1

$y' - 4y = 6e^{6t}$

$y(0) = -2$

I can't seem to get the right answer. I believe it is just my arithmetic.

@T.Bongers
u(t) = e^(-4t)

-y*e^(-4t) = int e^(-4t) * 6e^(6t)

= 3e^(2t) + c

y = - 3 e^(6t) + c

-2 = -3 e^(6(0)) + c

c = 1

Chris
  • 25

1 Answers1

1

We are given:

$$\tag 1 y' - 4y = 6e^{6t}, ~y(0) = -2$$

Our integrating factor is:

$$\mu(t) = e^{\int -4~dt} = e^{-4t}$$

Multiplying this I.F. with $(1)$, yields:

  • $e^{-4t}y' -(4e^{-4t})y = 6 e^{2t}$, so
  • $e^{-4t}y' + \dfrac{d}{dt}(e^{-4t})y = 6 e^{2t}$, so
  • $\dfrac{d}{dt}(e^{-4t}y) = 6 e^{2t}$, so we integrate each side and get
  • $\displaystyle \int \dfrac{d}{dt}(e^{-4t}y)~dt = 6 \int e^{2t}~ dt$, or
  • $e^{-4t} y = 3e^{2t} + c$, so

$$y(t) = e^{4t}(3 e^{2t}+c)$$

Now, using our initial condition to solve for $c$, we get:

$$y(0) = 3 + c = -2 \rightarrow c = -5$$

Our final solution is:

$$y(t) = e^{4t}(3 e^{2t} - 5)$$

Amzoti
  • 56,093