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AP Calculus AB : Solving separable differential equations and using them in modeling

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Example Questions

Example Question #21 : Solving Separable Differential Equations And Using Them In Modeling

Solve the separable differential equation:

$\frac{\mathrm{d} y}{\mathrm{d} x}=y\cos(x)$ $\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x}=y\cos(x)$

given the condition at

$x=\pi, y=2$ $\displaystyle x=\pi, y=2$

Possible Answers:

$y=e^{\sin(x)}$ $\displaystyle y=e^{\sin(x)}$

y=x $\displaystyle y=x$

$y=Ce^{\sin(x)}$ $\displaystyle y=Ce^{\sin(x)}$

$y=2e^{\sin(x)}$ $\displaystyle y=2e^{\sin(x)}$

Correct answer:

$y=2e^{\sin(x)}$ $\displaystyle y=2e^{\sin(x)}$

Explanation:

To solve the separable differential equation, we must separate x and y to the same sides as their respective derivatives:

$\frac{dy}{y}=\cos(x)dx$ $\displaystyle \frac{dy}{y}=\cos(x)dx$

Next, we integrate both sides:

$\int \frac{dy}{y}=\ln\left | y\right |+C$ $\displaystyle \int \frac{dy}{y}=\ln\left | y\right |+C$ ; $\int \cos(x)dx=\sin(x)+C$ $\displaystyle \int \cos(x)dx=\sin(x)+C$

The integrals were solved using the identical rules.

The two constants of integration are now combined to make a single one:

$\ln \left | y\right |= \sin(x)+C$ $\displaystyle \ln \left | y\right |= \sin(x)+C$

Now, exponentiate both sides of the equation to solve for y, and use the properties of exponents to rearrange C:

$y=e^{\sin(x)+C}=e^{\sin(x)}\cdot e^C=Ce^{\sin(x)}$ $\displaystyle y=e^{\sin(x)+C}=e^{\sin(x)}\cdot e^C=Ce^{\sin(x)}$

Finally, we solve for the integration constant using the given condition:

$2=Ce^{\sin(\pi)}=Ce^0=C$ $\displaystyle 2=Ce^{\sin(\pi)}=Ce^0=C$

Our final answer is

$y=2e^{\sin(x)}$ $\displaystyle y=2e^{\sin(x)}$

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AP Calculus AB : Solving separable differential equations and using them in modeling

Study concepts, example questions & explanations for AP Calculus AB

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Example Question #21 : Solving Separable Differential Equations And Using Them In Modeling

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