1. Find the general solution of the equation y '' * (1 + y ^ 2) = 2Y * y '^ 2. Find the general solution of the equation y ^ 3 * y' '+ 1 = 0

1. Find the general solution of the equation y '' * (1 + y ^ 2) = 2Y * y '^ 2. Find the general solution of the equation y ^ 3 * y' '+ 1 = 0

1. Let P = y '= dy / DX
Then y '' = D (y ') / DX = DP / DX = (DP / dy) * (dy / DX) = P * (DP / dy)
The original equation is changed into:
p*(dp/dy)*(1+y^)=2y*p^
P = 0 or (1 / P) DP = [2Y / (1 + y ^)] dy
If P = 0 holds, then y '= 0, y = C, which is a part of the general solution of the equation;
When (1 / P) DP = [2Y / (1 + y ^)] dy is true:
∫dp/p = ∫d(1+y^)/(1+y^)
lnp = ln(1+y^) + C
p=C1*(1+y^)
That is dy / DX = C1 * (1 + y ^)
dy/(1+y^) = C1*dx
∫dy/(1+y^) = C1∫dx
arctany=C1x+C2
y=tan(C1x+C2)
The two general solutions of the original equation are y = C and y = Tan (c1x + C2)
2. As in the previous question, let P = y '= dy / DX
∴y''=p*dp/dy
So the original equation is changed into:
(y^3)*p*dp/dy + 1=0
p*dp = -dy/(y^3)
∫p*dp=∫[-y^(-3)]*dy
p^/2 = y^(-2)/2 + C
p^=C1 + (1/y^)
p^= (C1y^+1)/y^
p=±[√(C1y^+1)/y]
That is dy / DX = ± [√ (c1y ^ + 1) / y]
[y/√(C1y^+1)] *dy=±dx
∫(1/C1)*d(C1y^+1)/[2√(C1y^+1)] =±∫dx
(1/C1)*√(C1y^+1)= ±x + C3
C1y^+1=(C1*C3±C1x)^
y^= [(C2±C1x)^-1]/C1
If you have to write the form of "y =", that is:
y=±√{[C2±C1x]^-1]/C1}