Q46E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Show that the boundary value problem $\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}{\mathbf{\lambda }}^2\mathbf{y}\mathbf{=}\mathbf{sint}\mathbf{;}\mathbf{y}\left(\mathbf{0}\right)\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{y}\left(\mathbf{\pi }\right)\mathbf{=}\mathbf{1}$has a solution if and only if  $\mathbf{\lambda }\ne \mathbf{\pm }\mathbf{1}\mathbf{,}\mathbf{\pm }\mathbf{2}\mathbf{,}\mathbf{\pm }\mathbf{3}\mathbf{,}......$

Step-by-Step Solution

Answer

The solution to the boundary value problem is:

 $\mathrm{y}=\frac{\sin \left(\mathrm{\lambda t}\right)}{\sin \left(\mathrm{\lambda \pi }\right)}+\frac{1}{{\mathrm{\lambda }}^2-1}\sin \left(\mathrm{t}\right)$

Step 1: Use the given information to write the homogeneous differential equation.

Given that, 

 

The differential equation is,

 

 $\mathrm{y}\text{'}\text{'}+{\mathrm{\lambda }}^2\mathrm{y}=\mathrm{sint} ......\left(1\right)$

 

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

 

$\mathrm{y}\text{'}\text{'}+{\mathrm{\lambda }}^2\mathrm{y}=0$ 

 

Step 2: Find the complementary solution of the given equation.

The auxiliary equation for the above equation,

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

 

 

The roots of the auxiliary equation are, 

 

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

 

The complementary solution of the given equation is,

 

 ${\mathrm{y}}_{\mathrm{c}}={\mathrm{c}}_1\cos \left(\mathrm{\lambda t}\right)+{\mathrm{c}}_2\sin \left(\mathrm{\lambda t}\right)$

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

Assume, the particular solution of equation (1),

${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)=\mathrm{Asin}\left(\mathrm{t}\right)+\mathrm{Bcos}\left(\mathrm{t}\right) ......\left(2\right)$

Now find the first and second derivatives of the above equation,

 $\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\text{'}\left(\mathrm{t}\right)=\mathrm{Acos}\left(\mathrm{t}\right)-\mathrm{Bsin}\left(\mathrm{t}\right)\\ {\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}\left(\mathrm{t}\right)=-\mathrm{Asin}\left(\mathrm{t}\right)-\mathrm{Bcos}\left(\mathrm{t}\right)\end{array}$

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

$\begin{array}{c}\mathrm{y}\text{'}\text{'}+{\mathrm{\lambda }}^2\mathrm{y}=\mathrm{sint}\\ -\mathrm{Asin}\left(\mathrm{t}\right)-\mathrm{Bcos}\left(\mathrm{t}\right)+{\mathrm{\lambda }}^2[\mathrm{Asin}\left(\mathrm{t}\right)+\mathrm{Bcos}\left(\mathrm{t}\right)]=\mathrm{sint}\\ (-1+{\mathrm{\lambda }}^2)\mathrm{Asin}\left(\mathrm{t}\right)+(-1+{\mathrm{\lambda }}^2)\mathrm{Bcos}\left(\mathrm{t}\right)=\mathrm{sint}\end{array}$

Comparing all coefficients of the above equation,

$\begin{array}{c}(-1+{\mathrm{\lambda }}^2)\mathrm{A}=1\Rightarrow \mathrm{A}=\frac{1}{{\mathrm{\lambda }}^2-1}\\ (-1+{\mathrm{\lambda }}^2)\mathrm{B}=0\Rightarrow \mathrm{B}=0\end{array}$

Substitute the value of A and B in the equation (2),

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)=\mathrm{Asin}\left(\mathrm{t}\right)+\mathrm{Bcos}\left(\mathrm{t}\right)\\ {\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)=\frac{1}{{\mathrm{\lambda }}^2-1}\sin \left(\mathrm{t}\right)+\left(0\right)\cos \left(\mathrm{t}\right)\\ {\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)=\frac{1}{{\mathrm{\lambda }}^2-1}\sin \left(\mathrm{t}\right)\end{array}$

