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?

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- Jan 13th 2010, 01:38 PMharveyonon-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? - Jan 13th 2010, 02:09 PMChris L T521
Let $\displaystyle u=y^{\prime}\implies u^{\prime}=y^{\prime\prime}$.

The equation becomes $\displaystyle u^{\prime}+2xu^2=0\implies u^{\prime}=-2xu^2$. Apply separation of variables to solve for u, where $\displaystyle y^{\prime}\!\left(0\right)=1\implies u\!\left(0\right)=1$.

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

Can you take it from here? - Jan 13th 2010, 03:09 PMharveyo
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 - Jan 13th 2010, 03:26 PMJhevon
- Jan 15th 2010, 04:03 PMharveyo
du/dx = -2x u^2

du/u = -2x dx

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

-u ^-1 = -x^+c - Jan 16th 2010, 03:33 AMHallsofIvy
You mean du/u^2

Quote:

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

-u ^-1 = -x^+c

And now, since $\displaystyle u= \frac{dy}{dx}$, you still need to solve

$\displaystyle \frac{dy}{dx}= \frac{1}{x- c}$

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