Q5 E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

In Problems 3–8, determine whether the given function is a solution to the given differential equation.

$\mathbf{\theta }\mathbf{=}\mathbf{2}{\mathbf{e}}^{\mathbf{3}\mathbf{t}}\mathbf{-}{\mathbf{e}}^{\mathbf{2}\mathbf{t}}$, $\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{\theta }}{{\mathbf{dt}}^{\mathbf{2}}}\mathbf{-}\mathbf{\theta }\frac{\mathbf{d\theta }}{\mathbf{dt}}\mathbf{+}\mathbf{3}\mathbf{\theta }\mathbf{=}\mathbf{-}\mathbf{2}{\mathbf{e}}^{\mathbf{2}\mathbf{t}}$

Step-by-Step Solution

Answer

The given function is not a solution to the given differential equation.

Step 1: Differentiating the given equation w.r.t. (with respect to) t

Firstly, we will differentiate $\mathrm{\theta }=2{\mathrm{e}}^{3\mathrm{t}}- {\mathrm{e}}^{2\mathrm{t}}$ with respect to t,

$\frac{\mathrm{d\theta }}{\mathrm{dt}}=6{\mathrm{e}}^{3\mathrm{t}}-2{\mathrm{e}}^{2\mathrm{t}}$

Again, differentiating with respect to t,

$\frac{{\mathrm{d}}^2\mathrm{\theta }}{{\mathrm{dt}}^2}=18{\mathrm{e}}^{3\mathrm{t}}-4{\mathrm{e}}^{2\mathrm{t}}$

Step 2: Simplification

Putting the values from step 1 in the L.H.S. (Left-hand side) of the given differential equation,

 $$\begin{gathered} \frac{{\mathrm{d}}^2\theta }{{\mathrm{dt}}^2}-\theta \frac{\mathrm{d}\theta }{\mathrm{dt}}+3\theta =18{\mathrm{e}}^{3\mathrm{t}}-4{\mathrm{e}}^{2\mathrm{t}}-\left(2{\mathrm{e}}^{3\mathrm{t}}-{\mathrm{e}}^{2\mathrm{t}}\right)\times \left(6{\mathrm{e}}^{3\mathrm{t}}-2{\mathrm{e}}^{2\mathrm{t}}\right)+3\times \left(2{\mathrm{e}}^{3\mathrm{t}}-{\mathrm{e}}^{2\mathrm{t}}\right) \\ \frac{{\displaystyle {\mathrm{d}}^2\theta }}{{\displaystyle {\mathrm{dt}}^2}}-\theta \frac{{\displaystyle \mathrm{d}\theta }}{{\displaystyle \mathrm{dt}}}+3\theta =18{\mathrm{e}}^{3\mathrm{t}}-4{\mathrm{e}}^{2\mathrm{t}}-\left(12{\mathrm{e}}^{6\mathrm{t}}-4{\mathrm{e}}^{5\mathrm{t}}-6{\mathrm{e}}^{5\mathrm{t}}+2{\mathrm{e}}^{4\mathrm{t}}\right)+6{\mathrm{e}}^{3\mathrm{t}}-3{\mathrm{e}}^{2\mathrm{t}} \\ \frac{{\displaystyle {\mathrm{d}}^2\mathrm{\theta }}}{{\displaystyle {\mathrm{dt}}^2}}-\mathrm{\theta }\frac{{\displaystyle \mathrm{d\theta }}}{{\displaystyle \mathrm{dt}}}+3\mathrm{\theta }=24{\mathrm{e}}^{3\mathrm{t}}-7{\mathrm{e}}^{2\mathrm{t}}-\left(12{\mathrm{e}}^{6\mathrm{t}}-4{\mathrm{e}}^{5\mathrm{t}}-6{\mathrm{e}}^{5\mathrm{t}}+2{\mathrm{e}}^{4\mathrm{t}}\right) \end{gathered}$$

 

which is not the same as the R.H.S. (Right-hand side) of the given differential equation.

Hence, $\mathbf{\theta }\mathbf{=}\mathbf{2}{\mathbf{e}}^{\mathbf{3}\mathbf{t}}\mathbf{-}{\mathbf{e}}^{\mathbf{2}\mathbf{t}}$ is not a solution to the differential equation $\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{\theta }}{{\mathbf{dt}}^{\mathbf{2}}}\mathbf{-}\mathbf{\theta }\frac{\mathbf{d\theta }}{\mathbf{dt}}\mathbf{+}\mathbf{3}\mathbf{\theta }\mathbf{=}\mathbf{-}\mathbf{2}{\mathbf{e}}^{\mathbf{2}\mathbf{t}}$.