求微分方程的通解 y✀✀=e^y

e^y=y'' = d(y')/dx = d(y')/dy * dy/dx = y'd(y')/dy,
y'd(y') = e^ydy,
d[(y')^2] = d[2e^y],
2e^y + C^2 = (y')^2, C 为任意常数.

y' = [2e^y + C^2]^(1/2)或y'=-[2e^y+C^2]^(1/2),

(1)y' = [2e^y + C^2]^(1/2),
C=0时, dy/dx = y' = [2e^y]^(1/2) = 2^(1/2)e^(y/2),
e^(-y/2)dy = 2^(1/2)dx,
d[e^(-y/2)] = d[-2^(-1/2)x],
e^(-y/2) = c- x*2^(-1/2), c为任意常数.

C^2>0时, u=[2e^y+C^2]^(1/2)>0, u^2 - C^2 = 2e^y, e^y = (u^2-C^2)/2, e^ydy=udu,
dy = udu/e^y = 2udu/(u^2-C^2),
2udu/[(u^2-C^2)dx]=dy/dx = y' = [2e^y + C^2]^(1/2) = u,
2du/[u^2-C^2] = dx = (1/C)du/(u-C) - (1/C)du/(u+C) =d[ (1/C)[ln|u-C| - ln|u+C|] ],
x+c = (1/C)[ ln|u-C| - ln|u+C|] = (1/C) [ ln| [2e^y + C^2]^(1/2) - C| - ln| [2e^y + C^2]^(1/2) + C| ]
C为任意非0常数, c为任意常数.

(2)y' = -[2e^y + C^2]^(1/2),
C=0时, dy/dx = y' = -[2e^y]^(1/2) = -2^(1/2)e^(y/2),
e^(-y/2)dy = -2^(1/2)dx,
d[e^(-y/2)] = d[2^(-1/2)x],
e^(-y/2) = c+x*2^(-1/2), c为任意常数.

C^2>0时, u=[2e^y+C^2]^(1/2)>0, u^2 - C^2 = 2e^y, e^y = (u^2-C^2)/2, e^ydy=udu,
dy = udu/e^y = 2udu/(u^2-C^2),
2udu/[(u^2-C^2)dx]=dy/dx = y' = -[2e^y + C^2]^(1/2) = -u,
2du/[u^2-C^2] = -dx = (1/C)du/(u-C) - (1/C)du/(u+C) =d[ (1/C)[ln|u-C| - ln|u+C|] ],
-x+c = (1/C)[ ln|u-C| - ln|u+C|] = (1/C) [ ln| [2e^y + C^2]^(1/2) - C| - ln| [2e^y + C^2]^(1/2) + C| ]
C为任意非0常数, c为任意常数.

若是2C1e^(-y/2)=sin(C1x+C2),则
2C1e^(-y/2)*(-1/2)y' = cos(C1x+C2)*C1,
-y'e^(-y/2)=cos(C1x+C2),
y' = e^(y/2)cos(C1x+C2).
y''=e^(y/2)cos(C1x+C2)*(y'/2) - e^(y/2)*sin(C1x+C2)*C1
=(1/2)e^(y)[cos(C1x+C2)]^2 - C1e^(y/2)*sin(C1x+C2),
不等于e^(y)啊?
所以,那个答案不对哈.

p=y' pdp/dy=y'' pdp=e^ydy 解得：p^2=2e^y+C1
y'=±√(2e^y+C1) dy/(√(2e^y+C1))=±dx

1.C1=0，通解为：e^(-y/2)=C2±x/√2

2.C1<0, 对于积分：∫dy/(√(2e^y-|C1|))，
设√(2e^y-|C1|))=u，2e^y-|C1|=u^2 e^ydy=udu,
∫dy/(√(2e^y-|C1|))=2∫du/(u^2+|C1|)=(2/√|C1|)arctanu/√|C1|
通解为：(2/√|C1|)arctan[√(2e^y-|C1|)/√|C1|]=±x+C2

3 C1>0, 对于积分：∫dy/(√(2e^y+C1)
设√(2e^y+C1)=u，2e^y+C1=u^2 e^ydy=udu,
∫dy/(√(2e^y+C1)=2∫du/(u^2-C1)=(1/√C1)ln|(u-√C1)/(u+√C1)|
通解为：(1/√C1)ln|(√(2e^y+C1)-√C1)/(√(2e^y+C1)+√C1)|=±x+C2

注：上述2,3还可作适当的简化。

令P=y'，则
y''=dP/dx=(dP/dy)·(dy/dx)= P dP/dy
则
P dP/dy =e^y
→P dP =e^y·dy
两边积分得
(1/2)P² =e^y +(1/2)C1²;
P² =2e^y +C1²
P=√(2e^y +C1²)
则
dy/dx=√(2e^y +C1²)
→dx=(1/√(2e^y +C1²) ) dy
积分得
x=∫1/√(2e^y +C1²) dy
=∫e^y/[e^y·√(2e^y +C1²)] dy
=∫1/[e^y·√(2e^y +C1²)] d e^y
令e^y=t,则
x=∫1/[t·√(2t +C1²)] d t
再令u=√(2t +C1²)，则t=(u²-C1²)/2
dt=udu
则x=∫u / [u(u²-C1²)/2] d u
=2∫1 / (u²-C1²) d u
=(1/C1)·∫[1/(u-C1) - 1/(u+C1)] d u
=(1/C1)· ln|(u-C1)/(u+C1)| +C2
=(1/C1)· ln|(√(2t +C1²)-C1) / (√(2t +C1²)+C1)| +C2
=(1/C1)· ln|(√(2e^y +C1²) - C1) / (√(2e^y +C1²) + C1)| + C2

原微分方程的解是
x=(1/C1)· ln|(√(2e^y +C1²) - C1) / (√(2e^y +C1²) + C1)| + C2

y=e^y+Cy

y=e^y+c1y+c2