Question

How to solve?
x"' +(x^2-c)x' = 0?
Please, help

Answer 1

thought about this way?

\(\dddot x + x^2 \dot x - c \dot x = 0\)

\(\ddot x + \dfrac{x^3}{3} - c x = d\)

\(\dot x \dfrac{d \dot x}{dx} = d + c x- \dfrac{x^3}{3} \)

\(\dfrac{\dot x^2}{2} = dx + c \dfrac{x^2}{2} - \dfrac{x^4}{12} + e\)

\(\int \; \dfrac{ dx }{\sqrt{2 dx + c x^2 - \dfrac{x^4}{6} + 2e}} = \int \; dt\)

just typed this out and unfortunate use of d as a integration constant but think it's still readable .... looks horrrible :-(

Answer 2

@IrishBoy123 @Loser66 use Laplace transform , is better method to solve it

Answer 3

Show me, please.I didn't think of it. I'll try now.

Answer 4

No, we can't since we don't have initial values

Answer 5

Are you sure about this.. I just tried plugging it into wolfram alpha and it seems pretty unsolvable: http://www.wolframalpha.com/input/?i=x'''(t)+%2B(x(t)%5E2-c)*x'(t)+%3D+0

Answer 6

@Loser66 yup that we cant solve with laplace

Answer 7

my try