- #1
MathewsMD
- 433
- 7
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.)
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.)