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## 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
- Particular solutions to differential equations: exponential function
- Particular solutions to differential equations
- Worked example: finding a specific solution to a separable equation
- Worked example: separable equation with an implicit solution
- Particular solutions to separable differential equations

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# 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?(7 votes)
- 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(2 votes)

- 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).(4 votes)

- 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.(3 votes)

- at0:31, 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?(2 votes)

- Shouldn't the constant be multiplied by 24, if you multiply the indefinite integral of x^-3 by the scalar?(1 vote)
- Because the negative exponent which is -2 will going to be positive beacause of the numerator that we add then solve(1 vote)

- Isn't this calc 1 material? I remember learning this in calc 1.(0 votes)
- Calc ab covers most of calc 1 they are both calculus anyway so the material will be similar, they are just different curriculums(1 vote)

## Video transcript

- [Voiceover] So we're told
that F of two is equal to 12, F prime of X is equal to
24 over X to the third and what we want to figure out
is what is F of negative one. Alright, so they give us the
derivative in terms of X. So maybe we could take the anti-derivative of the derivative to find
our original function. So let's do that. So we could say that F of X, F of X is going to be equal
to the anti-derivative or we could say the indefinite
integral of F prime of X, which is equal to 24 over X to the third. I could write it over like
this, 24 over X to the third. But to help me process
it a little bit more, I'm gonna write this as
24 X to the negative three 'cause then it'll become how
to take that anti-derivative. D, DX. And so what is the anti-derivative of 24X to the negative three? Well we're just going to do
the power rule in reverse. So what we're going to do is we're going to increase the exponent. So let me just rewrite it. It's going to be 24X to the, we're going to increase
the exponent by one, so it's gonna be X to the
negative three plus one and then we're going to divide
by that increased exponent. So negative three plus one and so that is going to be negative three plus one
is X to the negative two and then we divide by negative two and if you're in doubt
about what we just did where we're kind of doing
the power rule in reverse, now take the derivative of
this using the power rule. Negative two times 24 over negative two is just gonna be 24 and then
you decrement that exponent, you go to negative three. So are we done here, is this F of X? Well F of X might involve a constant. So let's put a constant out here because notice if you were to take the derivative of this thing here, the derivative of 24X to the
negative two over negative two we already established is
24X to the negative three. But then if you take the
derivative of a constant, well that just disappears. So you don't see it when
you look at the derivative. So we have to make sure that
there might be a constant and I have a feeling based
on the information that they've given us, we're going
to make use of that constant. So let me rewrite F of X. So we know that F of X can be expressed as 24 divided by negative two. It's negative 12X to the
negative two plus some constant. So how do we figure out that constant? Well they have told us what F of two is. F of two is equal to 12. So let's write this down. So F of two is equal to 12, which is equal to well we
just have to put two in everywhere we see an X. That's going to be negative two times two to the negative two power plus C and so 12 is equal to, what is this, two to the negative two. Two to the negative two is
equal to one over two squared, which is equal to 1/4. So this is negative 12 times 1/4. Negative 12 times 1/4 is negative three. So it's negative three plus C. Now we can add three to
both sides to solve or C. We get 15 is equal to C. So or C is equal to 15. That is equal to 15. And so now we can write our F of X as we get F of X is equal to negative 12 and I could even write that as negative 12 over X squared if we like. Negative 12 over X squared plus 15 and now using that we can
evaluate F of negative one. F of negative one. Everywhere we see an X we
put a negative one there. So this is going to be
negative one squared. So F of negative one is
equal to 12 divided by, negative 12 divided by
negative one squared. Well negative one squared is just one. So it's gonna be negative 12 plus 15, which is equal to three and we're done. This thing is equal to three.