dy/dx +6y = 3 satisfying initial condition y(0)=0?

Question

what is y= to

Answer 1

First, write dy/dx = y' , so you want y' + 6y = 3.

First, you would like to get rid of that "3" there. To do so, you can do a change of variables:

Let y = z + 1/2.

Then y' = z' and 6y = 6z + 3, so z' +6z = 0.

Now, solving, you get z'/z = -6

Integrating both sides, you get ln(z) = -6x + C and so z = Ce^(-6x)

Finally, y = Ce^(-6x) + 1/2.

Using the y(0) = 0 constraint, you get 0 = C + 1/2, so C = -1/2

SO y = (-1/2)e^(-6x) + 1/2

Answer 2

That is saying that if you plugged in 0 for y, the answer would become 0 =3. That is impossible. It pretty much is a domain error.

Answer 3

dy/dx = 3 - 6y
∫ [1 / (3 - 6y)].dy =∫ dx
log(3 - 6y) = x + k
From boundary conditions:-
log3 = k, thus:-
log(3 - 6y) = x + log3
log(3 - 6y) - log3 = x
log(1 - 2y) = x
1 - 2y = e^(x)
2y = 1 - e^(x)
y = (1/2).[1 - e^(x) ]