Question

Find the general solution of the following differential equations (complementary function + particular solution). Find the particular solution by inspection or by or .Also find a computer solution and reconcile differences if necessary, noticing especially whether the particular solution is in simplest form [seeand the discussion after].

$\mathbf{5}{\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{+}\mathbf{12}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{+}\mathbf{20}\mathit{y}\mathbf{=}\mathbf{120}\mathit{s}\mathit{i}\mathit{n}\mathbf{2}\mathit{x}$

Short answer

The general solution given by differential equation is$y\left(x\right)=(C_1\sin \frac{16}{10}x+C_2\cos \frac{16}{10}x)e^{\frac{-12}{10}x}-5\cos 2x$

Step-by-step solution

Given data. 

Given equation is$(5y^{\text{'}\text{'}}+12y^{\text{'}}+20y=120\sin 2x)$

General solution of differential equation

A general solution to the nth order differential equation is one that incorporates a significant number of arbitrary constants. If oneuses the variable approach to solve a first-order differential equation, one must insert an arbitrary constant as soon as integration is completed.

Step 3:Find the general solution of given differential equation.5y''+12y'+20y=120sin2x

Substitute the value as,

$D=y^{\text{'}},D^2=y^{\text{'}\text{'}}$

$\begin{array}{c}5y^{\text{'}\text{'}}+12y^{\text{'}}+20y=120\sin 2x\\ (5D^2+12D+20)y=120\sin 2x\end{array}$

The auxiliary equation can be written as

$5m^2+12m+20=0$

Solve value of m using discriminant method

$\begin{array}{l}\Rightarrow m=\frac{-12\pm \sqrt{144-400}}{10}\\ \Rightarrow m=\frac{-12\pm \sqrt{-256}}{10}\\ \Rightarrow m=\frac{-12\pm 16i}{10}\end{array}$

$\begin{array}{c}\text{C.}F=(C_1\sin \frac{16}{10}x+C_2\cos \frac{16}{10}x)e^{\frac{-12}{10}x}\\ P.I=\frac{1}{5D^2+12D+20}120\sin 2x\end{array}$

Solve the equation further

$\begin{array}{c}\Rightarrow \frac{1}{5(-4)+12D+20}120\sin 2x({\text{PutD}}^2=-a^2,\text{where}a=2)\\ \Rightarrow \frac{1}{12D}120\sin 2x\\ \Rightarrow \frac{-10\cos 2x}{2}\\ \Rightarrow -5\cos 2x\end{array}$

Solve the problem further as,

$\begin{array}{c}\text{P.}I=-5\cos 2x\\ \text{C.}S=(C_1\sin \frac{16}{10}x+C_2\cos \frac{16}{10}x)e^{\frac{-12}{10}x}-5\cos 2x\end{array}$

The solution of the differential equation can be written as

_{$y\left(x\right)=(C_1\sin \frac{16}{10}x+C_2\cos \frac{16}{10}x)e^{\frac{-12}{10}x}-5\cos 2x$}