Question

Solve the following equations using method (d) above.

$x^2{y}^{\text{'}\text{'}}+y=3x^2$

Short answer

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

Step-by-step solution

Given information

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

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 the auxiliary equation

Consider the given equation.

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

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,

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

The auxiliary equation of the above equation is,

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

The solution of the auxiliary equation is,

$m=\frac{1}{2}\pm \frac{\sqrt{3}}{2}i$

Solve for yp

Now,

$Q=3e^{2z}$

So,

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

Complete solution

Thus, the complete solution is given as,

$y=y_c+y_p$

$=e^{z/2}\left(c_{1}cos\frac{\sqrt{3}z}{2}+c_{2}sin\frac{\sqrt{3}z}{2}\right)+e^{2z}$

$=\sqrt{x}\left(c_{1}cos\frac{\sqrt{3}(lnx)}{2}+c_{2}sin\frac{\sqrt{3}(lnx)}{2}\right)+x^2$

Therefore, the general solution of the equation is $y=\sqrt{x}\left(c_{1}cos\frac{\sqrt{3}(lnx)}{2}+c_{2}sin\frac{\sqrt{3}(lnx)}{2}\right)+x^2$.