AP Calculus AB Exam Question:
Consider the differential equation:
dxdy=xsin(y)(a)(b) Given the initial condition y(0)=4π, find the particular solution to the differential equation.
Answer:
(a) To solve the differential equation dxdy=xsin(y), we can separate the variables and integrate both sides of the equation.
Starting with the original equation:
sin(y)dy=xdxTo separate the variables, we can multiply both sides of the equation by sin(y):
dy=xsin(y)dxWe can then divide both sides by sin(y):
sin(y)dy=xdxNow, we integrate both sides of the equation:
∫sin(y)dy=∫xdxTo integrate sin(y)dy, we can use the substitution u=cos(y), which gives us du=−sin(y)dy. Therefore, the equation becomes:
−∫u1du=∫xdxSimplifying the integrals leads to:
−ln∣u∣=21x2+C1Substituting back u=cos(y):
−ln∣cos(y)∣=21x2+C1Now, we can solve for y:
ln∣cos(y)∣=−21x2−C1Exponentiating both sides:
cos(y)=e−21x2−C1Since e−21x2−C1 is always positive, we can write it as:
cos(y)=C2e−21x2Finally, we can take the inverse cosine of both sides to solve for y:
y=cos−1(C2e−21x2)+2πnTherefore, the general solution to the given differential equation is:
y=cos−1(C2e−21x2)+2πnwhere C2 and n are constants.
(b) Given the initial condition y(0)=4π, we can substitute x=0 and y=4π into the general solution:
4π=cos−1(C2e−21⋅02)+2πnSimplifying the equation:
4π=cos−1(C2)+2πnSince 4π is in the range of the inverse cosine function, we can rewrite the equation as:
C2=cos(4π)Simplifying:
C2=22Substituting this back into the general solution:
y=cos−1(22e−21x2)+2πnTherefore, the particular solution to the differential equation with the initial condition y(0)=4π is:
y=cos−1(22e−21x2)+2πnwhere n is an integer.