If y′= x(1+ y) and y > -1 , then y =
The differential equation dy/dx=y/x^2 has a solution given by:
3 answers
http://www.wolframalpha.com/input/?i=Solve++y%27(x)+%3D+x%2Bxy
dy/y = dx/x^2
ln y = -1/x + c
y = C e^-1/x
http://www.wolframalpha.com/input/?i=Solve++y%27(x)+%3Dy%2Fx%5E2
ln y = -1/x + c
y = C e^-1/x
http://www.wolframalpha.com/input/?i=Solve++y%27(x)+%3Dy%2Fx%5E2
more like
dy/(1+y) = x dx
ln(1+y) = x^2/2
1+y = c e^(x^2/2)
y = c e^(x^2/2) - 1
dy/(1+y) = x dx
ln(1+y) = x^2/2
1+y = c e^(x^2/2)
y = c e^(x^2/2) - 1