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### 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

# Solve the given initial value problem. ${\mathbf{y}}{\text{'}}{\text{'}}{\mathbf{+}}{\mathbf{9}}{\mathbf{y}}{\mathbf{=}}{\mathbf{0}}{\mathbf{;}}$${\mathbf{y}}\left(\mathbf{0}\right){\mathbf{=}}{\mathbf{1}}{,}$${\mathbf{y}}{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}$

The solution of the given initial value $\mathrm{y}\text{'}\text{'}+9\mathrm{y}=0$ is $\mathrm{y}\left(\mathrm{t}\right)=\mathrm{cos}3\mathrm{t}+\frac{}{}\mathrm{sin}3\mathrm{t}$ when $\mathrm{y}\left(0\right)=1$ and $\mathrm{y}\text{'}\left(0\right)=1$ .

See the step by step solution

## Step 1: Complex conjugate roots.

If the auxiliary equation has complex conjugate roots ${\mathbf{\alpha }}{\mathbf{±}}{\mathbf{i\beta }}$ , then the general solution is given as:

${\mathrm{y}}\left(\mathrm{t}\right){=}{{\mathrm{c}}}_{{1}}{{\mathrm{e}}}^{{\mathrm{\alpha t}}}{\mathrm{cos\beta t}}{+}{{\mathrm{c}}}_{{2}}{{\mathrm{e}}}^{{\mathrm{\alpha t}}}{\mathrm{sin\beta t}}$

## Step 2: Finding the roots of the auxiliary equation.

Given differential equation is $\mathrm{y}\text{'}\text{'}+9\mathrm{y}=0$

Then the auxiliary equation is;

$\begin{array}{c}{\mathrm{r}}^{2}+9=0\\ {\mathrm{r}}^{2}=-9\\ \mathrm{r}=±\sqrt{-9}\\ \mathrm{r}=±3\mathrm{i}\end{array}$

Therefore, the general solution is:

$\begin{array}{l}\mathrm{y}\left(\mathrm{t}\right)={\mathrm{e}}^{0×\mathrm{t}}\left({\mathrm{c}}_{1}\mathrm{cos}\left(3\mathrm{t}\right)+{\mathrm{c}}_{2}\mathrm{sin}\left(3\mathrm{t}\right)\right)\\ ={\mathrm{c}}_{1}\mathrm{cos}\left(3\mathrm{t}\right)+{\mathrm{c}}_{2}\mathrm{sin}\left(3\mathrm{t}\right)\right)\end{array}$

## Step 3: Finding the values of c1 and c2

Given initial conditions are $\mathrm{y}\left(0\right)=1$ and $\mathrm{y}\text{'}\left(0\right)=1$

$\begin{array}{l}\mathrm{y}\left(0\right)=\left({\mathrm{c}}_{1}\mathrm{cos}\left(3×0\right)+{\mathrm{c}}_{2}\mathrm{sin}\left(3×0\right)\right)\\ {\mathrm{c}}_{1}=1\end{array}$

And $\mathrm{y}\text{'}\left(\mathrm{t}\right)=\left(-{\mathrm{c}}_{1}\mathrm{sin}\left(3\mathrm{t}\right)+{\mathrm{c}}_{2}\mathrm{cos}\left(3\mathrm{t}\right)\right)$

Then $\mathrm{y}\text{'}\left(0\right)=\left(-3{\mathrm{c}}_{1}\mathrm{sin}\left(3×0\right)+3{\mathrm{c}}_{2}\mathrm{cos}\left(3×0\right)\right)$

role="math" localid="1654843939522" $\begin{array}{c}3{\mathrm{c}}_{2}=1\\ {\mathrm{c}}_{2}=\frac{1}{3}\end{array}$

Therefore, the solution is role="math" localid="1654843966518" $\mathrm{y}\left(\mathrm{t}\right)=\mathrm{cos}3\mathrm{t}+\frac{1}{3}\mathrm{sin}3\mathrm{t}$