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course MTH 279
3.4.1. Solve the equation y ' = 2 t y ( 1 - y), with y(0) = -1, as a Bernoulli equation.
dy/dt = 2 t y - 2 t y^2
y^-2 dy/dt - 2 t y^-1 = -2t
v = y^-1
dv/dt = -y^-2 dy/dt
-dv/dt - 2 t v = -2t
dv/dt + 2 t v = 2t
f(t) = 2t g(t) = 2t
Integrating Factor = e^(t^2)
e^(t^2) dv/dt + 2 e^(t^2) t v = 2t*e^(t^2)
Integrate
e^(t^2) v = e^(t^2) + C
v = 1 + C/ (e^(t^2))
1/y = 1 + C/ (e^(t^2))
-1 = 1 + C/ (e^(0))
C = -2
1/y = 1 + -2/(e^(t^2))
2. Solve the equation y ' - y = t y^(1/3), y(0) = -9 as a Bernoulli equation.
y^(-1/3) * dy/dt - y^(2/3) = t
v = y^(2/3)
dv/dt = 2/3 y^(-1/3) dy/dt
2/3 y^(-1/3) * dy/dt - 2/3 y^(2/3) = 2/3 t
dv/dt - 2/3 v= 2/3 t
f(t) = 2/3 g(t) = 2/3 t
Integrating factor:
e^(2/3 t)
e^(2/3 t) * dv/dt - 2/3 e^(2/3 t) v= 2/3t e^(2/3 t)
Integrate
e^(2/3 t) * v = t e^(2/3 t) - 3/2 e^(2/3 t) + C
v = t - 3/2 + C
y^(2/3) = t - 3/2 + C
-9 ^ (2/3) = 0 - 3/2 + C
C = -2.83
y^(2/3) = t - 4.33
3. Solve the equation y ' = - (y + 1) + t ( y + 1)^(-2) as a Bernoulli equation.
dy/dt + (y+1) = t (y +1) ^(-2)
(y+1)^2 dy/dt + (y+1)^3 = t
v = (y+1)^3
dv/dt = 3(y+1)^2 dy/dt
dv/dt + 3v = 3t
f(t) = 3 g(t) = 3t
Integrating factor:
e^(3t)
e^(3t) dv/dt + 3v* e^(3t) = 3t e^(3t)
Integrate
e^(3t) v = 1/3 e^(3t) * (3t-1) + C
v = t - 1/3 + C / (e^(3t))
(y+1)^3 = t - 1/3 + C / (e^(3t))
C cannot be found as no initial condition is given
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