Find the function f(x)f\left(x\right)f(x) that satisfies the following differential equation.f′′(x)=3x2f^{\prime\prime}\left(x\right)=3x^2f′′(x)=3x2; f′(0)=1f^{\prime}\left(0\right)=1f′(0)=1; f(1)=3f\left(1\right)=3f(1)=3

f ( x ) = x 4 4 + x + 7 4 f\left(x\right)=\frac{x^4}{4}+x+\frac74 f ( x ) = 4 x 4 + x + 4 7

Find the function f ( x ) f\left(x\right) f ( x ) that satisfies the following differential equation. f ′ ′ ( x ) = 3 x 2 f^{\prime\prime}\left(x\right)=3x^2 f ′′ ( x ) = 3 x 2 ; f ′ ( 0 ) = 1 f^{\prime}\left(0\right)=1 f ′ ( 0 ) = 1 ; f ( 1 ) = 3 f\left(1\right)=3 f ( 1 ) = 3

this problem, we're asked to find the function f of x that satisfies the following differential equation. Now this is the differential equation that we're given, f double prime of x equals 3 x squared. Now you might notice something interesting about this differential equation. Notice we're dealing with 2 derivatives here. And because of that, we have 2 initial conditions given to us. So what we're going to need to do is work our way down so that we were only dealing with f of x. So we're going to start by dealing with these two situations, this this second derivative and then this first derivative initial condition, and then we're gonna work down to this one and eventually just end up with our function f of x. So let's get started. Now first off, we'll start with this f double prime of x here. And what I need to do is recognize that integrating this function is, in essence, taking the reverse process of the derivative. Since we're, in essence, just reversing a derivative, then that means that it would turn our f double prime of x here into f prime of x. And then what that means is I need to integrate this right side of the equation here. So I'm trying to find the integral of 3x squared with respect to x. Now I could just use the power rule here for integrals to solve this, but what I recognize is 3x squared is actually a function we've seen before. This looks very similar to the derivative of x cubed. In fact, that's exactly what it is. So I can say that f prime of x is simply equal to x cubed. And, again, you could use the power rule to verify this, but going back here, if we did the derivative power rule, we would get 3x squared. So this is going to be the correct answer for this integral right here. But don't forget you also have to add the plus c constant at the end of any integral that you do that's indefinite. So an indefinite integral gives you plus c. Now from here, what I'm going to do is apply this first initial condition that we have, f prime of 0 is equal to 1. So we have the f prime of 0 equals 1, meaning I can take every place that I see x in my f prime equation and replace it with 0. So if the f prime of 0 is equal to and in this case, we're going to have 0 cubed plus the constant c. Now this 0 is just gonna go away completely, meaning we'll have that c is equal to f prime of 0, which we said f prime of 0 is 1. So that means going back up to the c, we can replace it with 1. So this is going to be our function f prime of x. Now in problems we've seen in the past, we typically would stop after we do this process once. But notice that this reduced us down to just f prime, and we're trying to get to f of x. So what that means is I'm going to need to do this a second time. So let's go over to this right side of the screen here. Let's do this again. So on this right side of the screen here, we have that f prime of x is equal to x cubed plus 1. Now what we've already done is we've used this equation right here and this initial condition. So what I'm going to do is move over to this initial condition. So we can see here that we have that f of 1 is equal to 3. Now I can't apply it just yet because we have f prime, and our initial condition says f of 1. But what I can do is integrate this function. Because if I integrate f prime of x, that's going to give me the function f of x. That's the function we're looking for. I noticed this is also f of 1. So we have f here, f there, so we're good. But I need to figure out what f of x is by integrating the right side of this function. To do that, well, I'm first going to go ahead and apply the rules for integrals that I know. I can integrate each one of these terms separately. So I'll have the integral of x cubed with respect to x plus the integral of 1 with respect to x. Now doing this, I am gonna use the power rule here. I'm gonna add 1 to the exponent and then divide by that new exponent. So we're going to have x to the 4th power over 4. You're going to have plus the integral of 1 with respect to x, which is just gonna give us x because, again, the derivative of x is 1. So integrating 1, we get x, then we'll have plus the constant c. This right here is the general solution for f of x. What I need to do is apply my initial condition, f of 1. So doing that, I can take every place that I see x in the function f of x and replace it with 1. So we're going to have 1 to the 4th power over 4. So we're gonna have 1 to the 4th over 4 plus 1 plus c. Now simplifying this, 1 to the 4th power is just gonna be 1. So we'll have 1 fourth, and it's gonna be plus 1, then plus c. And 1 fourth plus 1, well, that's saving as 1 fourth plus 4 over 4. So I can rewrite 1 as just 4 divided by 4. And what that means so that's the fraction that we have right here. Then we can combine these fractions. Now also keep in mind that f of 1 is 3, so we're gonna have that 3 or f of 1 that's going to be equal to and then we have 1 fourth plus 4 over 4, which is gonna give us 5 over 4, and then we're gonna have plus c. Then what I need to do is subtract this 5 over 4 from both sides of the equation. So subtract the 5 over 4 from here, subtract the 5 over 4 from there. That's going to get things to cancel on the right side here, leaving me with just the constant c. So we'll have c is equal to 3 minus 5 4ths. Now 3 is the same thing as 12 over 4. Dividing 4 into 12 will give you 3. And we're subtracting off 5 over 4, and I did this to get a common denominator just like I did over here. So doing this, we're gonna have 12 minus 5, which is going to give us 7. So we're gonna have 7 over 4 as this fraction. So c is equal to 7 over 4, meaning I can go back up to this general solution of being calculated, and I can replace that c with 7 over 4. So this right here would be the solution for our function, and that is how you can solve this problem and deal with this initial value problem where we have multiple derivatives taken to start with. So So that's how you can deal with these situations with with differential equations, where you have these initial value problems, where you have multiple initial values and multiple derivatives. Hope you found this video helpful, and let's move on and get some more practice.

