Q20RP - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Find a general solution to the given differential equation.

$\mathbf{2}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{y}\mathbf{=}\mathbf{tsint}$

Step-by-Step Solution

Answer

The general solution to the given differential equation is;

$\mathbf{y}\mathbf{=}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\frac{\mathbf{1}}{\sqrt{\mathbf{2}}}\left(\mathbf{t}\right)}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{-}\frac{\mathbf{1}}{\sqrt{\mathbf{2}}}\left(\mathbf{t}\right)}\mathbf{-}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{tsint}\mathbf{-}\frac{\mathbf{4}}{\mathbf{9}}\mathbf{cost}$

Firstly, write the auxiliary equation of the given differential equation

The given differential equation is,

$\mathbf{2}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{y}\mathbf{=}\mathbf{tsint} ......\left(\mathbf{1}\right)$

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

$\mathbf{2}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{y}\mathbf{=}\mathbf{0}$

The auxiliary equation for the above equation $\mathbf{2}{\mathbf{m}}^{\mathbf{2}}\mathbf{-}\mathbf{1}\mathbf{=}\mathbf{0}$.

Now find the roots of the auxiliary equation

Solve the auxiliary equation,

$\begin{array}{rcl}\mathbf{2}{\mathbf{m}}^{\mathbf{2}}\mathbf{-}\mathbf{1}& \mathbf{=}& \mathbf{0}\\ \mathbf{2}{\mathbf{m}}^{\mathbf{2}}& \mathbf{=}& \mathbf{1}\\ {\mathbf{m}}^{\mathbf{2}}& \mathbf{=}& \frac{\mathbf{1}}{\mathbf{2}}\\ \mathbf{m}& \mathbf{=}& \mathbf{\pm }\frac{\mathbf{1}}{\sqrt{\mathbf{2}}}\\ & & \end{array}$

The roots of the auxiliary equation are ${\mathbf{m}}_{\mathbf{1}}\mathbf{=}\frac{\mathbf{1}}{\sqrt{\mathbf{2}}}\mathbf{,} {\mathbf{m}}_{\mathbf{2}}\mathbf{=}\mathbf{-}\frac{\mathbf{1}}{\sqrt{\mathbf{2}}}$.

The complementary solution of the given equation is ${\mathbf{y}}_{\mathbf{c}}\mathbf{=}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\frac{\mathbf{1}}{\sqrt{\mathbf{2}}}\left(\mathbf{t}\right)}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{-}\frac{\mathbf{1}}{\sqrt{\mathbf{2}}}\left(\mathbf{t}\right)}$.

Now, find the particular solution to find a general solution for the equation.

Assume, the particular solution of equation (1),

${\mathbf{y}}_{\mathbf{p}}\left(\mathbf{t}\right)\mathbf{=}\mathbf{Atsint}\mathbf{+}\mathbf{Btcost}\mathbf{+}\mathbf{Ccost}\mathbf{+}\mathbf{Dsint} ......\left(\mathbf{2}\right)$

Now find the derivative of the above equation,

$$\begin{gathered} {\mathbf{y}}_{\mathbf{p}}\mathbf{\text{'}}\left(\mathbf{t}\right)\mathbf{=}\mathbf{Atcost}\mathbf{+}\mathbf{Asint}\mathbf{-}\mathbf{Btsint}\mathbf{+}\mathbf{Bcost}\mathbf{-}\mathbf{Csint}\mathbf{+}\mathbf{Dcost} \\ {\mathbf{y}}_{\mathbf{p}}\mathbf{\text{'}}\left(\mathbf{t}\right)\mathbf{=}\left(\mathbf{At}\mathbf{+}\mathbf{B}\mathbf{+}\mathbf{D}\right)\mathbf{cost}\mathbf{+}\left(\mathbf{A}\mathbf{-}\mathbf{Bt}\mathbf{-}\mathbf{C}\right)\mathbf{sint} \\ {\mathbf{y}}_{\mathbf{p}}\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{t}\right)\mathbf{=}\mathbf{-}\left(\mathbf{At}\mathbf{+}\mathbf{B}\mathbf{+}\mathbf{D}\right)\mathbf{sint}\mathbf{+}\left(\mathbf{A}\right)\mathbf{cost}\mathbf{+}\left(\mathbf{A}\mathbf{-}\mathbf{Bt}\mathbf{-}\mathbf{C}\right)\mathbf{cost}\mathbf{+}\left(\mathbf{-}\mathbf{B}\right)\mathbf{sint} \\ {\mathbf{y}}_{\mathbf{p}}\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{t}\right)\mathbf{=}\mathbf{-}\left(\mathbf{At}\mathbf{+}\mathbf{2}\mathbf{B}\mathbf{+}\mathbf{D}\right)\mathbf{sint}\mathbf{+}\left(\mathbf{2}\mathbf{A}\mathbf{-}\mathbf{Bt}\mathbf{-}\mathbf{C}\right)\mathbf{cost} \end{gathered}$$

