Answer:
The solution to the differential equation y'(y + 7)y'' = (y')²
y = Ae^(Kx) - 7
Step-by-step explanation:
Given the differential equation
y'(y + 7)y'' = (y')² ..................(1)
We want to solve using the substitution u = y'.
Let u = y'
The u' = y''
Using these, (1) becomes
u(y + 7)u' = u²
u' = u²/u(y + 7)
u' = u/(y + 7)
But u' = du/dy
So
du/dy = u/(y + 7)
Separating the variables, we have
du/u = dy/(y + 7)
Integrating both sides, we have
ln|u| = ln|y + 7| + ln|C|
u = e^(ln|y + 7| + ln|C|)
= K(y + 7)
But u = y' = dy/dx
dy/dx = K(y + 7)
Separating the variables, we have
dy/(y + 7) = Kdx
Integrating both sides
ln|y + 7| = Kx + C1
y + 7 = e^(Kx + C1) = Ae^(Kx)
y = Ae^(Kx) - 7