Solving for nonlinear 2nd order DE

[MathewsMD]

Given a DE in the general form of either y'' = y^2 or y'' = (y-1)^2, is there a general method to solve these?

I separated the equations to get y''(y^-2) = 1 and then integrated, which left me with (-y^-1)dy = (t + c)dt, and then integrated once more.

Is this correct so far? I have essentially split the d^2y/dt^2 terms and integrated w/ respect to each variable (i.e. dt on one side and dy on the other, leaving me w/ another dy and dt on each respective side still).

Honestly, this all just seems like poor mathematics and faulty reasoning since d^2y and dt^2 are single terms themselves, and must be integrated w/ respect to d^2y and dt^2 respectively, and not dy and dt, to my knowledge. This seems like a fundamental concept but besides splitting up the d^2y/dt^2 terms to differentiate, i don't quite see another method to solve for y in this case. Any methods or hints on ways to approach the question would be greatly appreciated! (I have not quite tackled second order, non linear DEs in my studies yet, but any information needed to answer the above problem is always welcome.)

 

[HallsofIvy]

I Yes, there is a method for solving second order d.e s where the independent variable, "x", does not appear explicitly. It's called "quadrature". We can define $v= dy/dt$ and then write $d^2y/dx^2$ as $dv/dx$ and then use the "chain rule" to write $dv/dt= (dv/dy)(dy/dx)= v(dv/dy)$. So $y''= -y$ becomes $v(dv/dy)= -y^2$. That can be written as $vdv= -y^2dy$ and integrated for v= dy/dx as a function of y and so a first order differential equation. Similarly, $d^2y/dx^2= (y- 1)^2$ becomes $v dv/dy= (y- 1)^2$ so that $vdv= (y- 1)^2 dy$.

 

[bigfooted]

In the case y"=y^2 will lead to the Weierstrass elliptic function as a solution, see for instance chapter 6 of H.T. Davis, introduction to nonlinear differential and integral equations

 

[MathewsMD]

I Yes, there is a method for solving second order d.e s where the independent variable, "x", does not appear explicitly. It's called "quadrature". We can define $v= dy/dt$ and then write $d^2y/dx^2$ as $dv/dx$ and then use the "chain rule" to write $dv/dt= (dv/dy)(dy/dx)= v(dv/dy)$. So $y''= -y$ becomes $v(dv/dy)= -y^2$. That can be written as $vdv= -y^2dy$ and integrated for v= dy/dx as a function of y and so a first order differential equation. Similarly, $d^2y/dx^2= (y- 1)^2$ becomes $v dv/dy= (y- 1)^2$ so that $vdv= (y- 1)^2 dy$.

So in the solution above that I posted, I did do that, and ended up getting:

$$2t + k = \int [(y-1)^3 + C]^{-1/2} dy$$

My main problem is going from here. Trigonometric substitution seems like one method, though I did mess up somewhere (I think) and it's a fairly long process. I feel like I'm overlooking a simpler method and if you could help hint at another way to tack this, that would be great. But if there is not another method, thank you for the help.

 

[MathewsMD]

I've revised my solution since I realized I made a blatant error. I also added an extra image to hopefully make it more clear. So I've arrived at a solution, but I've also been given initial conditions (a y' and y value at specified t) but I can't exactly plug it into solve for the exact solution. If anyone has any ideas on how to solve for the constants and also if there's another better method, that would be very helpful! Thank you!