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Q4.3-4E

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Found in: Page 172

### 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

# The auxiliary equation for the given differential equation has complex roots. Find a general solution. ${\mathbf{y}}{\mathbf{\text{'}}}{\mathbf{\text{'}}}{\mathbf{-}}{\mathbf{10}}{\mathbf{y}}{\mathbf{\text{'}}}{\mathbf{+}}{\mathbf{26}}{\mathbf{y}}{\mathbf{=}}{\mathbf{0}}$

The auxiliary equation for the given differential equation $\mathrm{y}\text{'}\text{'}-10\mathrm{y}\text{'}+26\mathrm{y}=0$ has complex roots and its general solution is $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{e}}^{5\mathrm{t}}\left({\mathrm{c}}_{1}\mathrm{cos}\left(\mathrm{t}\right)+{\mathrm{c}}_{2}\mathrm{sin}\left(\mathrm{t}\right)\right)$.

See the step by step solution

## Step 1: Complex conjugate roots.

If the auxiliary equation has complex conjugate roots ${\mathrm{\alpha }}{±}{\mathrm{i\beta }}$, 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: The auxiliary equation.

Assume that $\mathrm{y}={\mathrm{e}}^{\mathrm{rt}}$ is a solution to the given equation.

Since the given equation is of order two, differentiate role="math" localid="1654061985491" $y$ with respect to role="math" localid="1654061981089" $x$ twice:

$\mathrm{y}\text{'}\left(\mathrm{t}\right)=\mathrm{r}{\mathrm{e}}^{\mathrm{rt}}\phantom{\rule{0ex}{0ex}}{\mathrm{y}}^{"}\left(\mathrm{t}\right)={\mathrm{r}}^{2}{\mathrm{e}}^{\mathrm{rt}}\phantom{\rule{0ex}{0ex}}$

Substitute $\mathrm{y}",\mathrm{y}\text{'}$ and $y$ in the given equation to obtain: ${\mathrm{e}}^{\mathrm{rt}}\left({\mathrm{r}}^{2}-10\mathrm{r}+26\right)=0.$

Then the auxiliary equation is ${\mathrm{r}}^{2}-10\mathrm{r}+26=0.$

## Step 3: Finding the roots.

Solve for the roots of the auxiliary equation

${\mathrm{r}}_{1/2}=\frac{10±\sqrt{{10}^{2}-4×1×26}}{2×1}\phantom{\rule{0ex}{0ex}}=\frac{10±\sqrt{100-104}}{}\phantom{\rule{0ex}{0ex}}=\frac{10±2\mathrm{i}}{}\phantom{\rule{0ex}{0ex}}=5±\mathrm{i}$

Since it has a complex conjugate of the form $\mathrm{r}=\mathrm{\alpha }±\mathrm{i\beta }$ for $\alpha =5$ and $\beta =1$
Thus, the general solution is $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{e}}^{5\mathrm{t}}\left({\mathrm{c}}_{1}\mathrm{cos}\left(\mathrm{t}\right)+{\mathrm{c}}_{2}\mathrm{sin}\left(\mathrm{t}\right)\right)$.