# Thread: non-linear 2nd order differential

1. ## 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?

2. 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 $u=y^{\prime}\implies u^{\prime}=y^{\prime\prime}$.

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

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

Can you take it from here?

3. 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

4. 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. $\int \frac 1{u^2}~du ~{\color{red} \ne }~ \ln u^2 + C$ (!)

Note that $\int \frac 1{u^2}~du = \int u^{-2}~du$

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

5. du/dx = -2x u^2

du/u = -2x dx

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

-u ^-1 = -x^+c

6. 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 $u= \frac{1}{x- c}$

And now, since $u= \frac{dy}{dx}$, you still need to solve
$\frac{dy}{dx}= \frac{1}{x- c}$
That should be an easy integral and will introduce a second "constant of integration".