Hey all, i just lost a TON of points on a test for solving a problem in a way that is apparently invalid.

The problem was verify that y(x) = x+1 is a solution for dy/dx = y*y-x*x-2x; y(0) = 1. i plugged y = x+1 into the right side of the second equation, got dy/dx = 1, integrated to get y=x+c, used y(0) = 1 to get c= 1, therefore y = x + 1

My professor's annoyed 2-second explanation about why my method is invalid was that I assumed that it worked to prove that it worked. I sorta buy it, but I'm not completely convinced, could someone give me a counter example to prove that my method is not legit? [to clarify, my method is to plug in y(x) into the DE, then integrate, then use the given initial conditions to solve for c to get a new y(x) and make sure that my new y(x) is the same as the old one].

The counter example I am requesting would take a form that is similar to the problem above, except that y(x) would not be a legit solution to dy/dx, BUT my method would falsely show that y(x) does work. Obviously, if no such counter example exists, that my method proves that the DE works and I should not have lost any points

2. $\frac{dy}{dx}=y^{2}-x^{2}-2x$ and $y'=1$ besides $y^{2}-x^{2}-2x$ becomes 1 so there's the equality, and that's all you need to do, nothing about integration, you can't do that, that's not what the problem is asking.

3. I feel like I did a logically equivalent thing, just working backwards:

You took the derivative of x+1 to get 1, I took the integral of 1 (and used the initial conditions) to get x+1, logically equivalent things, I just did it in reverse (but the converse is logically equivalent because its obviously an "if and only if" situation)

If anyone has that counter example to prove that I'm wrong I would be VERY happy, but thanks for your reply!

4. Originally Posted by Jaevko

The problem was verify that y(x) = x+1 is a solution for dy/dx = y*y-x*x-2x; y(0) = 1. i plugged y = x+1 into the right side of the second equation, got dy/dx = 1, integrated to get y=x+c, used y(0) = 1 to get c= 1, therefore y = x + 1
If that states the problem, then you just contemplate its derivative and to the algebra on the RHS.

You can't integrate, that's not what the problem is asking. You're not meant to solve the ODE so that you can plug initial conditions.

Of course in the right side of the equation you got $1$ because you just substituted $y=x+1$ on it, and that's what is not right, because you're told that $y=x+1$ satisfies the equation (you need to verify that, so read again my first line of this post), then, when you tried to use it on the right side, you're actually assuming that $y=x+1$ is the solution.