Question

resolva a seguinte equacao diferencial y'=1+y/1+x

Answer 1

Resposta:

dy/dx =(1+y)/(1+x)

1/(y+1)  dy  = 1/(1+x)  dx

∫1/(y+1)  dy  = ∫1/(1+x)  dx

u=y+1       ==>du=dy

e  

w=1+x    ==>dw=dx

∫1/u  du  = ∫1/w dw

ln(u) = ln(w)  + c

ln(y+1) = ln(x+1)  + c

y+1 = e^[ln(x+1) +c]

y = e^[ln(x+1) +c]  -1

y=e^[ln(x+1] * e^c  - 1        ...fazendo e^c=k  ..constante

y=k * e^(ln(x+1)  -  1                x >1