Question

Solve the given differential equation by undetermined coefficients.

$\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathit{y}\mathbf{=}\mathbf{2}\mathit{x}\mathit{s}\mathit{i}\mathit{n}\mathit{x}$

Short answer

The general solution of the given differential equation is

$y=k_1\cos x+k_2\sin x-\frac{1}{2}{x}^2\cos x+\frac{1}{2}x\sin x$

Step-by-step solution

Note the given data

Given the second order differential equation$y\text{'}\text{'}+y=2x\sin x$

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$y\text{'}\text{'}+y=0$

Corresponding auxiliary equation is$m^2+1=0$

Solving,

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

m₁ =-i and m₂ = i, which are complex and conjugate.

So the complementary function is

$y_c=c_1{e}^{ix}+c_2{e}^{-ix}=k_1\cos x+k_2\sin x$

Finding the particular solution

Given non homogeneous differential equation is $y\text{'}\text{'}+y=2x\sin x$

Assume that$y_p=(A{x}^2+Bx)\cos x+(C{x}^2+Dx)\sin x$……(1) is a solution of thenon homogeneous differential equation

Differentiate with respect to x

$y_p\text{'}=\left[\right(2Ax+B)\cos x+(2Cx+D)\sin x-(A{x}^2+Bx)\sin x+(C{x}^2+Dx)\cos x$

Or$y_p\text{'}=[C{x}^2+(2A+D)x+B]\cos x+[-A{x}^2+(-B+2C)x+D]\sin x$

$$\begin{gathered} y_p\text{'}\text{'}=[2Cx+(2A+D)+]\cos x+[-2Ax+(-B+2C\left)\right]\sin x-[C{x}^2+(2A+D)x+B]\sin x \\ +[-A{x}^2+(-B+2C)x+D]\cos x \end{gathered}$$

or $y_p\text{'}\text{'}=[-A{x}^2+(-B+4C)x+(2A+2D\left)\right]\cos x+[-C{x}^2+(-4A-D)x+(2C-2B\left)\right]\sin x$……(2)

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

$$\begin{gathered} [-A{x}^2+(-B+4C)x+(2A+2D\left)\right]\cos x+[-C{x}^2+(-4A-D)x+(2C-2B\left)\right]\sin x+(A{x}^2+Bx)\cos x+ \\ (C{x}^2+Dx)\sin x=2x\sin x \end{gathered}$$

or

$[4Cx+(2A+2D)\cos x+[-4Ax+(2C-2B]\sin x=2x\sin x$

comparing the coefficients,

$4C=0\Rightarrow C=0$

$-4A=2\Rightarrow A=\frac{-1}{2}$

$2A+2D=0\Rightarrow D=\frac{1}{2}$

$2C-2B=0\Rightarrow B=0$

Hence the particular solution is (1) becomes

$y_p=\frac{-1}{2}{x}^2\cos x+\frac{1}{2}x\sin x$

Finding the solution

Hence the solution of given differential equation is

y=y_c+y_p

$y=k_1\cos x+k_2\sin x-\frac{1}{2}{x}^2\cos x+\frac{1}{2}x\sin x$