A first order linear equation in the form y'+p(x)y=f(x) can be solved by finding an integrating factor mu (x) = exp( int (p(x) dx))
(1) Given the equation y' + 3 y = 2 find mu(x) =
(2) Then find an explicit general solution with arbitrary constant C.
y = .
(3) Then solve the initial value problem with y(0)=2
y = .

Question

A first order linear equation in the form y'+p(x)y=f(x) can be solved by finding an integrating factor  mu (x) = exp( int (p(x) dx))
(1) Given the equation y' + 3 y = 2 find mu(x) =
(2) Then find an explicit general solution with arbitrary constant C.
 y  = .
(3) Then solve the initial value problem with y(0)=2
 y  = .

myininaya · Answer

what is considered mu?

ipwnbunnies · Answer

Your integrating factor should be:
$$\mu (x) = e^{\int\limits_{}^{} p(x) dx}$$
This form of differential equation is written as:
y' + p(x)*y = g(x)

myininaya · Answer

oh integrating factor is mu ?

anonymous · Answer

$$\mu (x)$$ the integrating factor. I don't know how to start.

myininaya · Answer

@ipwnbunnies gave you the formula