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