Question

Solve the given differential equation by undetermined coefficients.

$\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{25}\mathit{y}\mathbf{=}\mathbf{20}\mathit{s}\mathit{i}\mathit{n}\mathbf{5}\mathit{x}$

Short answer

$y=k_1\cos 5x+k_2\sin 5x-2x\cos 5x$

Step-by-step solution

Note the given data

Given the second order differential equation $y\text{'}\text{'}+25y=20\sin 5x$

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$y\text{'}\text{'}+25y=0$

Corresponding auxiliary equation is$m^2+25=0$

Solving,

m₁ =+5i and m₂ = -5i, which are complex and conjugate

So the homogeneous solution of the given differential equation is

$y_h=k_1\cos 5x+k_2\sin 5x$

Obtain the particular solution of the non-homogeneous differential equation  

Given non homogeneous differential equation is$y\text{'}\text{'}+25y=20\sin 5x$

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

$(D^2+25)y=20\sin 5x$

On applying the annihilator $(D^2+25)$ (comparing with $20\sin \beta x$, $\alpha =0,\beta =5,n=1$,$20\sin 5x$is annihilated$(D^2-2\alpha D+{\alpha }^2+{\beta }^2{)}^n$ )

$$\begin{gathered} (D^2+25)(D^2+25)y=(D^2+25)20\sin 2x \\ \Rightarrow (D^2+25)(D^2+25)y=0 \end{gathered}$$

Then the new auxiliary equation is

$\Rightarrow (m^2+25)(m^2+25)=0$

The roots are m₁= +5i, m₂= -5i, m₃= +5i, m₄= -5i.which are complex and repeated

Finding the particular solution

We can obtain the general solution of the given differential equation$y\text{'}\text{'}+25y=20\sin 5x$ as $y=k_1\cos 5x+k_2\sin 5x-k_{3}x\cos 5x+k_{4}x\sin 5x$

Particular solution will be $y_p=k_{3}x\cos 5x+k_{4}x\sin 5x$

We can find this particular solution by assuming $y_p=Ax\cos 5x+Bx\sin 5x$…(1)

Differentiate with respect to x

$$\begin{gathered} y_p\text{'}=Ax\cos 5x-5Ax\sin 5x+B\sin 5x+5Bx\cos 5x \\ {y}_p\text{'}\text{'}=5B\cos 5x-5(5Bx+A)\sin 5x-5A\sin 5x+5(-5Ax+B)\cos 5x \end{gathered}$$

Or $y_p\text{'}\text{'}=(-25Ax+10B)\cos 5x-(25Bx+10A)\sin 5x$……(2)

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

$(-25Ax+10B)\cos 5x-(25Bx+10A)\sin 5x-25\left\{\right(Ax\cos 5x+Bx\sin 5x\}=20\sin 5x$

Or$\{10B\cos 5x-10A\sin 5x\}=20\sin 5x$

comparing the coefficients,

10B = 0$\Rightarrow B=0$

$-10A=20\Rightarrow A=-2$

Hence the particular solution is

1. becomes

$y_p=-2x\cos 5x$

Finding the solution

Hence the solution of given differential equation is

y = y_h+ y_p

$y=k_1\cos 5x+k_2\sin 5x-2x\cos 5x$