Question

Solve the given differential equation by undetermined coefficients.

$\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathit{y}\mathbf{\text{'}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{4}}\mathit{y}\mathbf{=}{\mathit{e}}^{\mathbf{x}}\mathbf{(}\mathit{s}\mathit{i}\mathit{n}\mathbf{3}\mathit{x}\mathbf{-}\mathit{c}\mathit{o}\mathit{s}\mathbf{3}\mathit{x}\mathbf{)}$

Short answer

$y=k_1{e}^{-\frac{1}{2}x}+k_{2}x{e}^{-\frac{1}{2}x}-\frac{4}{225}{e}^x\cos 3x-\frac{28}{225}{e}^x\sin 3x$

Step-by-step solution

Note the given data

Given the second order differential equation $y\text{'}\text{'}+y\text{'}+\frac{1}{4}y=e^x(\sin 3x-\cos 3x)$

Obtain the general solution of the homogeneous differential equation  

Weknow ahomogeneous differential equation is an equation containing

differentiation and a function with a set of variables.

First, we need to solve the associated homogeneous equation$y\text{'}\text{'}+y\text{'}+\frac{1}{4}y=0$

Corresponding auxiliary equation is$m^2+m+\frac{1}{4}=0$

Solving,$4m^2+4m+1=0$

$(2m+1{)}^2=0$

m₁ =$-\frac{1}{2}$and m₂ = $-\frac{1}{2}$, which are real and equal.

So the homogeneous solution of the given differential equation is

$y_h=k_1{e}^{-\frac{1}{2}x}+k_{2}x{e}^{-\frac{1}{2}x}$

Obtain the particular solution of the non-homogeneous differential equation  

Given non homogeneous differential equation is$y\text{'}\text{'}+y\text{'}+\frac{1}{4}y=e^x(\sin 3x-\cos 3x)$

Writing the differential equation in terms of its linear operator D, we have

$(D^2+D+\frac{1}{4})y=e^x(\sin 3x-\cos 3x)$

On applying the annihilator $(D^2-2D+10)$ (comparing with $e^{\alpha x}(\sin \beta x-\cos \beta x)$,$\alpha =1,\beta =3,n=1$ ,$e^x\sin x$is annihilated$(D^2-2\alpha D+{\alpha }^2+{\beta }^2{)}^n$ )

$$\begin{gathered} (D^2-2D+10)(D^2+D+\frac{1}{4})y=(D^2-2D+10)e^x(\sin 3x-\cos 3x) \\ \Rightarrow (D^2-2D+10)(D^2+D+\frac{1}{4})y=0 \end{gathered}$$

Then the new auxiliary equation is

$$\begin{gathered} \Rightarrow (m^2-2m+10)(m^2+m+\frac{1}{4})=0 \\ \Rightarrow (m^2-2m+10)(m+\frac{1}{2}{)}^2=0 \end{gathered}$$

Solving,

The roots are m_1,2= , m₃= 1+3i, m₄= 1-3i, which are combination of real and complex conjugate.

Finding the particular solution

We can obtain the general solution of the given differential equation

$y\text{'}\text{'}+y\text{'}+\frac{1}{4}y=e^x(\sin 3x-\cos 3x)$as $y=k_1{e}^{-\frac{1}{2}x}+k_{2}x{e}^{-\frac{1}{2}x}+k_3{e}^x\cos 3x+k_4{e}^x\sin 3x$

Particular solution will be $y_p=k_3{e}^x\cos 3x+k_4{e}^x\sin 3x$

We can find this particular solution by assuming $y_p=A{e}^x\cos 3x+B{e}^x\sin 3x$…(1)

Differentiate with respect to x

$$\begin{gathered} y_p\text{'}=A{e}^x\cos 3x-3A{e}^x\sin 3x+B{e}^x\sin 3x+3B{e}^x\cos 3x \\ {y}_p\text{'}\text{'}=(A+3B)e^x\cos 3x-3(A+3B)e^x\sin 3x+(B-3A)e^x\sin 3x+(B-3A)e^x\cos 3x \end{gathered}$$

Or $y_p\text{'}\text{'}=(6B-8A)e^x\cos 3x-(6A+8B)e^x\sin 3x$……(2)

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

$$\begin{gathered} (6B-8A)e^x\cos 3x-(6A+8B)e^x\sin 3x)+(A+B)e^x\cos 3x+(B-A3\left)e^x\sin 3x\right\} \\ \frac{1}{4}\{A{e}^x\cos 3x+B{e}^x\sin 3x\}=e^x(\sin 3x-\cos 3x) \end{gathered}$$

Or$(9B-\frac{27}{4}A)e^x\cos 3x+(-9A-\frac{27}{4}B)e^x\sin 3x\}=e^x(\sin 3x-\cos 3x)$

comparing the coefficients,

$\begin{array}{rcl}(9B-\frac{27}{4}A)& =& -1\\ (-9A-\frac{27}{4}B& =& 1\end{array}$

Solving these two equations, we get

$$\begin{gathered} A=-\frac{4}{225} \\ B=-\frac{28}{225} \end{gathered}$$

Hence the particular solution is

1. becomes

$y_p=-\frac{28}{225}{e}^x\sin 3x-\frac{4}{225}{e}^x\cos 3x$

Finding the solution

Hence the solution of given differential equation is

y = y_h+ y_p

$y=k_1{e}^{-\frac{1}{2}x}+k_{2}x{e}^{-\frac{1}{2}x}-\frac{4}{225}{e}^x\cos 3x-\frac{28}{225}{e}^x\sin 3x$