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# 8.3: Charged Up

Difficulty Level: 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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Feb 23, 2012
Nov 04, 2014
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