# 8.3: Charged Up

**At Grade**Created by: CK-12

*This activity is intended to supplement Calculus, Chapter 7, Lesson 7.*

## Part 1 – Separable Differential Equations Introduced

1. A capacitor, like one used for a camera flash, is charged up. When it discharges rapidly the rate of change of charge, \begin{align*}q\end{align*}, with respect to time, \begin{align*}t\end{align*}, is directly proportional to the charge. Write this as a differential equation.

The first step is to separate the variables, and then integrate and solve for \begin{align*}y\end{align*}.

2. Find \begin{align*}y(0)\end{align*}, if \begin{align*}\frac{dy}{dx} = \sin (x) \cos^2\ (x)\end{align*} and \begin{align*}y \left (\frac{\pi}{2} \right ) = 0\end{align*}. After integrating, use the initial condition \begin{align*}y = 0\end{align*} when \begin{align*}x =\frac{\pi}{2}\end{align*} to find the constant of integration. Then, substitute \begin{align*}x = 0\end{align*} to find \begin{align*}y(0)\end{align*}.

Let’s return to the capacitor. Now that it is discharged, we need to get it charged up again. \begin{align*}A\ 9V\end{align*} battery is connected to a \begin{align*}100k\Omega\end{align*} resister, \begin{align*}R\end{align*}, and \begin{align*}100\mu F\end{align*} capacitor, \begin{align*}C\end{align*}.

The conservation of energy gives us the differential equation \begin{align*}\frac{dq}{dt} \cdot R = V - \frac{q}{C} \rightarrow \frac{dq}{dt} \cdot R \cdot C = V \cdot C - q\end{align*}.

After substituting the given information and simplifying, we get the differential equation \begin{align*}10 \frac{dq}{dt} = 0.9 - q\end{align*}.

3. For the differential equation \begin{align*}10 \frac{dq}{dt} = 0.9 - q\end{align*}, separate the variables and integrate.

4. Apply the initial condition when \begin{align*}\text{time} = 0\end{align*}, the charge \begin{align*}q = 0\end{align*} and solve for \begin{align*}q\end{align*}.

The syntax for **deSolve** is **deSolve**\begin{align*}(y'=f(x,y),x,y)\end{align*} where \begin{align*}x\end{align*} is the independent and \begin{align*}y\end{align*} is the dependent variable. The **deSolve** command can be found in the **HOME** screen by pressing \begin{align*}\Box\end{align*} and selecting **C:deSolve(**.

5. On the **HOME** screen, type **deSolve(**\begin{align*}10q'=0.9-q\end{align*} **and** \begin{align*}q(0)=0,t,q\end{align*}**)**. Write down this answer and reconcile it with your previous solution.

6. In the **HOME** screen enter **deSolve**\begin{align*}(y'=y/x,x,y)\end{align*} to find the general solution of \begin{align*}\frac{dy}{dx} = \frac{y}{x}\end{align*}. Write the answer. Show your work to solve this differential equation by hand and apply the initial condition \begin{align*}y(1) = 1\end{align*} to find the particular solution.

## Part 2 – Homework/Extension – Practice with deSolve and Exploring DEs

Find the general solution for the following separable differential equations. Write the solution in an acceptable format, (for example, use \begin{align*}C\end{align*} instead of **@**** 7**). Show all the steps by hand if your teacher instructs you to do so.

1. \begin{align*}y' = k \cdot y\end{align*}

2. \begin{align*}y' = \frac{x}{y}\end{align*}

3. \begin{align*}y' = \frac{2x}{y^2}\end{align*}

4. \begin{align*}y' = \frac{3x^2}{y}\end{align*}

Open the **GDB** graph labeled *diffq1*. Observe the family of solutions to the differential equation from Question 4, \begin{align*}y' = \frac{3t^2}{y}\end{align*}. Many particular solutions can come from a general solution. When you are finished viewing the family of functions, go to the \begin{align*}Y=\end{align*} screen and delete the function.

5. Not all differential equations are separable. Use **deSolve** to find the solution to the non-separable differential equation \begin{align*}x \cdot y' = 3x^2 + 2 - y\end{align*}. What does this graph look like if the integration constant is 0? Explain. Open picture *diffq2* to view graph.

Find the particular solution for the following equations. Show your work. Solve for \begin{align*}y\end{align*}. Explore other DEs on your own. Do you get any surprising results?

6. \begin{align*}y' = x \cdot y^2\end{align*} and \begin{align*}y(0) = 1\end{align*}

7. \begin{align*}y' = 1+ y^2\end{align*} and \begin{align*}y(0) = 1\end{align*}

8. \begin{align*}y' = 7y\end{align*} and \begin{align*}y(0) = \ln(e)\end{align*}

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