Q7RP - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Decide whether the statement made is True or False. The function  $\mathbf{x}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{t}}^{\mathbf{-}\mathbf{3}}\mathbf{sin} \mathbf{t}\mathbf{+}\mathbf{4}{\mathbf{t}}^{\mathbf{-}\mathbf{3}}$ is a solution to ${\mathbf{t}}^{\mathbf{3}}\frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{=}\mathbf{cos} \mathbf{t}\mathbf{-}\mathbf{3}{\mathbf{t}}^{\mathbf{2}}\mathbf{x}$ .

Step-by-Step Solution

Answer

The statement is true.

Finding the derivative of the equation x ( t ) = t - 3 sin   t + 4 t - 3 with respect to t.

Consider,  $\mathbf{x}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{t}}^{\mathbf{-}\mathbf{3}}\mathbf{sin} \mathbf{t}\mathbf{+}\mathbf{4}{\mathbf{t}}^{\mathbf{-}\mathbf{3}}$

Then,

$$\begin{gathered} \frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{=}\frac{\mathbf{d}}{\mathbf{dt}}\mathbf{(}{\mathbf{t}}^{\mathbf{-}\mathbf{3}}\mathbf{sint}\mathbf{+}\mathbf{4}{\mathbf{t}}^{\mathbf{-}\mathbf{3}}\mathbf{)} \\ \mathbf{=}{\mathbf{t}}^{\mathbf{-}\mathbf{3}}\mathbf{.}\frac{\mathbf{d}}{\mathbf{dt}}\mathbf{\left(}\mathbf{sint}\mathbf{\right)}\mathbf{+}\mathbf{sint}\frac{\mathbf{d}}{\mathbf{dt}}\mathbf{\left(}{\mathbf{t}}^{\mathbf{-}\mathbf{3}}\mathbf{)}\mathbf{+}\frac{\mathbf{d}}{\mathbf{dt}}\mathbf{(}\mathbf{4}{\mathbf{t}}^{\mathbf{-}\mathbf{3}}\mathbf{\right)} \\ \mathbf{=}{\mathbf{t}}^{\mathbf{-}\mathbf{3}}·\mathbf{cos} \mathbf{t}\mathbf{+}\mathbf{sin} \mathbf{t} \mathbf{(}\mathbf{-}\mathbf{3}{\mathbf{t}}^{\mathbf{-}\mathbf{4}}\mathbf{)}\mathbf{-}\mathbf{12}{\mathbf{t}}^{\mathbf{-}\mathbf{4}} \end{gathered}$$

Putting the value of dx dt in the LHS of the equation t 3 dx dt = cos   t - 3 t 2 x to check if RHS is obtained

 Equating Left and Right-hand side,

$$\begin{gathered} \mathbf{LHS}\mathbf{=}{\mathbf{t}}^{\mathbf{3}}\frac{\mathbf{dx}}{\mathbf{dt}} \\ \mathbf{=}{\mathbf{t}}^{\mathbf{3}}\mathbf{(}{\mathbf{t}}^{\mathbf{-}\mathbf{3}}·\mathbf{cos} \mathbf{t}\mathbf{+}\mathbf{sin} \mathbf{t} \mathbf{(}\mathbf{-}\mathbf{3}{\mathbf{t}}^{\mathbf{-}\mathbf{4}}\mathbf{)}\mathbf{-}\mathbf{12}{\mathbf{t}}^{\mathbf{-}\mathbf{4}}\mathbf{)} \\ \mathbf{=}\mathbf{cos} \mathbf{t}\mathbf{-}\mathbf{3}{\mathbf{t}}^{\mathbf{-}\mathbf{1}}·\mathbf{sin} \mathbf{t}\mathbf{-}\mathbf{12}{\mathbf{t}}^{\mathbf{-}\mathbf{1}} \end{gathered}$$

$$\begin{gathered} \mathbf{RHS}\mathbf{=}\mathbf{cos} \mathbf{t}\mathbf{-}\mathbf{3}{\mathbf{t}}^{\mathbf{2}}\mathbf{x} \\ \mathbf{=}\mathbf{cost}\mathbf{-}\mathbf{3}{\mathbf{t}}^{\mathbf{2}}\mathbf{(}{\mathbf{t}}^{\mathbf{-}\mathbf{3}}\mathbf{sin} \mathbf{t}\mathbf{+}\mathbf{4}{\mathbf{t}}^{\mathbf{-}\mathbf{3}}\mathbf{)} \\ \mathbf{=}\mathbf{cost}\mathbf{-}\mathbf{3}{\mathbf{t}}^{\mathbf{-}\mathbf{1}}·\mathbf{sin} \mathbf{t}\mathbf{-}\mathbf{12}{\mathbf{t}}^{\mathbf{-}\mathbf{1}} \end{gathered}$$

Hence, LHS = RHS

Therefore, this shows that  $\mathbf{x}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{t}}^{\mathbf{-}\mathbf{3}}\mathbf{sint}\mathbf{+}\mathbf{4}{\mathbf{t}}^{\mathbf{-}\mathbf{3}}$ is a solution to  ${\mathbf{t}}^{\mathbf{3}}\frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{=}\mathbf{cost}\mathbf{-}\mathbf{3}{\mathbf{t}}^{\mathbf{2}}\mathbf{x}$.