Question

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

$\mathbf{4}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{9}\mathit{y}\mathbf{=}\mathbf{15}\mathbf{ }$

Short answer

$y=c_1 \cos \left(\frac{3}{2}x\right)+c_2 \sin \left(\frac{3}{2}x\right)+\frac{5}{3}$

Step-by-step solution

Solving equation

We have the non-homogeneous differential equation

$4y\text{'}\text{'}+9y=15$

Consider as the solution of the differential equation,

$y=e^{mx}$

Substitute $y=e^{mx},y\text{'}=m{e}^{mx}$,$y\text{'}\text{'}=m^2{e}^{mx}$ into$4y\text{'}\text{'}+9y=0$ to obtain the auxiliary equation.

$\begin{array}{c}4m^2 e^{mx}+9e^{mx}=0\\ e^{mx}\left(4m^2+9\right)=0\end{array}$

For any real$x,e^{mx}\ne 0$, we have

$\begin{array}{c}4m^2+9=0\\ 4m^2=-9\end{array}$

Then the eigenvalues are,

${\lambda }_1=\frac{3}{2}i , {\lambda }_2=-\frac{3}{2}i$

which are conjugate complex

Then the solution of the corresponding homogeneous equation $4y\text{'}\text{'}+9y=0$can be given as $y=c_1 \cos \left(\frac{3}{2}x\right)+c_2 \sin \left(\frac{3}{2}x\right)$

Solution of the non-homogeneous differential equation

Second, we have to find the particular solution of the non-homogeneous differential equation$4y\text{'}\text{'}+9y=15$as the following technique:

Assume that$y_p=A{e}^x+Ba$is a solution for the non-homogeneous differential equation.

After that, differentiate the assumption with respect to x, and substitute $y{\text{'}}_p=A{e}^x,y\text{'}{\text{'}}_p=A{e}^x$into $y\text{'}\text{'}+3y\text{'}+2y=6$to obtain the auxiliary equation.

$\begin{array}{c}4A{e}^x+9A{e}^x+9B=15 \\ 13A{e}^x+9B=15\\ A=0 B=\frac{5}{3}\end{array}$

Then the particular solution becomes

role="math" localid="1663845703205" width="242" height="74">y=yh+yp    =c1  cos  32x+c2  sin  32x+53

Then we obtain

Hence the final solution is$y=c_1 \cos \left(\frac{3}{2}x\right)+c_2 \sin \left(\frac{3}{2}x\right)+\frac{5}{3}$