If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

## AP®︎/College Calculus BC

### Course: AP®︎/College Calculus BC>Unit 7

Lesson 7: Finding particular solutions using initial conditions and separation of variables

# Particular solutions to differential equations: rational function

Sal finds f(-1) given that f'(x)=24/x³ and f(2)=12.

## Want to join the conversation?

• Couldn't we technically approach this by changing the question to 24ln(x^3)? Or does that not work when there are exponents in the variable?
• it doesn't work since the 1/x doesn't follow the other exponential laws of calculus, so if it was 1/x^3, then you would change it to x^-3 and use the exponent laws
• Could this problem have been solved using the method mentioned in the video called "Worked example: finding a specific solution to a separable equation" (link: https://www.khanacademy.org/math/ap-calculus-bc/bc-differential-equations-new/bc-7-7/v/particular-solution-to-differential-equation-example)?
(1 vote)
• Yes, because 𝑓 '(𝑥) = 24∕𝑥³ is a separable equation.

This becomes apparent if we instead write
𝑑𝑦∕𝑑𝑥 = 24∕𝑥³

Multiplying both sides by 𝑑𝑥, we get
𝑑𝑦 = (24∕𝑥³)𝑑𝑥

Then we integrate both sides, which is the same thing as finding the antiderivative of 𝑓 '(𝑥).
• where are the videos to finding derivatives? I want to know how to find derivatives as well as antiderivatives.
(1 vote)
• Well you can find the videos in which they tell how to find derivatives in the Differential Calculus section. https://www.khanacademy.org/math/differential-calculus Here is the link to it. First there is limit and then you can find derivative.