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my Calculus II class is nearing the end of the quarter and we've just started differential equations to get ready for Calculus III. In my homework, I came upon these problems.

One of the problems was:

Find the general solution to the differential equation $$\frac{dy}{dt} = t^3 + 2t^2 - 8t.$$

The teacher just said to integrate. So I did. Then in question 8a it gives the differential equation:

$$\frac{dy}{dt} = y^3 + 2y^2 - 8y.$$

and asks "Why can't we find a solution like we did to the previous problem? My guess was: "In 7 we were integrating with respect to t. Since this equation is the highest order derivative, we can't solve it like # 7". Although, I have no confidence in that answer and I'm not sure it makes total sense even to me.

Also, part 8B. asks: Show that the constant function $y(t) = 0$ is a solution.

I've done a problem like this before, except that it wasn't a constant function. This problem seems like a question that asks: "show that every member of the family of functions $y = (\ln x + C)/x$ is a solution to the differential equation (some diff. equation)" except it seems a little bit different.

Any hints on how I can solve this?

Thank you.

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3 Answers 3

up vote 3 down vote accepted

Parts of problems that say "show that [explicitly given function] is a solution" can be solved by simply "plugging in". E.g., if you're supposed to show that $y(t)=\frac{1}{1-t}$ is a solution to the differential equation $dy/dt=y^2$, then you should compute $dy/dt=\frac{1}{(1-t)^2}$, compute $y^2=\frac{1}{(1-t)^2}$, and "plug them in" to check that the differential equation is satisfied for this choice of $y$.

You're correct that the difference in the second problem is that you no longer have the derivative with respect to $t$ expressed as a function of $t$. When the equation involves both a function $y$ and its derivative (or derivatives), it can generally be more difficult to solve. In this case, a hint I'll give you is that $dy/dt=1/(dt/dy)$ (e.g., see Wikipedia). You can first turn things around and think of $t$ as a function of $y$. You can find $t(y)$ using similar methods to the first problem (but with a more complicated function), and then find the inverse function to get $y(t)$.

A somewhat more general method of solving differential equations that applies in this case is separation of variables.

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The difference is that the main variable is $y$ and you separate the variables to integrate and obtain the solution

Hint 1:

The first equation is very easy, so :

$$\frac{dy}{dt}=t^3 + 2t^2 - 8t \Rightarrow dy = (t^3 + 2t^2 - 8t) dt \Rightarrow $$

$$ \int dy = \int (t^3 + 2t^2 - 8t) dt $$ ( which is the left side ? )

Hint2:

For the other. $$\frac{dy}{dt}=y^3 + 2y^2 - 8y \Rightarrow \frac{1}{y^3 + 2y^2 - 8y}dy = dt $$

Then proceed integrating.

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In the first question, you are given the derivative in terms of the variable. But in the second question, you are given an expression for the derivative that involves the function. For instance, it would be one thing if you were told $\frac{dy}{dx} = x$ (which would mean that $y = \frac{1}{2}x^2 + C$), and a completely different thing if you are told $\frac{dy}{dx}=y$ (this tells you that the function $y$ is equal to its derivative; which means that $y=Ae^x$ for some constant $A$).

Actually, we can solve the second differential equation; but we don't solve it by simple integration, just like we don't solve the equation $y' = y$ by integrating.

Showing that $y(t)=0$ is a solution to the second equation is just a question of plugging in $0$ for $y$ and verifying that you get a solution by verifying that the resulting equality is true.

You never said what question 8a was, through; were you asked to solve the differential equation, or just to say what is the difference between it and the previous one?

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