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Q37E

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Found in: Page 7

### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974

# Find the function${\mathbit{f}}{\mathbf{\left(}}{\mathbit{t}}{\mathbf{\right)}}$ of the form ${\mathbit{f}}{\mathbf{\left(}}{\mathbit{t}}{\mathbf{\right)}}{\mathbf{=}}{\mathbit{a}}{{\mathbit{e}}}^{\mathbf{3}\mathbf{t}}{\mathbf{+}}{\mathbit{b}}{{\mathbit{e}}}^{\mathbf{2}\mathbf{t}}$ such that ${\mathbit{f}}{\mathbf{\left(}}{\mathbf{0}}{\mathbf{\right)}}{\mathbf{=}}{\mathbf{1}}$and ${\mathbit{f}}{\mathbf{\text{'}}}{\mathbf{\left(}}{\mathbf{0}}{\mathbf{\right)}}{\mathbf{=}}{\mathbf{4}}$.

The function of the form $f\left(t\right)=a{e}^{3t}+b{e}^{2t}$ such that $f\left(0\right)=1$ and $f\text{'}\left(0\right)=4$ is $f\left(t\right)=2{e}^{3t}-{e}^{2t}.$

See the step by step solution

## Step 1: Consider the points and substitute these in the standard equation

Consider the function $f\left(t\right)=a{e}^{3t}+b{e}^{2t}$. Put $f\left(0\right)=1$ in $f\left(t\right)=a{e}^{3t}+b{e}^{2t}$

$f\left(0\right)=a{e}^{3\left(0\right)}+b{e}^{2\left(0\right)}\phantom{\rule{0ex}{0ex}}1=a+b$

## Step 2: Consider the derivative of the polynomial

Take the derivative of the polynomial $f\left(t\right)=a{e}^{3t}+b{e}^{2t}$ and put $f\text{'}\left(0\right)=4$ into its derivative.

${f}^{\text{'}}=\left(t\right)=3a{e}^{3t}+2b{e}^{2t}\phantom{\rule{0ex}{0ex}}{f}^{\text{'}}\left(0\right)=3a{e}^{3×0}+2b{e}^{2×0}\phantom{\rule{0ex}{0ex}}4=3a+2b$

## Step 3: Solve the above equations.

Arrange the equations obtained in above steps into the matrix form.

$\left(\begin{array}{cc}1& 1\\ 3& 2\end{array}\right)\left(\begin{array}{c}a\\ b\end{array}\right)=\left(\begin{array}{c}1\\ 4\end{array}\right)$

Upon solving the values of $a,b$ are obtained as $a=2,b=-1$

Substitute these values in the polynomial equation.

role="math" localid="1659340662448" $f\left(t\right)=a{e}^{3t}+b{e}^{2t}\phantom{\rule{0ex}{0ex}}f\left(t\right)=2{e}^{3t}-1{e}^{2t}$

The function $f\left(t\right)$ of the form $f\left(t\right)=a{e}^{3t}+b{e}^{2t}$ such that $f\left(0\right)=1$ and $f\text{'}\left(0\right)=4$ is $f\left(t\right)=2{e}^{3t}-{e}^{2t}$.