Solving exact 2nd order differential equations?

Question

The equation P(x)y” + Q(x)y’ + R(x)y= 0 is said to be exact if it can be written in the form [P(x)y’]’ + [f(x)y]’=0, where f(x) is to be determined in terms of P, Q, and R. The latter equation can be integrated once immediately, resulting in a first order linear eq for y that can be solved. By equating the coefficients for the preceding equations and then eliminating f(x), show that a necessary condition for exactness is P”(x) – Q’(x) + R(x) =0 . It can be shown that this is also a sufficient condition.

You don’t actually have to do this problem, but it’s supposed to help with the one I’m supposed to do.

1) x2y” + xy’ – y = 0, x>0.

How would you do this problem… or solve any exact equation

Answer 1

If you carry out the differentiation of the exact form, you get

P(x)y'' + P'(x)y' + f'(x)y + f(x)y' =
P(x)y'' + (P'(x)+f(x))y' + f'(x)y.

By matching coefficients, you obtain
f'(x) = R(x)
P'(x) + f(x) = Q(x)
If you differentiate the second equation and substitute the first, you obtain the necessary condition for exactness above.

To solve an exact equation, find f(x). Using the second equation above, you have f(x) = Q(x)-P'(x). In the equation you gave, P(x)=x^2, Q(x)=x, and R(x)=-1, so f(x) = x - 2x = -x. Sure enough, f'(x) = R(x).

Now you have the equation in the form
[x^2y']' + [-xy]' = 0.
Integrate, and you have
x^2 y' - xy = 0,
or, after dividing by x,
xy' - y = 0.
It follows that y' = y/x, so y = x.

Answer 2

could desire to assert i don't comprehend Garvit's answer. P = cos(x)cos(y) + 2x, Q = -sinx siny -2y Exactness: dP/dy = -- cosx siny = dQ/dx discover f(x,y) such that df/dx = P, df/dy = Q df/dx = cosx comfortable + 2x ==> f(x,y) = sin x cos y + x^2 + g(y) Then df/dy = -- sinx siny + g'(y) subsequently g'(y) could desire to = -2y; g(y) = -- y^2 + C f(x,y) = sin(x) cos(y) + x^2 - y^2 + C all suggestions of the DE: y^2 - x^2 -- sin(x) cos(y) = C, any consistent