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

$\begin{array}{rcl}\mathbf{2}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{y}& \mathbf{=}& \mathbf{tsint}\\ \mathbf{2}\left[\mathbf{-}\left(\mathbf{At}\mathbf{+}\mathbf{2}\mathbf{B}\mathbf{+}\mathbf{D}\right)\mathbf{sint}\mathbf{+}\left(\mathbf{2}\mathbf{A}\mathbf{-}\mathbf{Bt}\mathbf{-}\mathbf{C}\right)\mathbf{cost}\right]\mathbf{-}\left[\mathbf{Atsint}\mathbf{+}\mathbf{Btcost}\mathbf{+}\mathbf{Ccost}\mathbf{+}\mathbf{Dsint}\right]& \mathbf{=}& \mathbf{tsint}\\ \mathbf{-}\mathbf{3}\mathbf{Atsint}\mathbf{-}\left(\mathbf{4}\mathbf{B}\mathbf{+}\mathbf{3}\mathbf{D}\right)\mathbf{sint}\mathbf{+}\left(\mathbf{4}\mathbf{A}\mathbf{-}\mathbf{3}\mathbf{C}\right)\mathbf{cost}\mathbf{-}\mathbf{3}\mathbf{Btcost}& \mathbf{=}& \mathbf{tsint}\end{array}$

Comparing the all coefficients of the above equation,

$\begin{array}{rcl}\mathbf{-}\mathbf{3}\mathbf{A}& \mathbf{=}& \mathbf{1} \Rightarrow \mathbf{A}\mathbf{=}\mathbf{-}\frac{\mathbf{1}}{\mathbf{3}}\\ \mathbf{-}\mathbf{3}\mathbf{Bt}& \mathbf{=}& \mathbf{0} \Rightarrow \mathbf{B}\mathbf{=}\mathbf{0}\\ \mathbf{4}\mathbf{B}\mathbf{+}\mathbf{3}\mathbf{D}& \mathbf{=}& \mathbf{0} ......\left(\mathbf{3}\right)\\ \mathbf{4}\mathbf{A}\mathbf{-}\mathbf{3}\mathbf{C}& \mathbf{=}& \mathbf{0} ......\left(\mathbf{4}\right)\end{array}$

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

$\begin{array}{rcl}\mathbf{4}\mathbf{B}\mathbf{+}\mathbf{3}\mathbf{D}& \mathbf{=}& \mathbf{0}\\ \mathbf{4}\left(\mathbf{0}\right)\mathbf{+}\mathbf{3}\mathbf{D}& \mathbf{=}& \mathbf{0}\\ \mathbf{D}& \mathbf{=}& \mathbf{0}\\ & & \end{array}$

Substitute the value of A in the equation (4),

 $\begin{array}{rcl}\mathbf{4}\mathbf{A}\mathbf{-}\mathbf{3}\mathbf{C}& \mathbf{=}& \mathbf{0}\\ \mathbf{4}\left(\mathbf{-}\frac{\mathbf{1}}{\mathbf{3}}\right)\mathbf{-}\mathbf{3}\mathbf{C}& \mathbf{=}& \mathbf{0}\\ \mathbf{3}\mathbf{C}& \mathbf{=}& \mathbf{4}\left(\mathbf{-}\frac{\mathbf{1}}{\mathbf{3}}\right)\\ \mathbf{C}& \mathbf{=}& \frac{\mathbf{-}\mathbf{4}}{\mathbf{9}}\end{array}$

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

$$\begin{gathered} {\mathbf{y}}_{\mathbf{p}}\left(\mathbf{t}\right)\mathbf{=}\mathbf{Atsint}\mathbf{+}\mathbf{Btcost}\mathbf{+}\mathbf{Ccost}\mathbf{+}\mathbf{Dsint} \\ {\mathbf{y}}_{\mathbf{p}}\left(\mathbf{t}\right)\mathbf{=}\left(\mathbf{-}\frac{\mathbf{1}}{\mathbf{3}}\right)\mathbf{tsint}\mathbf{+}\left(\mathbf{0}\right)\mathbf{tcost}\mathbf{+}\left(\mathbf{-}\frac{\mathbf{4}}{\mathbf{9}}\right)\mathbf{cost}\mathbf{+}\left(\mathbf{0}\right)\mathbf{sint} \\ {\mathbf{y}}_{\mathbf{p}}\left(\mathbf{t}\right)\mathbf{=}\mathbf{-}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{tsint}\mathbf{-}\frac{\mathbf{4}}{\mathbf{9}}\mathbf{cost} \end{gathered}$$

Therefore, the particular solution of equation (1),

${\mathbf{y}}_{\mathbf{p}}\left(\mathbf{t}\right)\mathbf{=}\mathbf{-}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{tsint}\mathbf{-}\frac{\mathbf{4}}{\mathbf{9}}\mathbf{cost}$

Final conclusion, the general solution

Therefore, the general solution is,

$$\begin{gathered} \mathbf{y}\mathbf{=}{\mathbf{y}}_{\mathbf{c}}\left(\mathbf{t}\right)\mathbf{+}{\mathbf{y}}_{\mathbf{p}}\left(\mathbf{t}\right) \\ \mathbf{y}\mathbf{=}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\frac{\mathbf{1}}{\sqrt{\mathbf{2}}}\left(\mathbf{t}\right)}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{-}\frac{\mathbf{1}}{\sqrt{\mathbf{2}}}\left(\mathbf{t}\right)}\mathbf{-}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{tsint}\mathbf{-}\frac{\mathbf{4}}{\mathbf{9}}\mathbf{cost} \end{gathered}$$