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Are the statements in Problems 46-54 true or false?

  1. If $F(x)$ is an antiderivative of $f(x)$, then $y=F(x)$ is a solution to the differential equation $\frac{dy}{dx}=f(x)$.

  2. If $y=F(x)$ is a solution to the differential equation $\frac{dy}{dx}=f(x)$, then $F(x)$ is an antiderivative of $f(x)$.

  3. There is only one solution $y(t)$ to the initial value problem $\frac{dy}{dx} = 3t^2, y(1) = \pi$

My answer:

The first two questions seems to be both true, I am 100% that the 2nd is true but the first makes me not sure.

The third question I would say yes, why? because I solved for C

$$ y=t^3+C $$

$$ \pi=1^3+C $$

$$ C=\pi-1 $$ Therefore, I would say it has only one solution. Am I right ?

Alice Ryhl
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user157908
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2 Answers2

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For questions 1 and 2, I would think your answers are correct. In order to be doubly sure, check your textbook for the definition of an anti-derivative.

For your last question, your answer is right; there is indeed a unique solution to the initial value problem.

However, strictly speaking, the reasoning you've provided is insufficient. You've shown that there is an answer, but not that the answer you've provided is the only answer. You should have some statement in the textbook that initial value problems such as these have a unique solution.

Sometimes this result is bunched in with the Picard-Lindelöf theorem. In order to be strictly correct, check your textbook, and cite the relevant result.

Ben Grossmann
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Your answers are all correct.

1) If F(x) is an antiderivative of f(x), then y=F(x) is a solution to the differential equation dy/dx=f(x).

True because:

$$ \frac{dy}{dx} = f(x) \rightarrow dy = f(x)dx \rightarrow \int dy = \int f(x)dx \Rightarrow y = \int f(x)dx$$

$$\text{Since they've told us }F(x) = \int f(x) \text{, we can substitute } F(x) \text{ for } \int f(x)dx \text{,}$$

$$ \text{ which leaves: } y = F(x) $$

2) If y=F(x) is a solution to the differential equation dy/dx=f(x), then F(x) is an antiderivative of f(x).

True because we can also work backwards from above. When we solve for $y$ in the differential equation $\frac{dy}{dx} = f(x)$ we get $y = \int f(x)$, since we already know that $ y = F(x) $ based on the given information, we can say $y = y$ and $F(x) = \int f(x)$.

3) There is only one solution y(t) to the initial value problem dy dt = 3t2 , y(1) = π.

True. The authors were trying to lead you to this solution with the previous two questions. If we set the problem up:

$$ \frac{dy}{dt} = 3t^2 $$

$$ dy = 3t^2dt $$

$$ \int dy = \int 3t^2dt $$

$$ y = t^3 + C $$

We know that $y(1) = \pi$, so we can solve for C as you did.

$$ \pi = (1)^3 + C \rightarrow C = \pi - 1 \Rightarrow y = t^3 + \pi - 1$$

This means there's only one solution for any $t$ in the function $y(t)$.

MrCMedlin
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