How to find the general solution of first order linear differential equation

How to find the general solution of first order linear differential equation

As shown in the figure, remember the formula first, and then apply it

Given two special solutions of second order nonhomogeneous linear differential equation, how to find the general solution?

If two linearly independent special U (x), V (x) of: Y "- P (x) * y '- Q (x) * y = 0 are obtained, then the general solution formula of the non-homogeneous equation: Y" - P (x) * y' - Q (x) * y = f (x) is: y = C1 * u (x) + C2 * V (x) + ∫ [u (s) * V (x) - U (x) * V (s)] / [u (s) * V '(x) - V (s)

The general solution of the second order differential equation y '' = 1 + (y ') ^ 2
Using the method of PDP / dy = y ", P = y '
Just this way!
I do it this way
P ^ 2 = e ^ (2Y + 2C) - 1 and then we can't do it

Since we know it will be very troublesome, why do we have to do it
Next, it can be written as
p^2=(y')^2=C1e^(2y)-1
So y '= √ (C1E ^ (2Y) - 1)
So dy / √ (C1E ^ (2Y) - 1) = DX
Then let u = √ (C1E ^ (2Y) - 1)
So y = (1 / 2) ln (1 + u ^ 2) - C '
dy=udu/(1+u^2)
therefore
∫dy/√(C1e^(2y)-1)=∫du/(1+u^2)=arctanu=arctan√(C1e^(2y)-1)
So the integral on both sides of ^ (dy / √) - E is obtained
arctan√(C1e^(2y)-1)=x+C2