Post

Created by @nathanedwards
 at November 1st 2023, 10:06:23 am.

AP Calculus AB Exam Question:

Consider the differential equation:

dydx=xsin(y) \frac{dy}{dx} = x \sin(y) (a)

(b) Given the initial condition y(0)=π4y(0) = \frac{\pi}{4}, find the particular solution to the differential equation.

Answer:

(a) To solve the differential equation dydx=xsin(y)\frac{dy}{dx} = x \sin(y), we can separate the variables and integrate both sides of the equation.

Starting with the original equation:

dysin(y)=xdx \frac{dy}{\sin(y)} = x \, dx

To separate the variables, we can multiply both sides of the equation by sin(y)\sin(y):

dy=xsin(y)dx dy = x \sin(y) \, dx

We can then divide both sides by sin(y)\sin(y):

dysin(y)=xdx \frac{dy}{\sin(y)} = x \, dx

Now, we integrate both sides of the equation:

dysin(y)=xdx \int \frac{dy}{\sin(y)} = \int x \, dx

To integrate dysin(y)\frac{dy}{\sin(y)}, we can use the substitution u=cos(y)u = \cos(y), which gives us du=sin(y)dydu = -\sin(y) \, dy. Therefore, the equation becomes:

1udu=xdx -\int \frac{1}{u} \, du = \int x \, dx

Simplifying the integrals leads to:

lnu=12x2+C1 -\ln|u| = \frac{1}{2}x^2 + C_1

Substituting back u=cos(y)u = \cos(y):

lncos(y)=12x2+C1 -\ln|\cos(y)| = \frac{1}{2}x^2 + C_1

Now, we can solve for yy:

lncos(y)=12x2C1 \ln|\cos(y)| = -\frac{1}{2}x^2 - C_1

Exponentiating both sides:

cos(y)=e12x2C1 \cos(y) = e^{-\frac{1}{2}x^2 - C_1}

Since e12x2C1e^{-\frac{1}{2}x^2 - C_1} is always positive, we can write it as:

cos(y)=C2e12x2 \cos(y) = C_2e^{-\frac{1}{2}x^2}

Finally, we can take the inverse cosine of both sides to solve for yy:

y=cos1(C2e12x2)+2πn y = \cos^{-1}(C_2e^{-\frac{1}{2}x^2}) + 2\pi n

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

y=cos1(C2e12x2)+2πn y = \cos^{-1}(C_2e^{-\frac{1}{2}x^2}) + 2\pi n

where C2C_2 and nn are constants.

(b) Given the initial condition y(0)=π4y(0) = \frac{\pi}{4}, we can substitute x=0x = 0 and y=π4y = \frac{\pi}{4} into the general solution:

π4=cos1(C2e1202)+2πn \frac{\pi}{4} = \cos^{-1}(C_2e^{-\frac{1}{2} \cdot 0^2}) + 2\pi n

Simplifying the equation:

π4=cos1(C2)+2πn \frac{\pi}{4} = \cos^{-1}(C_2) + 2\pi n

Since π4\frac{\pi}{4} is in the range of the inverse cosine function, we can rewrite the equation as:

C2=cos(π4) C_2 = \cos\left(\frac{\pi}{4}\right)

Simplifying:

C2=22 C_2 = \frac{\sqrt{2}}{2}

Substituting this back into the general solution:

y=cos1(22e12x2)+2πn y = \cos^{-1}\left(\frac{\sqrt{2}}{2} e^{-\frac{1}{2}x^2}\right) + 2\pi n

Therefore, the particular solution to the differential equation with the initial condition y(0)=π4y(0) = \frac{\pi}{4} is:

y=cos1(22e12x2)+2πn y = \cos^{-1}\left(\frac{\sqrt{2}}{2} e^{-\frac{1}{2}x^2}\right) + 2\pi n

where nn is an integer.