Question

Solve the following equations using method (d) above.

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

Short answer

General solution of the equation is $y=c_1\cos \left(\ln x\right)+c_2\sin \left(\ln x\right)+x$.

Step-by-step solution

Given information

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

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.

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

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,

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

Rewrite the given differential equation and solve

The given differential equation can be written as,

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

The auxiliary equation of the above equation is,

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

The solution of the auxiliary equation is,

$m=\pm 1i$

Now,

data-custom-editor="chemistry" $Q=2e^z$

So,

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

The general solution of the equation

Thus, the complete solution is given as,

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

Therefore, the general solution of the equation is $y=c_1\cos \left(\ln x\right)+c_2\sin \left(\ln x\right)+x$.