Find the function f ( x ) f\left(x\right) f ( x ) that satisfies the following differential equation. f ′ ′ ( x ) = 3 x 2 f^{\prime\prime}\left(x\right)=3x^2 f ′′ ( x ) = 3 x 2 ; f ′ ( 0 ) = 1 f^{\prime}\left(0\right)=1 f ′ ( 0 ) = 1 ; f ( 1 ) = 3 f\left(1\right)=3 f ( 1 ) = 3

f ( x ) = x 4 4 + x + 7 4 f\left(x\right)=\frac{x^4}{4}+x+\frac74 f ( x ) = 4 x 4 + x + 4 7

f ( x ) = 3 x 4 + x − 1 f\left(x\right)=3x^4+x-1 f ( x ) = 3 x 4 + x − 1

f ( x ) = 3 x 4 + x + 7 4 f\left(x\right)=3x^4+x+\frac74 f ( x ) = 3 x 4 + x + 4 7

f ( x ) = x 4 4 + x − 1 f\left(x\right)=\frac{x^4}{4}+x-1 f ( x ) = 4 x 4 + x − 1

7. Antiderivatives & Indefinite Integrals

Initial Value Problems

Calculus

7. Antiderivatives & Indefinite Integrals

Initial Value Problems

Struggling with Calculus?

Join thousands of students who trust us to help them ace their exams!Watch the first video

Multiple Choice

Find the function $f\left(x\right)$ that satisfies the following differential equation.
$f^{\prime\prime}\left(x\right)=3x^2$ ; $f^{\prime}\left(0\right)=1$ ; $f\left(1\right)=3$

$f\left(x\right)=3x^4+x-1$

$f\left(x\right)=\frac{x^4}{4}+x+\frac74$

$f\left(x\right)=3x^4+x+\frac74$

$f\left(x\right)=\frac{x^4}{4}+x-1$

Verified step by step guidance

Start by integrating the given second derivative f''(x) = 3x^2 to find the first derivative f'(x). The integration of 3x^2 with respect to x is ∫3x^2 dx = x^3 + C, where C is the constant of integration.

Use the initial condition f'(0) = 1 to solve for the constant C. Substitute x = 0 into the equation f'(x) = x^3 + C, which gives 1 = 0^3 + C, so C = 1. Therefore, f'(x) = x^3 + 1.

Integrate f'(x) = x^3 + 1 to find the function f(x). The integration of x^3 + 1 with respect to x is ∫(x^3 + 1) dx = (1/4)x^4 + x + D, where D is another constant of integration.

Use the initial condition f(1) = 3 to solve for the constant D. Substitute x = 1 into the equation f(x) = (1/4)x^4 + x + D, which gives 3 = (1/4)(1)^4 + 1 + D. Simplify to find D = 3 - 1/4 - 1 = 7/4.

Substitute the value of D back into the equation for f(x) to get the final function: f(x) = (1/4)x^4 + x + 7/4.

5:03m

Watch next

Master Initial Value Problems with a bite sized video explanation from Patrick

Start learning

Initial Value Problems practice set

Problem sets built by lead tutors

Expert video explanations

Go to Practice