Q18RP - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Find a general solution to the given differential equation.

${\mathbf{y}}^{\mathbf{4}}\mathbf{=}\mathbf{120}\mathbf{t}$

Step-by-Step Solution

Answer

$\mathbf{y}\mathbf{=}{\mathbf{c}}_{\mathbf{1}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}\mathbf{t}\mathbf{+}{\mathbf{c}}_{\mathbf{3}}{\mathbf{t}}^{\mathbf{2}}\mathbf{+}{\mathbf{c}}_{\mathbf{4}}{\mathbf{t}}^{\mathbf{3}}\mathbf{+}{\mathbf{t}}^{\mathbf{5}}$

Write the auxiliary equation of the given differential equation

The given differential equation is,

${\mathbf{y}}^{\mathbf{4}}\mathbf{=}\mathbf{120}\mathbf{t} ......\left(\mathbf{1}\right)$

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

$y^4=0$

The auxiliary equation for the above equation, $m^4=0$.

Now find the roots of the auxiliary equation.

The roots of the auxiliary equation are, 

${\mathbf{m}}_{\mathbf{1}}\mathbf{=}\mathbf{0}\mathbf{,} {\mathbf{m}}_{\mathbf{2}}\mathbf{=}\mathbf{0}\mathbf{,} {\mathbf{m}}_{\mathbf{3}}\mathbf{=}\mathbf{0}\mathbf{,} {\mathbf{m}}_{\mathbf{4}}\mathbf{=}\mathbf{0}$

The complementary solution of the given equation is,

$$\begin{gathered} {\mathbf{y}}_{\mathbf{c}}\mathbf{=}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{0}\left(\mathbf{t}\right)}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{te}}^{\mathbf{0}\left(\mathbf{t}\right)}\mathbf{+}{\mathbf{c}}_{\mathbf{3}}{\mathbf{t}}^{\mathbf{2}}{\mathbf{e}}^{\mathbf{0}\left(\mathbf{t}\right)}\mathbf{+}{\mathbf{c}}_{\mathbf{4}}{\mathbf{t}}^{\mathbf{3}}{\mathbf{e}}^{\mathbf{0}\left(\mathbf{t}\right)} \\ {\mathbf{y}}_{\mathbf{c}}\mathbf{=}{\mathbf{c}}_{\mathbf{1}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}\mathbf{t}\mathbf{+}{\mathbf{c}}_{\mathbf{3}}{\mathbf{t}}^{\mathbf{2}}\mathbf{+}{\mathbf{c}}_{\mathbf{4}}{\mathbf{t}}^{\mathbf{3}} \end{gathered}$$

Now, find the particular solution.

Assume, the particular solution of equation (1),

$$\begin{gathered} {\mathbf{y}}_{\mathbf{p}}\left(\mathbf{t}\right)\mathbf{=}\left(\mathbf{A}\mathbf{+}\mathbf{Bt}\right){\mathbf{t}}^{\mathbf{3}}\mathbf{\times }\mathbf{t} \\ {\mathbf{y}}_{\mathbf{p}}\left(\mathbf{t}\right)\mathbf{=}{\mathbf{At}}^{\mathbf{4}}\mathbf{+}{\mathbf{Bt}}^{\mathbf{5}} ......\left(\mathbf{2}\right) \end{gathered}$$

Now find the derivative of the above equation,

$$\begin{gathered} {\mathbf{y}}_{\mathbf{p}}\mathbf{\text{'}}\left(\mathbf{t}\right)\mathbf{=}\mathbf{4}{\mathbf{At}}^{\mathbf{3}}\mathbf{+}\mathbf{5}{\mathbf{Bt}}^{\mathbf{4}} \\ {\mathbf{y}}_{\mathbf{p}}\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{t}\right)\mathbf{=}\mathbf{12}{\mathbf{At}}^{\mathbf{2}}\mathbf{+}\mathbf{20}{\mathbf{Bt}}^{\mathbf{3}} \\ {\mathbf{y}}_{\mathbf{p}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{t}\right)\mathbf{=}\mathbf{24}\mathbf{At}\mathbf{+}\mathbf{60}{\mathbf{Bt}}^{\mathbf{2}} \\ {{\mathbf{y}}_{\mathbf{p}}}^{\mathbf{4}}\left(\mathbf{t}\right)\mathbf{=}\mathbf{24}\mathbf{A}\mathbf{+}\mathbf{120}\mathbf{Bt} \end{gathered}$$

Substitute the value of  ${{\mathbf{y}}_{\mathbf{p}}}^{\mathbf{4}}\left(\mathbf{t}\right)$ in the equation (1),

$$\begin{gathered} {\mathbf{y}}^{\mathbf{4}}\mathbf{=}\mathbf{120}\mathbf{t} \\ \mathbf{24}\mathbf{A}\mathbf{+}\mathbf{120}\mathbf{Bt}\mathbf{=}\mathbf{120}\mathbf{t} \end{gathered}$$

Comparing all coefficients of the above equation,

$$\begin{gathered} \mathbf{120}\mathbf{B}\mathbf{=}\mathbf{120} \Rightarrow \mathbf{B}\mathbf{=}\mathbf{1} \\ \mathbf{24}\mathbf{A}\mathbf{=}\mathbf{0} \Rightarrow \mathbf{A}\mathbf{=}\mathbf{0} \end{gathered}$$

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

$$\begin{gathered} {\mathbf{y}}_{\mathbf{p}}\left(\mathbf{t}\right)\mathbf{=}{\mathbf{At}}^{\mathbf{4}}\mathbf{+}{\mathbf{Bt}}^{\mathbf{5}} \\ {\mathbf{y}}_{\mathbf{p}}\left(\mathbf{t}\right)\mathbf{=}\left(\mathbf{0}\right){\mathbf{t}}^{\mathbf{4}}\mathbf{+}\left(\mathbf{1}\right){\mathbf{t}}^{\mathbf{5}} \\ {\mathbf{y}}_{\mathbf{p}}\left(\mathbf{t}\right)\mathbf{=}{\mathbf{t}}^{\mathbf{5}} \end{gathered}$$

Therefore, the particular solution of equation (1),

${\mathbf{y}}_{\mathbf{p}}\left(\mathbf{t}\right)\mathbf{=}{\mathbf{t}}^{\mathbf{5}}$

Find the general solution.

Therefore, the general solution is,

 $$\begin{gathered} \mathbf{y}\mathbf{=}{\mathbf{y}}_{\mathbf{c}}\left(\mathbf{t}\right)\mathbf{+}{\mathbf{y}}_{\mathbf{p}}\left(\mathbf{t}\right) \\ \mathbf{y}\mathbf{=}{\mathbf{c}}_{\mathbf{1}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}\mathbf{t}\mathbf{+}{\mathbf{c}}_{\mathbf{3}}{\mathbf{t}}^{\mathbf{2}}\mathbf{+}{\mathbf{c}}_{\mathbf{4}}{\mathbf{t}}^{\mathbf{3}}\mathbf{+}{\mathbf{t}}^{\mathbf{5}} \end{gathered}$$