Question

Solve the given differential equation by undetermined coefficients.

$\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{4}\mathit{y}\mathbf{=}\mathbf{(}{\mathit{x}}^{\mathbf{3}}\mathbf{-}\mathbf{3}\mathbf{)}\mathit{s}\mathit{i}\mathit{n}\mathbf{2}\mathit{x}$

Short answer

The general solution of the given differential equation is

$y=c_1{e}^{2x}+c_2{e}^{-2x}-\frac{1}{8}x\cos 2x+(\frac{-1}{8}{x}^2+\frac{11}{32})\sin 2x$

Step-by-step solution

Note the given data

Given the second order differential equation $y\text{'}\text{'}-4y=(x^3-3)\sin 2x$

Obtain the general solution of the homogeneous differential equation  

We know a homogeneous differential equation is an equation containing

differentiation and a function with a set of variables.

First, we need to solve the associated homogeneous equation$y\text{'}\text{'}-4y=0$

Corresponding auxiliary equation is$m^2-4=0$

Solving,

$m^2-(2{)}^2=0$

m₁ =-2 and m₂ = 2, which are real and distinct.

So the complementary function is

$y_c=c_1{e}^{2x}+c_2{e}^{-2x}$,c₁ and c₂ are arbitrary constants.

Finding the particular solution

Given non homogeneous differential equation is $y\text{'}\text{'}-4y=(x^3-3)\sin 2x$

$y_p=A\sin 2x+B\cos 2x$

Assume that $y_p=(A_1{x}^2+B_{1}x+C_1)\cos 2x+(A_2{x}^2+B_{2}x+C_2)\sin 2x$……(1) is a solution of thr non homogeneous differential equation

Differentiate with respect to x

$y_p\text{'}=\left[\right(2A_2{x}^2+(2A_1+2B_2)x+(B_1+2C_2)]\cos 2x+[(-2A_1{x}^2+(2A_2-2B_1)x+(B_2-2C_1\left)\right]\sin 2x$

$$\begin{gathered} y_p\text{'}\text{'}=\left[\right(-4A_1{x}^2+(8A_2-2B_1)x+(2A_1+4B_2-4C_1)]\cos 2x+[(-4A_2{x}^2+(-8A_1-4B_2)x+ \\ (2A_1-4B_1-4C_2\left)\right]\sin 2x \end{gathered}$$……(2)

Substituting (1) and (2) in given differential equation

$$\begin{gathered} \left[\right(-4A_1{x}^2+(8A_2-2B_1)x+(2A_1+4B_2-4C_1)]\cos 2x+[(-4A_2{x}^2+(-8A_1-4B_2)x+ \\ (2A_1-4B_1-4C_2\left)\right]\sin 2x+-4(A_1{x}^2+B_{1}x+C_1)\cos 2x-4(A_2{x}^2+B_{2}x+C_2)\sin 2x=(x^3-3)\sin 2x \end{gathered}$$

or

$$\begin{gathered} \cos 2x[-8A_1{x}^2+(8A_2-8B_1)x+(2A_1+4B_2-8C_1\left)\right]+\sin 2x[-8A_2{x}^2+(-8A_1-8B_2)x+ \\ (2A_2-4B_2-8C_2\left)\right]=(x^3-3)\sin 2x \end{gathered}$$

comparing the coefficients,

$8A_2-8B_1=0$

$2A_1+4B_2-8C_1=0$

$2A_1+4B_2-8C_1=0$

$-8A_2=0$

$-8A_1-8B_2=0$

$2A_2-4B_2-8C_2=-3$

Solving these equations

$A_1=0,B_1=\frac{-1}{8},C_1=0$

$A_2=\frac{-1}{8},B_2=0,C_2=\frac{11}{32}$

Hence the particular solution is (1) becomes

$y_p=\frac{-1}{8}x\cos 2x+\left(\frac{-1}{8}{x}^2+\frac{11}{32}\right)\sin 2x$

Finding the solution

Hence the solution of given differential equation is

y=y_c+y_p

$y=c_1{e}^{2x}+c_2{e}^{-2x}-\frac{1}{8}x\cos 2x+(\frac{-1}{8}{x}^2+\frac{11}{32})\sin 2x$