AP Calculus AB Exam Question:
Consider the following differential equation:
dxdy=x1⋅sin(2x)⋅ya) Apply the separation of variables technique to solve the differential equation.
b) Determine the particular solution that satisfies the initial condition y(2π)=3.
Solution:
a) To solve the given differential equation using separation of variables, we first write it in the form:
ydy=x1⋅sin(2x)⋅dxNext, we integrate both sides of the equation with respect to their respective variables:
∫ydy=∫x1⋅sin(2x)⋅dxOn the left side, we obtain:
ln∣y∣=∫x1⋅sin(2x)⋅dxTo integrate the right side, we use integration by parts with u=sin(2x) and dv=x1⋅dx. This gives us:
∫x1⋅sin(2x)⋅dx=−2x1⋅cos(2x)−∫−2x1⋅cos(2x)⋅dxSimplifying further:
∫x1⋅sin(2x)⋅dx=−2x1⋅cos(2x)+21∫x1⋅cos(2x)⋅dxThe integral on the right side can be evaluated using integration by parts as well. Setting u=cos(2x) and dv=x1⋅dx, we have:
∫x1⋅cos(2x)⋅dx=2x1⋅sin(2x)+21∫x1⋅sin(2x)⋅dxWe substitute this result back into our previous equation:
∫x1⋅sin(2x)⋅dx=−2x1⋅cos(2x)+21(2x1⋅sin(2x)+21∫x1⋅sin(2x)⋅dx)We now have an equation involving the integral of x1⋅sin(2x). Rearranging this equation, we obtain:
∫x1⋅sin(2x)⋅dx=−4x1⋅cos(2x)+41(x1⋅sin(2x))Substituting this back into the initial integration equation, we have:
ln∣y∣=−2x1⋅cos(2x)+4x1⋅cos(2x)−41(x1⋅sin(2x))Simplifying further, we obtain:
ln∣y∣=−4x1⋅cos(2x)−41(x1⋅sin(2x))Finally, we exponentiate both sides to eliminate the natural logarithm:
∣y∣=e−4x1⋅cos(2x)−41(x1⋅sin(2x))And since ∣y∣ represents the absolute value of y, we can remove the absolute value sign:
y=±e−4x1⋅cos(2x)−41(x1⋅sin(2x))Therefore, the general solution to the given differential equation is:
y=C⋅e−4x1⋅cos(2x)−41(x1⋅sin(2x))where C is an arbitrary constant.
b) To determine the particular solution that satisfies the initial condition y(2π)=3, we substitute 2π for x and 3 for y in the general solution:
3=C⋅e−4(2π)1⋅cos(2(2π))−41(2π1⋅sin(2(2π)))Simplifying further:
3=C⋅e−21⋅cos(π)−π2⋅sin(π)Since cos(π)=−1 and sin(π)=0, we have:
3=C⋅e−21⋅(−1)−π2⋅0Simplifying further:
3=C⋅e21To solve for C, we divide both sides of the equation by e21:
C=e213Therefore, the particular solution that satisfies the initial condition y(2π)=3 is:
y=e213⋅e−4x1⋅cos(2x)−41(x1⋅sin(2x))