Use implicit differentiation and the product rule. What do you get as a first step?
For C, can someone show me step-by-step how to isolate the function y and/or its derivatives? I said "and/or" just in case, I cannot isolate the function y itself but can for its derivatives so that I can say that the equation C is the general solution of one the numbers.
Any help would be greatly appreciated!
Thanks in advance!
1. is a standard second order linear constant coefficient ODE. How do you solve those?
2. Rewrite it as
and then make the substitution ...
3. Another second order linear constant coefficient ODE...
4. The DE is separable.
5. The answer will be the whichever one is left over...
I'm attaching the pdf which contains that segment of my work but I am now stuck at the next step. I also don't see how it's useful in differential equations to have solutions with y not isolated even though the derivatives y' (and so on) are isolated or basically if at least one of the equations has a problem with the isolation.
So, divide top and bottom of the RHS of your final result by -3. Does that start to look more familiar?
You can still obtain useful information about the solutions, even if they're defined implicitly. For example, computers can graph implicit solutions.I also don't see how it's useful in differential equations to have solutions with y not isolated even though the derivatives y' (and so on) are isolated or basically if at least one of the equations has a problem with the isolation.
Ackbeet: I noticed that C matches #2 when I divide top and bottom by -3. Does this mean that the antiderivative of #2 is the general solution for differential equation #2?
Prove It: You mentioned a few stuff that I did not cover so I'll look at it when I learn the new stuff to get this question from a different perspective so thank you still.
I wouldn't quite phrase it that way. You can't integrate # 2 directly. Say, rather, that Equation C is the general solution of DE # 2. On the other hand, the process of solving differential equations is essentially integrating. So maybe you can say that the integral of DE # 2 is Equation C.