Question

Find a particular solution to the differential equation.

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{4}\mathbf{y}\mathbf{=}\mathbf{16}\mathbf{tsin}\left(\mathbf{2}\mathbf{t}\right)$

Short answer

The particular solution is${\mathrm{y}}_{\mathrm{p}}=\mathrm{t}[\sin \left(2\mathrm{t}\right)-2\mathrm{tcos}\left(2\mathrm{t}\right)].$

Step-by-step solution

Firstly, write the auxiliary equation of the above differential equation.

The differential equation is$\mathrm{y}\text{'}\text{'}+4\mathrm{y}=16\mathrm{tsin}\left(2\mathrm{t}\right)\dots \left(1\right)$

Write the homogeneous differential equation of the equation (1),

$\mathrm{y}\text{'}\text{'}+4\mathrm{y}=0$

The auxiliary equation for the above equation,

${\mathrm{m}}^2+4=0$

Now find the roots of the auxiliary equation

Solve the auxiliary equation,

$\begin{array}{c}{\mathrm{m}}^2+4=0\\ \mathrm{m}=\pm 2\mathrm{i}\end{array}$

The roots of the auxiliary equation are,

${\mathrm{m}}_1=2\mathrm{i},\&{\mathrm{m}}_2=-2\mathrm{i}$

The complementary solution of the given equation is,

${\mathrm{y}}_{\mathrm{c}}={\mathrm{c}}_1\cos 2\mathrm{t}+{\mathrm{c}}_2\sin 2\mathrm{t}$

Use the method of undetermined coefficients to find a particular solution to the differential equation.

According to the method of undetermined coefficients, assume, the particular solution of equation (1),

${\mathrm{y}}_{\mathrm{p}}=\mathrm{t}[(\mathrm{At}+\mathrm{B})\sin \left(2\mathrm{t}\right)+(\mathrm{Ct}+\mathrm{D})\cos \left(2\mathrm{t}\right)]\dots \left(2\right)$

Now find the derivative of the above equation,

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\text{'}=(2{\mathrm{At}}^2+2\mathrm{Bt}+2\mathrm{Ct}+\mathrm{D})\cos \left(2\mathrm{t}\right)+(2\mathrm{At}-2\mathrm{Dt}-2{\mathrm{Ct}}^2+\mathrm{B})\sin \left(2\mathrm{t}\right)\\ {\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}=(8\mathrm{At}-4{\mathrm{Ct}}^2-4\mathrm{tD}+4\mathrm{B}+2\mathrm{C})\cos \left(2\mathrm{t}\right)+(-4{\mathrm{At}}^2-4\mathrm{D}-8\mathrm{Ct}-4\mathrm{Bt}+2\mathrm{A})\sin \left(2\mathrm{t}\right)\end{array}$

From the equation (1), Substitute the value of${\mathbf{y}}_p\mathbf{\text{'}}\mathbf{\text{'}}$ and${\mathbf{y}}_p$ in the equation (1),

role="math" localid="1654868530352" $\begin{array}{l}\Rightarrow {\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}+4{\mathrm{y}}_{\mathrm{p}}=16\mathrm{tsin}\left(2\mathrm{t}\right)\\ \Rightarrow (8\mathrm{At}-4{\mathrm{Ct}}^2-4\mathrm{tD}+4\mathrm{B}+2\mathrm{C})\cos \left(2\mathrm{t}\right)+(-4{\mathrm{At}}^2-4\mathrm{D}-8\mathrm{Ct}-4\mathrm{Bt}+2\mathrm{A})\sin \left(2\mathrm{t}\right)\\ +4\left\{\mathrm{t}[(\mathrm{At}+\mathrm{B})\sin \left(2\mathrm{t}\right)+(\mathrm{Ct}+\mathrm{D})\cos \left(2\mathrm{t}\right)]\right\}=16\mathrm{tsin}\left(2\mathrm{t}\right)\\ \Rightarrow \mathrm{tsin}\left(2\mathrm{t}\right)(-8\mathrm{C})+\mathrm{tcos}\left(2\mathrm{t}\right)\left(8\mathrm{A}\right)+\cos \left(2\mathrm{t}\right)(4\mathrm{B}+2\mathrm{C})+\sin \left(2\mathrm{t}\right)(2\mathrm{A}-4\mathrm{D})=16\mathrm{tsin}\left(2\mathrm{t}\right)\end{array}$

Final conclusion.

Comparing all coefficients of the above equation;

$\begin{array}{c}8\mathrm{A}=0\Rightarrow \mathrm{A}=0\\ -8\mathrm{C}=16\Rightarrow \mathrm{C}=-2\\ 4\mathrm{B}+2\mathrm{C}=8\dots \left(3\right)\\ 2\mathrm{A}-4\mathrm{D}=0\dots \left(4\right)\end{array}$

Substitute the value of C in the equation (3),

$\begin{array}{c}4\mathrm{B}+2(-2)=8\\ \mathrm{B}=1\end{array}$

Substitute the value of C in the equation (3),

$\begin{array}{c}2\left(0\right)-4\mathrm{D}=0\\ \mathrm{D}=0\end{array}$

Therefore, the particular solution of equation (1),

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}=\mathrm{t}[(\mathrm{At}+\mathrm{B})\sin \left(2\mathrm{t}\right)+(\mathrm{Ct}+\mathrm{D})\cos \left(2\mathrm{t}\right)]\\ {\mathrm{y}}_{\mathrm{p}}=\mathrm{t}[\sin \left(2\mathrm{t}\right)-2\mathrm{tcos}\left(2\mathrm{t}\right)]\end{array}$