What is the general solution of the differential equation y '' + y '^ 2 = 2E ^ (- y)?

What is the general solution of the differential equation y '' + y '^ 2 = 2E ^ (- y)?

Conclusion: y = ln (x ^ 2 + C [1] x + C [2]) is a subscript
From Y & # 39; &# 39; + Y & # 39; ^ 2 = 2E ^ (- y), e ^ y.y & # 39; &# 39; + e ^ y. (Y & # 39;) ^ 2 = 2 & nbsp;
(e ^ y.y & # 39;) & # 39; = 2 & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; the solution is e ^ y.y & # 39; = 2x + C [1]
From e ^ y.y & # 39; = 2x + C [1] & nbsp;, we get (e ^ y) &# 39; = 2x + C [1]
The solution is e ^ y = x ^ 2 + C [1] x + C [2]
So y = ln (x ^ 2 + C [1] x + C [2])
 
Hope to help you!
 

Finding the general solution of linear differential equation y '+ y = 2E ^ x

y(x) = exp(x)+C*exp(-x)
It is known visually that exp (x) is a special solution of the equation; the general solution of the linear homogeneous equation y '+ y = 0 of the original equation is c * exp (- x), so the linear combination
Y (x) = exp (x) + C * exp (- x) is the general solution of the equation

The general solution of Y '' + y = cosx
Answer quickly

y=sin(x)*C2+cos(x)*C1+1/2*cos(x)+1/2*sin(x)*x