91Ó°ÊÓ

The differential equation is,

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{3}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{7}\mathbf{y}\mathbf{=}\mathbf{7}{\mathbf{t}}^{\mathbf{2}}\mathbf{-}{\mathbf{e}}^{\mathbf{t}}                              ......\left(\mathbf{1}\right)$

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

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{3}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{7}\mathbf{y}\mathbf{=}\mathbf{0}$

The auxiliary equation for the above equation,${\mathbf{m}}^{\mathbf{2}}\mathbf{-}\mathbf{3}\mathbf{m}\mathbf{+}\mathbf{7}\mathbf{m}\mathbf{=}\mathbf{0}$.

Solve the auxiliary equation,

$\begin{array}{rcl}{\mathbf{m}}^{\mathbf{2}}\mathbf{-}\mathbf{3}\mathbf{m}\mathbf{+}\mathbf{7}\mathbf{m}& \mathbf{=}& \mathbf{0}\\ \mathbf{m}& \mathbf{=}& \frac{\mathbf{3}\mathbf{Â\pm }\sqrt{\mathbf{9}\mathbf{-}\mathbf{28}}}{\mathbf{2}}\\ \mathbf{m}& \mathbf{=}& \frac{\mathbf{3}\mathbf{Â\pm }\sqrt{\mathbf{-}\mathbf{19}}}{\mathbf{2}}\\ \mathbf{m}& \mathbf{=}& \frac{\mathbf{3}\mathbf{Â\pm }\mathbf{i}\sqrt{\mathbf{19}}}{\mathbf{2}}\end{array}$

The roots of the auxiliary equation are ${\mathbf{m}}_{\mathbf{1}}\mathbf{=}\frac{\mathbf{3}\mathbf{+}\mathbf{i}\sqrt{\mathbf{19}}}{\mathbf{2}}\mathbf{,}  \&  {\mathbf{m}}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{3}\mathbf{-}\mathbf{i}\sqrt{\mathbf{19}}}{\mathbf{2}}$

The complementary solution of the given equation is,

${\mathbf{y}}_{\mathbf{c}}\mathbf{=}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\frac{\mathbf{3}}{\mathbf{2}}\left(\mathbf{t}\right)}\mathbf{cos}\left(\frac{\sqrt{\mathbf{19}}}{\mathbf{2}}\mathbf{t}\right)\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\frac{\mathbf{3}}{\mathbf{2}}\left(\mathbf{t}\right)}\mathbf{sin}\left(\frac{\sqrt{\mathbf{19}}}{\mathbf{2}}\mathbf{t}\right)$

Assume, the particular solution of equation (1),

${\mathbf{y}}_{\mathbf{p}}\left(\mathbf{t}\right)\mathbf{=}\mathbf{A}\mathbf{+}\mathbf{Bt}\mathbf{+}{\mathbf{Ct}}^{\mathbf{2}}\mathbf{-}{\mathbf{De}}^{\mathbf{t}}                          ......\left(\mathbf{2}\right)$

Now find the derivative of the above equation,

$$\begin{gathered} {\mathbf{y}}_{\mathbf{p}}\mathbf{\text{'}}\left(\mathbf{t}\right)\mathbf{=}\mathbf{B}\mathbf{+}\mathbf{2}\mathbf{Ct}\mathbf{-}{\mathbf{De}}^{\mathbf{t}} \\ {\mathbf{y}}_{\mathbf{p}}\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{t}\right)\mathbf{=}\mathbf{2}\mathbf{C}\mathbf{-}{\mathbf{De}}^{\mathbf{t}} \end{gathered}$$

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),

$\begin{array}{rcl}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{3}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{7}\mathbf{y}& \mathbf{=}& \mathbf{7}{\mathbf{t}}^{\mathbf{2}}\mathbf{-}{\mathbf{e}}^{\mathbf{t}}\\ \mathbf{2}\mathbf{C}\mathbf{-}{\mathbf{De}}^{\mathbf{t}}\mathbf{-}\mathbf{3}\left[\mathbf{B}\mathbf{+}\mathbf{2}\mathbf{Ct}\mathbf{-}{\mathbf{De}}^{\mathbf{t}}\right]\mathbf{+}\mathbf{7}\left[\mathbf{A}\mathbf{+}\mathbf{Bt}\mathbf{+}{\mathbf{Ct}}^{\mathbf{2}}\mathbf{-}{\mathbf{De}}^{\mathbf{t}}\right]& \mathbf{=}& \mathbf{7}{\mathbf{t}}^{\mathbf{2}}\mathbf{-}{\mathbf{e}}^{\mathbf{t}}\\ \mathbf{7}{\mathbf{Ct}}^{\mathbf{2}}\mathbf{+}\left(\mathbf{7}\mathbf{B}\mathbf{-}\mathbf{6}\mathbf{C}\right)\mathbf{t}\mathbf{+}\left(\mathbf{7}\mathbf{A}\mathbf{-}\mathbf{3}\mathbf{B}\mathbf{+}\mathbf{2}\mathbf{C}\right)\mathbf{-}\mathbf{5}{\mathbf{De}}^{\mathbf{t}}& \mathbf{=}& \mathbf{7}{\mathbf{t}}^{\mathbf{2}}\mathbf{-}{\mathbf{e}}^{\mathbf{t}}\end{array}$

