Question

Consider the initial value problem y' + 3y = 9t, y(0) = 7.

Answer 1

Answer:

$y(t) = 3x+8e^{-3x} -1$

Step-by-step explanation:

Recall that the following laplace transforms

$L(y') = sY(s)-y(0)$

$L(t) = \frac{1}{s^2}$

The laplace transform is linear, so, applying the laplace transform to the equation we get

$L(y'+3y) = sY(s)-7+3Y(s) = L(9t) = \frac{9}{s^2}$

By some algebraic manipulations, we get

$Y(s)(s+3) = \frac{9+7s^2}{s^2}$

which is equivalent to

$Y(s) = \frac{9+7s^2}{s^2(s+3)} = \frac{9}{s^2(s+3)}+\frac{7}{s+3}$

By using the partial fraction decomposition, we get

$\frac{9}{s^2(s+3)} = \frac{-1}{s} + \frac{3}{s^2} + \frac{1}{s+3}$

then

$Y(s) = \frac{-1}{s} + \frac{3}{s^2} + \frac{1}{s+3} + \frac{7}{s+3} = \frac{8}{s+3} + \frac{3}{s^2}-\frac{1}{s}$

Using that

$L(e^{-ax}) = \frac{1}{s+a}$

$L(1) = \frac{1}{s}$

by taking the inverse on both sides we get

$y(t) = L^{-1}(\frac{8}{s+3})+L^{-1}(\frac{3}{s^2})+L^{-1}(-\frac{1}{s}) = 8e^{-3x} + 3x-1$