \end{gathered}$$

Solve for Yp

Now,

$Q=8e^{4z}$

So,

Now,

$D$represent the first derivative and $D^2$represent the second derivative.

So,

Complete solution

Thus, the complete solution is given as,

Therefore, the general solution of the equation is $y=c_1{x}^{-4}+c_2{x}^4+(lnx)x^4$