Question

Solve the following equations using method (d) above $x^2{y}^{\text{'}\text{'}}-5x{y}^{\text{'}}+9y=2x^3$.

Short answer

The general solution of the equation is$y=\left(c_1\ln x+c_2\right)x^3+(\ln x{)}^2{x}^3$ .

Step-by-step solution

Given information

The given differential equation is$x^2{y}^{\text{'}\text{'}}-5x{y}^{\text{'}}+9y=2x^3$ .

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 auxiliary equation

Consider the given equation.

$$\begin{gathered} x^2{y}^{\text{'}\text{'}}-5x{y}^{\text{'}}+9y=2x^3 \\ \text{Let}x=e^z· \\ \text{So,}x=e^z \\ z=\ln x \end{gathered}$$

Solve further

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

Now,

$$\begin{gathered} \frac{dy}{dx}=\frac{dy}{dz}·\frac{dz}{dx} \\ =\frac{dy}{dz}\left(\frac{1}{x}\right) \\ \frac{d^{2}y}{d{x}^2}=\frac{1}{x^2}\left(\frac{d^{2}y}{d{z}^2}-\frac{dy}{dz}\right) \\ {x}^2\frac{d^{2}y}{d{x}^2}=\frac{d^{2}y}{d{z}^2}-\frac{dy}{dz} \end{gathered}$$

And,

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

The given differential equation can be written as,

$\left(D\right(D-1)-5D+9)y=2e^{3z}$

The auxiliary equation of the above equation is,

$m(m-1)-5m+9=0$

The solution of the auxiliary equation is,

$m=3,3$

Solve for yp

Now,

$Q=2e^{3z}$

So,

$$\begin{gathered} y_p=\frac{2e^{3z}}{\left(D^2\right)} \\ =2e^{3z}\frac{1}{D}\int 1dz \\ =2e^{3z}\int zdz \\ =z^2{e}^{3z} \end{gathered}$$

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

Complete solution

Thus, the complete solution is given as,

$$\begin{gathered} y=y_{\text{c}}+y_{\text{p}} \\ =\left(c_{1}z+c_2\right)e^{3z}+z^2{e}^{3z} \\ =\left(c_1\ln x+c_2\right)x^3+(\ln x{)}^2{x}^3 \end{gathered}$$

Therefore, the general solution of the equation is $y=\left(c_1\ln x+c_2\right)x^3+(\ln x{)}^2{x}^3$.