Question

Solve the given differential equation by undetermined coefficients.

$\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{2}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{5}\mathit{y}\mathbf{=}{\mathit{e}}^{\mathbf{x}}\mathit{c}\mathit{o}\mathit{s}\mathbf{2}\mathit{x}$

Short answer

The general solution of the given differential equation is

$y=e^x(k_1\cos 2x+k_2\sin 2x)+\frac{1}{4}x{e}^x\sin 2x$

Step-by-step solution

Note the given data

Given the second order differential equation $y\text{'}\text{'}-2y\text{'}+5y=e^x\cos 2x$

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$y\text{'}\text{'}-2y\text{'}+5y=0$

Corresponding auxiliary equation is$m^2-2m+5=0$

Solving,

m₁ =1+2i and m₂ = 1-2i, which are complex and conjugate.

So the solution of corresponding homogeneous equation is

$y_h=c_1{e}^{(1+2i)x}+c_2{e}^{(1-2i)x}=e^x[c_1{e}^{\left(2i\right)x}+c_2{e}^{(-2i)x}]=e^x\left[c_1\right(\cos 2x+\sin 2x)+c_2(\cos 2x-\sin 2x\left)\right]$where k₁=c₁+c₂ and k₂=c₁- c₂ are arbitrary constants

$y_h=e^x(k_1\cos 2x+k_2\sin 2x)$

Finding the particular solution

Given non homogeneous differential equation is $y\text{'}\text{'}-2y\text{'}+5y=e^x\cos 2x$

Assume that $y_p=Ax{e}^x\cos 2x+Bx{e}^x\sin 2x$…..(1) is a solution of the non-homogeneous differential equation

Differentiate with respect to x

$\backslash \left[\right\{y\_p\}\text{'}=A\{e^x\}\backslash \cos 2x+Ax\{e^x\}\backslash \cos 2x-2Ax\{e^x\}\backslash \sin 2x+Bx\{e^x\}\backslash \sin 2x+B\{e^x\}\backslash \sin 2x+2Bx\{e^x\}\backslash \cos 2x\backslash ]or\backslash \left[\right\{y\_p\}\text{'}=[(A+2B)x+A\left]\right\{e^x\}\backslash \cos 2x+[(B-2A)x+B\left]\right\{e^x\}\backslash \sin 2x\backslash ]$ ……(2)

$y_p\text{'}\text{'}=\left[\right(4B-3A)x+(2A+4B\left)\right]e^x\cos 2x+\left[\right(-4A-3B)x+(2B-4A\left)\right]e^x\sin 2x$……(3)

Substituting (1) and (2) and (3) in given differential equation

$$\begin{gathered} \left[\right(4B-3A)x+(2A+4B\left)\right]e^x\cos 2x+\left[\right(-4A-3B)x+(2B-4A\left)\right]e^x\sin 2x \\ -2\left[\right(A+2B)x+A]e^x\cos 2x+\left[\right(B-2A)x+B]e^x\sin 2x]+5[Ax{e}^x\cos 2x+Bx{e}^x\sin 2x]=e^x\cos 2x \end{gathered}$$

Or$4B{e}^x\cos 2x-4A{e}^x\sin 2x=e^x\cos 2x$

comparing the coefficients,

$4B=1\Rightarrow B=\frac{1}{4}$

$4A=0\Rightarrow A=0$

Hence the particular solution is (1) becomes

$y_p=\frac{1}{4}x{e}^x\sin 2x$

 Step 5 : Finding the solution

Hence the solution of given differential equation is

y=y_h+y_p

$y=e^x(k_1\cos 2x+k_2\sin 2x)+\frac{1}{4}x{e}^x\sin 2x$