# (probably stupid) Differential equation question

1. Feb 21, 2009

### Jaevko

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. Feb 22, 2009

### Staff: Mentor

What you did was sort of a cross between verifying that a given function was a solution and attempting to find the solution.

When you substituted y = x + 1 into the right side to get 1, why didn't do the same substitution on the left side? After all, if y = x + 1, dy/dx = 1.

When you got dy/dx = 1, that's not the same differential equation as the one you started with. dy/dx happens to be equal to 1 when y = x + 1. By treating dy/dx as a constant, you are eliminating all of the other potential soltutions of the DE dy/dx = y^2 - x^ - 2x. Off the top of my head I don't know what the other solutions to this DE might be, but such an equation (without the initial condition) generally has an infinite number of them.

There's a big difference between being asked to verify that a function is a solution of an initial value problem (a DE + a set of initial conditions), and finding the solutions to a DE. At this point, you probably don't have the tools to solve nonlinear DEs like this one, so make your life a little easier and do what is asked for. After you've done that, you can explore alternate techniques.