Question: Identify each of the differential equations in Problems 1to 24 as to type (for example, separable, linear first order, linear second order, etc.), and then solve it.

$\mathit{x}{\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{+}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{=}\mathbf{4}\mathit{x}$

The differential equation is $x{y}^{\text{'}\text{'}}+y^{\text{'}}=4x$.

When fand its derivatives are inserted into the equation, a solution is a function y = f(x) that solves the differential equation. The highest order of any derivative of the unknown function appearing in the equation is the order of a differential equation.

A differential equation of the form$\mathbf{(}\mathit{D}\mathbf{-}\mathit{a}\mathbf{)}\mathbf{(}\mathit{D}\mathbf{-}\mathit{b}\mathbf{)}\mathit{y}\mathbf{=}\mathbf{0}\mathbf{,}\mathit{a}\mathbf{\ne }\mathit{b}$ has general solution$\mathit{y}\mathbf{=}{\mathit{c}}_{\mathbf{1}}{\mathit{e}}^{\mathbf{a}\mathbf{x}}\mathbf{+}{\mathit{c}}_{\mathbf{2}}{\mathit{e}}^{\mathbf{b}\mathbf{x}}\mathbf{.}$

Consider the equation.

$x{y}^{\text{'}\text{'}}+y^{\text{'}}=4x$

Rearrange above equation.

$x^2{y}^{\text{'}\text{'}}+y^{\text{'}}=4x^2$

The above equation is a non-homogenous equation.

The solution of equation is,

$y=y_c+y_p\dots \dots \dots \dots \dots .\left(1\right)$

Assume x = e^z

$\begin{array}{c}z=\ln x\\ \frac{dz}{dx}=\frac{1}{x}\\ \frac{dy}{dx}=\frac{dy}{dx}\frac{1}{x}\\ x^2\frac{d^{2}y}{d{x}^2}=\frac{d^{2}y}{d{z}^2}-\frac{dy}{dz}\end{array}$

Substitute the values in above equation.

$\begin{array}{c}\left(D\right(D-1)+D)y=4e^{2x}\\ D^{2}y=4e^{2x}\end{array}$

The auxiliary equation is,

$\begin{array}{c}{m}^3=0\\ m=0,0\end{array}$

The solution for m = 0 is,

$\begin{array}{c}{y}_c=\left(c_{1}z+c_2\right)e^{0\times z}\\ =\left(c_{1}z+c_2\right)\end{array}$

The value of y_p is,

$\begin{array}{c}{y}_p=\frac{1}{P\left(D\right)}4e^{2x}\\ =\frac{1}{2^2}4e^{2x}\\ =e^{2x}\end{array}$

Substitute the values of y_p and y_c in Equation (1).

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

Thus, the solution of given differential equation is $c_1\ln x+c_2+x^2$.