Post

Created by @nathanedwards
 at November 1st 2023, 1:33:10 pm.

AP Calculus AB Exam Question - Separation of Variables

Consider the following differential equation:

dydx=xy \frac{dy}{dx} = xy (a)

(b) Use the solution from part (a) to find the particular solution that satisfies the initial condition y(0)=2 y(0) = 2 .


Solution:

(a) To solve the differential equation using separation of variables, we need to write it in the form dydx=g(x)h(y)\frac{dy}{dx} = g(x)h(y), where g(x)g(x) is a function of xx only and h(y)h(y) is a function of yy only.

Given equation: dydx=xy\frac{dy}{dx} = xy

So, let's rewrite it as:

dyy=xdx\frac{dy}{y} = x \, dx

Now, we can separate the variables:

1ydy=xdx\int \frac{1}{y} \, dy = \int x \, dx

Integrating both sides, we get:

lny=12x2+C1\ln |y| = \frac{1}{2}x^2 + C_1

Here, C1C_1 represents an arbitrary constant.

To get rid of the absolute value, we can rewrite it as:

y=±e12x2+C1y = \pm e^{\frac{1}{2}x^2 + C_1}

Introducing another arbitrary constant, C2=±eC1C_2 = \pm e^{C_1}, we have:

y=C2e12x2y = C_2e^{\frac{1}{2}x^2}

So, the solution to the given differential equation is:

y=C2e12x2(where C2 is an arbitrary constant)y = C_2e^{\frac{1}{2}x^2} \quad \text{(where } C_2 \text{ is an arbitrary constant)}(b)
2=C2e12(0)2    2=C22 = C_2e^{\frac{1}{2}(0)^2} \implies 2 = C_2

Therefore, the particular solution is:

y=2e12x2y = 2e^{\frac{1}{2}x^2}

And we're done!