### Ordinary differential equation examples

#### Example 1

Solve the ordinary differential equation (ODE) $$\diff{x}{t} = 5x -3$$ for $x(t)$.

**Solution**: Using the shortcut method outlined in the
introduction
to ODEs, we multiply through by $dt$ and divide through by $5x-3$:
$$\frac{dx}{5x-3} = dt.$$
We integrate both sides
\begin{align*}
\int \frac{dx}{5x-3} &= \int dt\\
\frac{1}{5} \log |5x-3| &= t + C_1\\
5x-3 &= \pm \exp(5t+5C_1)\\
x &= \pm \frac{1}{5}\exp(5t+5C_1) + 3/5 .
\end{align*}
Letting $C = \frac{1}{5}\exp(5C_1)$, we can write the solution as
$$x(t) = Ce^{5t}+ \frac{3}{5}.$$

We check to see that $x(t)$ satisfies the ODE: \begin{gather*} \diff{x}{t} = 5Ce^{5t}\\ 5x-3 = 5Ce^{5t}+ 3-3 = 5Ce^{5t}. \end{gather*} Both expressions are equal, verifying our solution.

#### Example 2

Solve the ODE combined with initial condition: \begin{align*} \diff{x}{t} &= 5x -3\\ x(2) &= 1. \end{align*}

**Solution**: This is the same ODE as example 1, with solution
$$x(t) = Ce^{5t}+ \frac{3}{5}.$$
We just need to
use the initial condition $x(2)=1$ to determine $C$.

$C$ must satisfy \begin{align*} 1 = Ce^{5\cdot 2}+ \frac{3}{5}, \end{align*} so it must be \begin{align*} C = \frac{2}{5} e^{-10}. \end{align*}

Our solution is $$x(t) = \frac{2}{5}e^{5(t-2)}+ \frac{3}{5}.$$ You can verify that $x(2)=1$.

#### Example 3

Solve the ODE with initial condition: \begin{align*} \diff{y}{x} &= 7y^2x^3\\ y(2) &= 3. \end{align*}

**Solution**: We multiply both sides of the ODE by $dx$, divide
both sides by $y^2$, and integrate:
\begin{align*}
\int y^{-2}dy &= \int 7x^3 dx\\
- y^{-1} &= \frac{7}{4}x^4 +C\\
y & = \frac{-1}{\frac{7}{4}x^4 +C}.
\end{align*}
The general solution is
\begin{align*}
y(x) & = \frac{-1}{\frac{7}{4}x^4 +C}.
\end{align*}

Verify the solution: \begin{align*} \diff{y}{x} &= \diff{}{x}\left(\frac{-1}{\frac{7}{4}x^4 +C}\right)\\ &=\frac{7x^3}{(\frac{7}{4}x^4 +C)^2}. \end{align*} Given our solution for $y$, we know that \begin{align*} y(x)^2 & = \left(\frac{-1}{\frac{7}{4}x^4 +C}\right)^2 = \frac{1}{(\frac{7}{4}x^4 +C)^2}. \end{align*} Therefore, we see that indeed \begin{align*} \diff{y}{x} &= \frac{7x^3}{(\frac{7}{4}x^4 +C)^2} = 7x^3y^2. \end{align*} The solution satisfies the ODE.

To determine the constant $C$, we plug the solution into the equation for the initial conditions $y(2) = 3$: \begin{align*} 3 & = \frac{-1}{\frac{7}{4}2^4 +C}. \end{align*} The constant $C$ is \begin{align*} C = -28\frac{1}{3}= -\frac{85}{3}, \end{align*} and the final solution is \begin{align*} y(x) & = \frac{-1}{\frac{7}{4}x^4 -\frac{85}{3}}. \end{align*}

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##### Math 5447, Fall 2017

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