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### Course: Calculus, all content (2017 edition)>Unit 4

Lesson 3: Indefinite integrals of common functions

# 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.
• at , Sal didn't add lower and upper bounds for the integral, so I assume the definite and indefinite integral properties are interchangeable here.

but if you add bounds for the integral, and solve the difference of the function's antiderivatives, the constant got cancelled out. in the above case, you still got the constant though.
(1 vote)
• The properties that apply to indefinite integrals are the same properties that apply to definite ones (though the definite ones do have some extra properties). Remember that definite integration is just indefinite integration, with an extra step of plugging in bounds. So, the properties really cannot be different. Plus, we don't really have bounds here, as we want to find a particular solution, which would be a function, not a number.

Kinda why you don't add bounds. We need to find f(-1). To find f(-1), we need an f(x) to plug in x = -1 into. How do we find f(x)? Well, we take the antiderivative of f'(x) and by the FToC, that should give us f(x). And how do we get a function's antiderivative? By taking an indefinite integral.

Also, the constant is necessary here, as on integrating f'(x), we get f(x) + c. Now, we need the value of c, as depending on it, the function changes, And how do we get c? We use the fact that f(2) is 1 and solve for c. Solving for c is essentially what finding a particular solution to a differential equation is.

Hope that made sense?