Comparing all coefficients of the above equation,

$\begin{array}{rcl}\mathbf{7}\mathbf{C}& \mathbf{=}& \mathbf{7}  ⇒\mathbf{C}\mathbf{=}\mathbf{1}\\ \mathbf{-}\mathbf{5}\mathbf{D}& \mathbf{=}& \mathbf{-}\mathbf{1}  ⇒\mathbf{D}\mathbf{=}\frac{\mathbf{1}}{\mathbf{5}}\\ \mathbf{7}\mathbf{B}\mathbf{-}\mathbf{6}\mathbf{C}& \mathbf{=}& \mathbf{0}                                     ......\left(\mathbf{3}\right)\\ \mathbf{7}\mathbf{A}\mathbf{-}\mathbf{3}\mathbf{B}\mathbf{+}\mathbf{2}\mathbf{C}& \mathbf{=}& \mathbf{0}                         ......\left(\mathbf{4}\right)\end{array}$

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

$\begin{array}{rcl}\mathbf{7}\mathbf{B}\mathbf{-}\mathbf{6}\left(\mathbf{1}\right)& \mathbf{=}& \mathbf{0}\\ \mathbf{7}\mathbf{B}& \mathbf{=}& \mathbf{6}\\ \mathbf{B}& \mathbf{=}& \frac{\mathbf{6}}{\mathbf{7}}\end{array}$

Substitute the value of B and C in the equation (4),

$\begin{array}{rcl}\mathbf{7}\mathbf{A}\mathbf{-}\mathbf{3}\left(\frac{\mathbf{6}}{\mathbf{7}}\right)\mathbf{+}\mathbf{2}\left(\mathbf{1}\right)& \mathbf{=}& \mathbf{0}\\ \mathbf{7}\mathbf{A}& \mathbf{=}& \frac{\mathbf{18}}{\mathbf{7}}\mathbf{-}\mathbf{2}\\ \mathbf{A}& \mathbf{=}& \frac{\mathbf{4}}{\mathbf{49}}\end{array}$

Therefore, the particular solution of equation (1),

$$\begin{gathered} {\mathbf{y}}_{\mathbf{p}}\left(\mathbf{t}\right)\mathbf{=}\mathbf{A}\mathbf{+}\mathbf{Bt}\mathbf{+}{\mathbf{Ct}}^{\mathbf{2}}\mathbf{-}{\mathbf{De}}^{\mathbf{t}} \\ {\mathbf{y}}_{\mathbf{p}}\left(\mathbf{t}\right)\mathbf{=}\frac{\mathbf{4}}{\mathbf{49}}\mathbf{+}\frac{\mathbf{6}}{\mathbf{7}}\mathbf{t}\mathbf{+}{\mathbf{t}}^{\mathbf{2}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{5}}{\mathbf{e}}^{\mathbf{t}} \end{gathered}$$

Therefore, the general solution is,

$$\begin{gathered} \mathbf{y}\mathbf{=}{\mathbf{y}}_{\mathbf{c}}\left(\mathbf{t}\right)\mathbf{+}{\mathbf{y}}_{\mathbf{p}}\left(\mathbf{t}\right) \\ \mathbf{y}\mathbf{=}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\frac{\mathbf{3}}{\mathbf{2}}\left(\mathbf{t}\right)}\mathbf{cos}\left(\frac{\sqrt{\mathbf{19}}}{\mathbf{2}}\mathbf{t}\right)\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\frac{\mathbf{3}}{\mathbf{2}}\left(\mathbf{t}\right)}\mathbf{sin}\left(\frac{\sqrt{\mathbf{19}}}{\mathbf{2}}\mathbf{t}\right)\mathbf{+}\frac{\mathbf{4}}{\mathbf{49}}\mathbf{+}\frac{\mathbf{6}}{\mathbf{7}}\mathbf{t}\mathbf{+}{\mathbf{t}}^{\mathbf{2}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{5}}{\mathbf{e}}^{\mathbf{t}} \end{gathered}$$