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.
Best Answer
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.