Therefore, the particular solution of equation (1),

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)=\mathrm{Asin}\left(\mathrm{t}\right)+\mathrm{Bcos}\left(\mathrm{t}\right)\\ {\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)=\frac{1}{{\mathrm{\lambda }}^2-1}\sin \left(\mathrm{t}\right)+\left(0\right)\cos \left(\mathrm{t}\right)\\ {\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)=\frac{1}{{\mathrm{\lambda }}^2-1}\sin \left(\mathrm{t}\right)\end{array}$

 

Step 4: Find the general solution and use the given initial condition.

Therefore, the general solution is,

 $\begin{array}{c}\mathrm{y}={\mathrm{y}}_{\mathrm{c}}\left(\mathrm{t}\right)+{\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)\\ \mathrm{y}={\mathrm{c}}_1\cos \left(\mathrm{\lambda t}\right)+{\mathrm{c}}_2\sin \left(\mathrm{\lambda t}\right)+\frac{1}{{\mathrm{\lambda }}^2-1}\sin \left(\mathrm{t}\right)......\left(3\right)\end{array}$

Given the initial condition

$\mathrm{y}\left(0\right)=0,\mathrm{y}\left(\mathrm{\pi }\right)=1$

Substitute the value of y = 0 and t = 0 in the equation (3),

  $\begin{array}{c}\mathrm{y}={\mathrm{c}}_1\cos \left(\mathrm{\lambda t}\right)+{\mathrm{c}}_2\sin \left(\mathrm{\lambda t}\right)+\frac{1}{{\mathrm{\lambda }}^2-1}\sin \left(\mathrm{t}\right)\\ 0={\mathrm{c}}_1\cos \left(0\right)+{\mathrm{c}}_2\sin \left(0\right)+\frac{1}{{\mathrm{\lambda }}^2-1}\sin \left(0\right)\\ {\mathrm{c}}_1=0\end{array}$

Substitute the value of y= 1 and $\mathrm{t}=\mathrm{\pi }$ in the equation (3),

$\begin{array}{c}1={\mathrm{c}}_1\cos \left(\mathrm{\lambda \pi }\right)+{\mathrm{c}}_2\sin \left(\mathrm{\lambda \pi }\right)+\frac{1}{{\mathrm{\lambda }}^2-1}\sin \left(\mathrm{\pi }\right)\\ {\mathrm{c}}_1\cos \left(\mathrm{\lambda \pi }\right)+{\mathrm{c}}_2\sin \left(\mathrm{\lambda \pi }\right)=1-\frac{1}{{\mathrm{\lambda }}^2-1}\sin \left(\mathrm{\pi }\right)\end{array}$

Substitute the value of  ${\mathrm{c}}_1$  in the above equation,

$\begin{array}{c}\left(0\right)\cos \left(\mathrm{\lambda \pi }\right)+{\mathrm{c}}_2\sin \left(\mathrm{\lambda \pi }\right)=1-\frac{1}{{\mathrm{\lambda }}^2-1}\sin \left(\mathrm{\pi }\right)\\ {\mathrm{c}}_2\sin \left(\mathrm{\lambda \pi }\right)=1\\ {\mathrm{c}}_2=\frac{1}{\sin \left(\mathrm{\lambda \pi }\right)}\end{array}$

Substitute the value of  ${\mathrm{c}}_1$ and  ${\mathrm{c}}_2$ in the equation (3),

$\begin{array}{c}\mathrm{y}=\left(0\right)\cos \left(\mathrm{\lambda t}\right)+\left(\frac{1}{\sin \left(\mathrm{\lambda \pi }\right)}\right)\sin \left(\mathrm{\lambda t}\right)+\frac{1}{{\mathrm{\lambda }}^2-1}\sin \left(\mathrm{t}\right)\\ \mathrm{y}=\frac{\sin \left(\mathrm{\lambda t}\right)}{\sin \left(\mathrm{\lambda \pi }\right)}+\frac{1}{{\mathrm{\lambda }}^2-1}\sin \left(\mathrm{t}\right)......\left(4\right)\end{array}$

Step 5: Check for the value of λ ≠ ± 1 ,   ± 2 ,   ± 3 ,   . .....

