Post

Created by @nathanedwards
 at October 31st 2023, 7:03:20 pm.

AP Calculus AB Exam Question:

Consider the following differential equation:

dydx=1xsin(2x)y\frac{dy}{dx} = \frac{1}{x} \cdot \sin(2x) \cdot y

a) Apply the separation of variables technique to solve the differential equation.

b) Determine the particular solution that satisfies the initial condition y(π2)=3y(\frac{\pi}{2}) = 3.

Solution:

a) To solve the given differential equation using separation of variables, we first write it in the form:

dyy=1xsin(2x)dx\frac{dy}{y} = \frac{1}{x} \cdot \sin(2x) \cdot dx

Next, we integrate both sides of the equation with respect to their respective variables:

dyy=1xsin(2x)dx\int \frac{dy}{y} = \int \frac{1}{x} \cdot \sin(2x) \cdot dx

On the left side, we obtain:

lny=1xsin(2x)dx\ln |y| = \int \frac{1}{x} \cdot \sin(2x) \cdot dx

To integrate the right side, we use integration by parts with u=sin(2x)u = \sin(2x) and dv=1xdxdv = \frac{1}{x} \cdot dx. This gives us:

1xsin(2x)dx=12xcos(2x)12xcos(2x)dx\int \frac{1}{x} \cdot \sin(2x) \cdot dx = -\frac{1}{2x} \cdot \cos(2x) - \int -\frac{1}{2x} \cdot \cos(2x) \cdot dx

Simplifying further:

1xsin(2x)dx=12xcos(2x)+121xcos(2x)dx\int \frac{1}{x} \cdot \sin(2x) \cdot dx = -\frac{1}{2x} \cdot \cos(2x) + \frac{1}{2} \int \frac{1}{x} \cdot \cos(2x) \cdot dx

The integral on the right side can be evaluated using integration by parts as well. Setting u=cos(2x)u = \cos(2x) and dv=1xdxdv = \frac{1}{x} \cdot dx, we have:

1xcos(2x)dx=12xsin(2x)+121xsin(2x)dx\int \frac{1}{x} \cdot \cos(2x) \cdot dx = \frac{1}{2x} \cdot \sin(2x) + \frac{1}{2} \int \frac{1}{x} \cdot \sin(2x) \cdot dx

We substitute this result back into our previous equation:

1xsin(2x)dx=12xcos(2x)+12(12xsin(2x)+121xsin(2x)dx)\int \frac{1}{x} \cdot \sin(2x) \cdot dx = -\frac{1}{2x} \cdot \cos(2x) + \frac{1}{2} \left(\frac{1}{2x} \cdot \sin(2x) + \frac{1}{2} \int \frac{1}{x} \cdot \sin(2x) \cdot dx\right)

We now have an equation involving the integral of 1xsin(2x)\frac{1}{x} \cdot \sin(2x). Rearranging this equation, we obtain:

1xsin(2x)dx=14xcos(2x)+14(1xsin(2x))\int \frac{1}{x} \cdot \sin(2x) \cdot dx = -\frac{1}{4x} \cdot \cos(2x) + \frac{1}{4} \left(\frac{1}{x} \cdot \sin(2x)\right)

Substituting this back into the initial integration equation, we have:

lny=12xcos(2x)+14xcos(2x)14(1xsin(2x))\ln |y| = -\frac{1}{2x} \cdot \cos(2x) + \frac{1}{4x} \cdot \cos(2x) - \frac{1}{4} \left(\frac{1}{x} \cdot \sin(2x)\right)

Simplifying further, we obtain:

lny=14xcos(2x)14(1xsin(2x))\ln |y| = -\frac{1}{4x} \cdot \cos(2x) - \frac{1}{4} \left(\frac{1}{x} \cdot \sin(2x)\right)

Finally, we exponentiate both sides to eliminate the natural logarithm:

y=e14xcos(2x)14(1xsin(2x))|y| = e^{-\frac{1}{4x} \cdot \cos(2x) - \frac{1}{4} \left(\frac{1}{x} \cdot \sin(2x)\right)}

And since y|y| represents the absolute value of yy, we can remove the absolute value sign:

y=±e14xcos(2x)14(1xsin(2x))y = \pm e^{-\frac{1}{4x} \cdot \cos(2x) - \frac{1}{4} \left(\frac{1}{x} \cdot \sin(2x)\right)}

Therefore, the general solution to the given differential equation is:

y=Ce14xcos(2x)14(1xsin(2x))\boxed{y = C \cdot e^{-\frac{1}{4x} \cdot \cos(2x) - \frac{1}{4} \left(\frac{1}{x} \cdot \sin(2x)\right)}}

where CC is an arbitrary constant.

b) To determine the particular solution that satisfies the initial condition y(π2)=3y(\frac{\pi}{2}) = 3, we substitute π2\frac{\pi}{2} for xx and 33 for yy in the general solution:

3=Ce14(π2)cos(2(π2))14(1π2sin(2(π2)))3 = C \cdot e^{-\frac{1}{4(\frac{\pi}{2})} \cdot \cos(2(\frac{\pi}{2})) - \frac{1}{4} \left(\frac{1}{\frac{\pi}{2}} \cdot \sin(2(\frac{\pi}{2}))\right)}

Simplifying further:

3=Ce12cos(π)2πsin(π)3 = C \cdot e^{-\frac{1}{2} \cdot \cos(\pi) - \frac{2}{\pi} \cdot \sin(\pi)}

Since cos(π)=1\cos(\pi) = -1 and sin(π)=0\sin(\pi) = 0, we have:

3=Ce12(1)2π03 = C \cdot e^{-\frac{1}{2} \cdot (-1) - \frac{2}{\pi} \cdot 0}

Simplifying further:

3=Ce123 = C \cdot e^{\frac{1}{2}}

To solve for CC, we divide both sides of the equation by e12e^{\frac{1}{2}}:

C=3e12C = \frac{3}{e^{\frac{1}{2}}}

Therefore, the particular solution that satisfies the initial condition y(π2)=3y(\frac{\pi}{2}) = 3 is:

y=3e12e14xcos(2x)14(1xsin(2x))\boxed{y = \frac{3}{e^{\frac{1}{2}}} \cdot e^{-\frac{1}{4x} \cdot \cos(2x) - \frac{1}{4} \left(\frac{1}{x} \cdot \sin(2x)\right)}}