I would rewrite the ODE as:
(y′)2+11d(y′)=dx
Integrating, we get:
tan−1(y′)=x+c1
or:
y′=tan(x+c1)
Integrating again, we find:
y=−ln∣cos(x+c1)∣+c2=−ln∣c2cos(x+c1)∣
That was very clever to see. Thanks for pointing that out! So short and simple if you see it.
And thanks for the alternative solution but I'm affraid it's a bit above my head to continue with the sec() formula's( we haven't seen those methods). There probably is a conventional solution though since my course exercices are all at the a certain level but still supposed to be solved('easily') with the substitution method i was trying to follow.
I've solved all my other 15 exercices given on 2nd order DE's(which is great news for me), only one left that I can't get the hang of it.
y'' + 2*x * y'² = 0
becomes after substitution y' = p , y'' = p' since x is in the DE
p' + 2*x*p² = 0
(1/p²)*dp = -2*x*dx
int => 1/p = x² - C1
[1/ (x² - C1)]*dx = dy
int(using the ln formula)=> gives a to complicated formula and made me think that i'm going the wrong again.
The problem is that I don't know the final answer for this DE(could have given me a hint what direction to work to).
They give however the particular solution(values of x and y to find C1 and C2) given y(1) = 1 and y'(1) = 1.
Filling in these values in the general solution would have to give y = - 1/x +2.
Thanks for your efforts so far.
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