From equation (4), we have:

$\begin{array}{c}\sin \left(\mathrm{\lambda \pi }\right)\ne 0,{\mathrm{\lambda }}^2-1\ne 0\\ \mathrm{\lambda \pi }\ne \mathrm{n\pi },\mathrm{\lambda }\ne \pm 1,\mathrm{n}\in \mathrm{\mathbb{Z}}\\ \mathrm{\lambda }\ne \mathrm{n},\mathrm{\lambda }\ne \pm 1,\mathrm{n}\in \mathrm{\mathbb{Z}}\\ \mathrm{\lambda }\ne 0,\pm 1,\pm 2,\pm 3,......\end{array}$

Substitute the value of  $\mathbf{\lambda }\mathbf{=}\mathbf{0}$ in the equation (1),

$\begin{array}{c}\mathrm{y}\text{'}\text{'}+{\mathrm{\lambda }}^2\mathrm{y}=\mathrm{sint}\\ \mathrm{y}\text{'}\text{'}+{\left(0\right)}^2\mathrm{y}=\mathrm{sint}\\ \mathrm{y}\text{'}\text{'}=\mathrm{sint}\end{array}$

Take integration of the above equation,

 $\begin{array}{c}\mathrm{y}\text{'}=-\mathrm{cost}+\mathrm{A}\\ \mathrm{y}=-\mathrm{sint}+\mathrm{At}+\mathrm{B} ......\left(5\right)\end{array}$

 

 

Given the initial condition,

 $\mathrm{y}\left(0\right)=0,\mathrm{y}\left(\mathrm{\pi }\right)=1$

Substitute the value of y = 0 and t = 0 in the equation (5),

 $\begin{array}{c}\mathrm{y}=-\mathrm{sint}+\mathrm{At}+\mathrm{B}\\ 0=-\sin \left(0\right)+\mathrm{A}\left(0\right)+\mathrm{B}\\ \mathrm{B}=0\end{array}$

Substitute the value of y = 1 and $\mathrm{t}=\mathrm{\pi }$  in the equation (5),

$\begin{array}{c}\mathrm{y}=-\mathrm{sint}+\mathrm{At}+\mathrm{B}\\ 1=-\mathrm{sin\pi }+\mathrm{A\pi }+\mathrm{B}\\ \mathrm{A\pi }+\mathrm{B}=1\end{array}$

Substitute the value of Bin in the above equation,

  $\begin{array}{c}\mathrm{A\pi }+0=1\\ \mathrm{A}=\frac{1}{\mathrm{\pi }}\end{array}$

Substitute the value of A and Bin the equation (5),

$\begin{array}{c}\mathrm{y}=-\mathrm{sint}+\mathrm{At}+\mathrm{B}\\ \mathrm{y}=-\mathrm{sint}+\frac{1}{\mathrm{\pi }}\mathrm{t}+0\\ \mathrm{y}=-\mathrm{sint}+\frac{\mathrm{t}}{\mathrm{\pi }}\end{array}$

This solution exists if and only if  $\mathrm{\lambda }=0$.

 

Therefore, the solution to the equation is:

  $\mathrm{y}=\frac{\sin \left(\mathrm{\lambda t}\right)}{\sin \left(\mathrm{\lambda \pi }\right)}+\frac{1}{{\mathrm{\lambda }}^2-1}\sin \left(\mathrm{t}\right)$  exits if and only if  $\mathrm{\lambda }\ne \pm 1,\pm 2,\pm 3,......$