Thread: non-linear 2nd order differential

y'' + 2x (y')^2= 0

show the general solution, then find an explicit formula for a solution y(x), which also satisfies intial conditions y'(0) =1 and y (0) = 0

where to start?

Originally Posted by harveyo

y'' + 2x (y')^2= 0

show the general solution, then find an explicit formula for a solution y(x), which also satisfies intial conditions y'(0) =1 and y (0) = 0

where to start?

Let .

The equation becomes . Apply separation of variables to solve for u, where .

Then back substitute and solve the resulting differential equation for y.

Can you take it from here?

seperation of variables =

u= -2x u^2

dy/dx = -2x u^2

dy/u^2 = -2x dx

integrate both sides

ln u^2 = -x^2 + C

am i doing this correct

Originally Posted by harveyo

seperation of variables =

u= -2x u^2

dy/dx = -2x u^2

dy/u^2 = -2x dx

integrate both sides

ln u^2 = -x^2 + C

am i doing this correct

no. (!)

Note that

by the way, why do you have dy? it should be du

du/dx = -2x u^2

du/u = -2x dx

intergrate u ^-2 du = intergrate -2x dx

-u ^-1 = -x^+c

Originally Posted by harveyo

du/dx = -2x u^2

du/u = -2x dx

You mean du/u^2

intergrate u ^-2 du = intergrate -2x dx

-u ^-1 = -x^+c

Good. so

And now, since , you still need to solve

That should be an easy integral and will introduce a second "constant of integration".