91Ó°ÊÓ

Given that,

The differential equation is,

$\mathrm{y}\text{'}\text{'}+{\mathrm{λ}}^2\mathrm{y}=\mathrm{sint}\text{ â¶Ä‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â¶Ä‰}......\left(1\right)$

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

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

The auxiliary equation for the above equation,

$\begin{array}{c}{\mathrm{m}}^2+{\mathrm{λ}}^2=0\\ \mathrm{m}=Â\pm \mathrm{λ¾\pm }\end{array}$

The roots of the auxiliary equation are,

${\mathrm{m}}_1=\mathrm{λ¾\pm },\text{ â¶Ä‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰}{\mathrm{m}}_2=-\mathrm{λ¾\pm }$

The complementary solution of the given equation is,

${\mathrm{y}}_{\mathrm{c}}={\mathrm{c}}_1\cos \left(\mathrm{λ³Ù}\right)+{\mathrm{c}}_2\sin \left(\mathrm{λ³Ù}\right)$

Assume, the particular solution of equation (1),

${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)=\mathrm{Asin}\left(\mathrm{t}\right)+\mathrm{Bcos}\left(\mathrm{t}\right)\text{ â¶Ä‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰}......\left(2\right)$

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),

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

Comparing all coefficients of the above equation,

role="math" localid="1655125629499" $\begin{array}{c}(-1+{\mathrm{λ}}^2)\mathrm{A}=1\text{ â¶Ä‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â€‰}⇒\mathrm{A}=\frac{1}{{\mathrm{λ}}^2-1}\\ (-1+{\mathrm{λ}}^2)\mathrm{B}=0\text{ â¶Ä‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰}⇒\mathrm{B}=0\end{array}$

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

role="math" localid="1655125669516" $\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)=\mathrm{Asin}\left(\mathrm{t}\right)+\mathrm{Bcos}\left(\mathrm{t}\right)\\ {\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)=\frac{1}{{\mathrm{λ}}^2-1}\sin \left(\mathrm{t}\right)+\left(0\right)\cos \left(\mathrm{t}\right)\\ {\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)=\frac{1}{{\mathrm{λ}}^2-1}\sin \left(\mathrm{t}\right)\end{array}$

Therefore, the particular solution of equation (1),

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

Therefore, the general solution is,

$\begin{array}{c}\mathrm{y}={\mathrm{y}}_{\mathrm{c}}\left(\mathrm{t}\right)+{\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)\\ \mathrm{y}={\mathrm{c}}_1\cos \left(\mathrm{λ³Ù}\right)+{\mathrm{c}}_2\sin \left(\mathrm{λ³Ù}\right)+\frac{1}{{\mathrm{λ}}^2-1}\sin \left(\mathrm{t}\right)\text{ â¶Ä‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â¶Ä‰â€‰}......\left(3\right)\end{array}$

Given the initial condition

$\mathrm{y}\left(0\right)=0,\text{ â¶Ä‰â¶Ä‰â¶Ä‰}\mathrm{y}\left(\mathrm{Ï€}\right)=1$

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

role="math" localid="1655125829275" $\begin{array}{c}\mathrm{y}={\mathrm{c}}_1\cos \left(\mathrm{λ³Ù}\right)+{\mathrm{c}}_2\sin \left(\mathrm{λ³Ù}\right)+\frac{1}{{\mathrm{λ}}^2-1}\sin \left(\mathrm{t}\right)\\ 0={\mathrm{c}}_1\cos \left(0\right)+{\mathrm{c}}_2\sin \left(0\right)+\frac{1}{{\mathrm{λ}}^2-1}\sin \left(0\right)\\ {\mathrm{c}}_1=0\end{array}$

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

role="math" localid="1655125901952" $\begin{array}{c}1={\mathrm{c}}_1\cos \left(\mathrm{λπ}\right)+{\mathrm{c}}_2\sin \left(\mathrm{λπ}\right)+\frac{1}{{\mathrm{λ}}^2-1}\sin \left(\mathrm{π}\right)\\ {\mathrm{c}}_1\cos \left(\mathrm{λπ}\right)+{\mathrm{c}}_2\sin \left(\mathrm{λπ}\right)=1-\frac{1}{{\mathrm{λ}}^2-1}\sin \left(\mathrm{π}\right)\end{array}$

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

role="math" localid="1655126722162" $\begin{array}{c}\left(0\right)\cos \left(\mathrm{λπ}\right)+{\mathrm{c}}_2\sin \left(\mathrm{λπ}\right)=1-\frac{1}{{\mathrm{λ}}^2-1}\sin \left(\mathrm{π}\right)\\ {\mathrm{c}}_2\sin \left(\mathrm{λπ}\right)=1\\ {\mathrm{c}}_2=\frac{1}{\sin \left(\mathrm{λπ}\right)}\end{array}$

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

role="math" localid="1655126830039" $\begin{array}{c}\mathrm{y}=\left(0\right)\cos \left(\mathrm{λ³Ù}\right)+\left(\frac{1}{\sin \left(\mathrm{λπ}\right)}\right)\sin \left(\mathrm{λ³Ù}\right)+\frac{1}{{\mathrm{λ}}^2-1}\sin \left(\mathrm{t}\right)\\ \mathrm{y}=\frac{\sin \left(\mathrm{λ³Ù}\right)}{\sin \left(\mathrm{λπ}\right)}+\frac{1}{{\mathrm{λ}}^2-1}\sin \left(\mathrm{t}\right)\text{ â¶Ä‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â¶Ä‰â€‰}......\left(4\right)\end{array}$

From equation (4), we have:

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

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

Take integration of the above equation,

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

Given the initial condition,

$\mathrm{y}\left(0\right)=0,\text{ â¶Ä‰â¶Ä‰â¶Ä‰}\mathrm{y}\left(\mathrm{Ï€}\right)=1$

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 $\mathrm{t}=\mathrm{Ï€}$ in the equation (5),

$\begin{array}{c}\mathrm{y}=-\mathrm{sint}+\mathrm{At}+\mathrm{B}\\ 1=-\mathrm{si²ÔÏ€}+\mathrm{´¡Ï€}+\mathrm{B}\\ \mathrm{´¡Ï€}+\mathrm{B}=1\end{array}$

Substitute the value of Bin in the above equation,

role="math" localid="1655127360327" $\begin{array}{c}\mathrm{´¡Ï€}+0=1\\ \mathrm{A}=\frac{1}{\mathrm{Ï€}}\end{array}$

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

role="math" localid="1655127422931" $\begin{array}{c}\mathrm{y}=-\mathrm{sint}+\mathrm{At}+\mathrm{B}\\ \mathrm{y}=-\mathrm{sint}+\frac{1}{\mathrm{Ï€}}\mathrm{t}+0\\ \mathrm{y}=-\mathrm{sint}+\frac{\mathrm{t}}{\mathrm{Ï€}}\end{array}$

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

Therefore, the solution to the equation is:

role="math" localid="1655127493667" $\mathrm{y}=\frac{\sin \left(\mathrm{λ³Ù}\right)}{\sin \left(\mathrm{λπ}\right)}+\frac{1}{{\mathrm{λ}}^2-1}\sin \left(\mathrm{t}\right)$exits if and only if $\mathrm{λ}≠Â\pm 1,\text{ }Â\pm 2,\text{ }Â\pm 3,\text{ }......$