How do you solve differential equations of the form dv/dt=f(t)+g(v)?

Question

The specific equation I'm trying to solve is:
dv/dt = ae^(bt) - cv
where a, b and c are positive constants.
I know how to solve differential equations of the form
dv/dt = f(t)g(v)
but not
dv/dt = f(t) + g(v).

Answer 1

dv/dt + cv = ae^(bt)

this is of the type dv/dt + P(t)v = Q(t)

multiply through by e^(∫P(t)dt), in this case e^(∫cdt) = e^(ct)

e^(ct).dv/dt + cv.dv/dt = ae^(bt).e^(ct)

the LHS is then an exact derivative

d/dt(ve^(ct)) = ae(bt + ct)

integrating both sides

ve^(ct) = ae(bt + ct)/(b + c) + K

v = ae^(bt) / (b + c) + Ke(-ct)

Answer 2

while the DE isn't = 0 you first sparkling up the auxiliary equation yet you may desire to function a particular answer (playstation ) which will supply the mind-blowing hand ingredient fee of 12. The playstation is of the type y=C and sub into to DE which provides -4C=12 so C=-3 and the final answer is y=Ae^(4x) + Be^(-x) - 3 there are distinctive techniques while the RHS isn't a continuing. Ask your instructor.

Answer 3

You could try

dv = [f(t) + g(v)] * dt --> dv = f(t) dt + g(v) dt

integrate both sides with respect to the variable there. In this case, g(v) acts like a constant from t's point of view.

v = F(t) + t * g(v). F(t) = integral of f(t), of course.

In this case, v = (a/b)*e^(bt) - cvt

v + cvt = v(1+ct) = (a/b)*e^(bt)

v = [(a/b)*e^bt)]/(1+ct)