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Q5 E

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Found in: Page 13

Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069

In Problems 3–8, determine whether the given function is a solution to the given differential equation.${\mathbf{\theta }}{\mathbf{=}}{\mathbf{2}}{{\mathbf{e}}}^{\mathbf{3}\mathbf{t}}{\mathbf{-}}{{\mathbf{e}}}^{\mathbf{2}\mathbf{t}}$, $\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{\theta }}{{\mathbf{dt}}^{\mathbf{2}}}{\mathbf{-}}{\mathbf{\theta }}\frac{\mathbf{d\theta }}{\mathbf{dt}}{\mathbf{+}}{\mathbf{3}}{\mathbf{\theta }}{\mathbf{=}}{\mathbf{-}}{\mathbf{2}}{{\mathbf{e}}}^{\mathbf{2}\mathbf{t}}$

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

See the step by step solution

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,

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