Q13RP - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Question: Find a general solution to the given differential equation.$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{16}\mathbf{y}\mathbf{=}{\mathbf{te}}^t$

Step-by-Step Solution

Answer

$\mathbf{y}\mathbf{=}\mathbf{Acos}\left(4t\right)\mathbf{+}\mathbf{Bsin}\left(4t\right)\mathbf{+}\left(\mathbf{-}\frac{2}{289}\mathbf{+}\frac{1}{17}\mathbf{t}\right){\mathbf{e}}^t$

Step 1: Complex conjugate roots.

If the auxiliary equation has complex conjugate roots , then the general solution is given as: $y\left(t\right)=c_1{e}^{\alpha t}cos\beta t+c_2{e}^{\alpha t}sin\beta t$

Step 1: Write the auxiliary equation of the given differential equation

The differential equation is,

 $\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{16}\mathbf{y}\mathbf{=}{\mathbf{te}}^t\mathrm{......}\left(1\right)$

 

 

The auxiliary equation for the above equation ${\mathbf{m}}^2\mathbf{+}\mathbf{16}\mathbf{=}\mathbf{0}.$

Step 2: Find the roots of the auxiliary equation.

Solve the auxiliary equation,

 $\begin{array}{c}{\mathbf{m}}^2\mathbf{+}\mathbf{16}\mathbf{=}\mathbf{0}\\ {\mathbf{m}}^2\mathbf{=}\mathbf{-}\mathbf{16}\\ m=\pm 4i\end{array}$

 

 

The roots of the auxiliary equation are  ${\mathbf{m}}_1\mathbf{=}\mathbf{4}\mathbf{i}\mathbf{,}\&{\mathbf{m}}_2\mathbf{=}\mathbf{-}\mathbf{4}\mathbf{i}.$

 

The complementary solution of the given equation is ${\mathbf{y}}_c\mathbf{=}\mathbf{Acos}\left(4t\right)\mathbf{+}\mathbf{Bsin}\left(4t\right).$

Step 3: Find the particular solution to find a general solution for the equation.

Assume, the particular solution of equation (1),

 ${\mathbf{y}}_p\left(t\right)\mathbf{=}\left({\mathbf{c}}_1\mathbf{+}{\mathbf{c}}_2\mathbf{t}\right){\mathbf{e}}^t\mathrm{......}\left(2\right)$

 

 

Now find the first and second derivatives of the above equation

$\begin{array}{c}{\mathbf{y}}_p\mathbf{\text{'}}\left(t\right)\mathbf{=}\left({\mathbf{c}}_1\mathbf{+}{\mathbf{c}}_2\mathbf{t}\right){\mathbf{e}}^t\mathbf{+}\left({\mathbf{c}}_2\right){\mathbf{e}}^t\\ {\mathbf{y}}_p\mathbf{\text{'}}\mathbf{\text{'}}\left(t\right)\mathbf{=}\left({\mathbf{c}}_1\mathbf{+}{\mathbf{c}}_2\mathbf{t}\right){\mathbf{e}}^t\mathbf{+}\left({\mathbf{c}}_2\right){\mathbf{e}}^t\mathbf{+}\left({\mathbf{c}}_2\right){\mathbf{e}}^t\\ {\mathbf{y}}_p\mathbf{\text{'}}\mathbf{\text{'}}\left(t\right)\mathbf{=}\left({\mathbf{c}}_1\mathbf{+}{\mathbf{c}}_2\mathbf{t}\right){\mathbf{e}}^t\mathbf{+}\mathbf{2}\left({\mathbf{c}}_2\right){\mathbf{e}}^t\end{array}$

Substitute the value of  ${\mathbf{y}}_p\left(t\right)\mathbf{,}{\mathbf{y}}_p\mathbf{\text{'}}\left(t\right)$ and  ${\mathbf{y}}_p\mathbf{\text{'}}\mathbf{\text{'}}\left(t\right)$ in the equation (1),

 $\begin{array}{c}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{16}\mathbf{y}\mathbf{=}{\mathbf{te}}^t\\ \left({\mathbf{c}}_1\mathbf{+}{\mathbf{c}}_2\mathbf{t}\right){\mathbf{e}}^t\mathbf{+}\mathbf{2}\left({\mathbf{c}}_2\right){\mathbf{e}}^t\mathbf{+}\mathbf{16}\left({\mathbf{c}}_1\mathbf{+}{\mathbf{c}}_2\mathbf{t}\right){\mathbf{e}}^t\mathbf{=}{\mathbf{te}}^t\\ \left(\mathbf{17}{\mathbf{c}}_1\mathbf{+}\mathbf{2}{\mathbf{c}}_2\right){\mathbf{e}}^t\mathbf{+}\left(\mathbf{17}{\mathbf{c}}_2\right){\mathbf{te}}^t\mathbf{=}{\mathbf{te}}^t\end{array}$

 

 

Comparing all coefficients of the above equation,

 $\begin{array}{c}\mathbf{17}{\mathbf{c}}_2\mathbf{=}\mathbf{1}\\ {\mathbf{c}}_2\mathbf{=}\frac{1}{17}\\ \mathbf{17}{\mathbf{c}}_1\mathbf{+}\mathbf{2}{\mathbf{c}}_2\mathbf{=}\mathbf{0}\mathrm{......}\left(3\right)\end{array}$

Substitute the value of  ${\mathbf{c}}_2$ in the equation (3),

 

 $\begin{array}{c}\mathbf{17}{\mathbf{c}}_1\mathbf{+}\mathbf{2}{\mathbf{c}}_2\mathbf{=}\mathbf{0}\\ \mathbf{17}{\mathbf{c}}_1\mathbf{+}\mathbf{2}\left(\frac{1}{17}\right)\mathbf{=}\mathbf{0}\\ {\mathbf{c}}_1\mathbf{=}\mathbf{-}\frac{2}{289}\end{array}$

 

Substitute the value of  ${\mathbf{c}}_2$ and ${\mathbf{c}}_1$  in the equation (2),

 

 $\begin{array}{c}{\mathbf{y}}_p\left(t\right)\mathbf{=}\left({\mathbf{c}}_1\mathbf{+}{\mathbf{c}}_2\mathbf{t}\right){\mathbf{e}}^t\\ {\mathbf{y}}_p\left(t\right)\mathbf{=}\left(\mathbf{-}\frac{2}{289}\mathbf{+}\frac{1}{17}\mathbf{t}\right){\mathbf{e}}^t\end{array}$

 

Therefore, the particular solution of equation (1),

 

 $\begin{array}{c}{\mathbf{y}}_p\left(t\right)\mathbf{=}\left({\mathbf{c}}_1\mathbf{+}{\mathbf{c}}_2\mathbf{t}\right){\mathbf{e}}^t\\ {\mathbf{y}}_p\left(t\right)\mathbf{=}\left(\mathbf{-}\frac{2}{289}\mathbf{+}\frac{1}{17}\mathbf{t}\right){\mathbf{e}}^t\end{array}$

Step 4: Final conclusion

Therefore, the general solution is:

 $\begin{array}{c}\mathbf{y}\mathbf{=}{\mathbf{y}}_c\left(t\right)\mathbf{+}{\mathbf{y}}_p\left(t\right)\\ \mathbf{y}\mathbf{=}\mathbf{Acos}\left(4t\right)\mathbf{+}\mathbf{Bsin}\left(4t\right)\mathbf{+}\left(\mathbf{-}\frac{2}{289}\mathbf{+}\frac{1}{17}\mathbf{t}\right){\mathbf{e}}^t\end{array}$