Question

In Problems 1–26 solve the given differential equation by undetermined coefficients.

$\mathbf{4}\mathit{y}\mathbf{-}\mathbf{4}\mathit{y}\mathbf{\text{'}}\mathbf{-}\mathbf{3}\mathit{y}\mathbf{=}\mathit{c}\mathit{o}\mathit{s}\mathbf{2}\mathit{x}$

Short answer

$y=c_1{e}^{-\frac{1}{2}x}+c_2{e}^{-\frac{3}{2}x}-\frac{19}{425}\cos 2x-\frac{8}{425}\sin 2x$

Step-by-step solution

General solution for homogeneous differential equation

We have the differential equation

$4y-4y\text{'}-3y=\cos 2x$

Consider$y=e^{mx}$as the solution of the differential equation,

Substitute$y=e^{mx},y\text{'}=m{e}^{mx},y\text{'}\text{'}=m^2{e}^{mx}$,into $4y-4y\text{'}-3y=0$to obtain the auxiliary equation.

$\begin{array}{c}4m^2{e}^{mx}-4m{e}^{mx}-3e^{mx}=0\\ e^{mx}\left(4m^2-4m-3\right)=0\end{array}$

Since$e^{mx}$cannot be equal to 0, then we have

$\begin{array}{c}4m^2-4m-3=0\\ \left(2m+1\right)\left(2m-3\right)=0\end{array}$

Then the roots are

$m_1=-\frac{1}{2} , m_2=\frac{3}{2}$

which are real and distinct roots.

Then the solution of the corresponding homogeneous equation$4y-4y\text{'}-3y=0$can be given by localid="1663832209347" ${\mathit{y}}_{\mathbf{h}}\mathbf{=}{\mathit{c}}_{\mathbf{1}}{\mathit{e}}^{\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{x}}\mathbf{+}{\mathit{c}}_{\mathbf{2}}{\mathit{e}}^{\frac{\mathbf{3}}{\mathbf{2}}\mathbf{x}}$

General solution for non-homogeneous differential equation

Second, we have to find the particular solution of the non-homogeneous differential

equation$4y-4y\text{'}-3y=\cos 2x$as the following technique :

Assume that$y_p=A\cos 2x+B\sin 2x$ (where )is a solution for the non-homogeneous differential equation.

After that, differentiate the assumption with respect to , then we obtain

$y_p\text{'}=-2A\sin 2x+2B\cos 2x$ and $y_p\text{'}\text{'}=-4A\cos 2x-4B\sin 2x$

Substitute withrole="math" localid="1663831921898" $y_p=A\cos 2x+B\sin 2x$, equation (5) and (6) into the non-homogeneous

differential equation, then we obtain

$\begin{array}{c}4\left(-4A\cos 2x-4B\sin 2x\right)-4\left(-2A\sin 2x+2B\cos 2x\right)-3\left(A\cos 2x+B\sin 2x\right)=\cos 2x\\ \left(-19A-8B\right)\cos 2x+\left(8A-19B\right)\sin 2x=\cos 2x\end{array}$

Then we have the two equations

$\begin{array}{c}-19A-8B=1\\ 8A-19B=0\end{array}$

By solving these equations,

$A=-\frac{19}{425}$ and $B=-\frac{8}{425}$

Then the particular solution becomes $y_p=-\frac{19}{425}\cos 2x-\frac{8}{425}\sin 2x$

Then we obtain

$\begin{array}{l}y=y_h+y_p\\ =c_1{e}^{-\frac{1}{2}x}+c_2{e}^{-\frac{3}{2}x}-\frac{19}{425}\cos 2x-\frac{8}{425}\sin 2x\end{array}$

is the general solution of the non-homogeneous differential equation.

Therefore,the generalsolution of the given differential equation by undetermined coefficients is found to be.

$y=c_1{e}^{-\frac{1}{2}x}+c_2{e}^{-\frac{3}{2}x}-\frac{19}{425}\cos 2x-\frac{8}{425}\sin 2x$