Question

Attached as a photo.

Answer 1

Applying the initial conditions, the specific solution is:

$y = e^{-2x}(-3\sin{3x} + 4\cos{3x})$

What is the general solution?

The general solution is given by:

$y = e^{-2x}(C_1\sin{3x} + C_2\cos{3x})$

How to find the specific solution?

We apply the initial conditions to find the specific solution.

First, we have that y(0) = 4, then:

$4 = e^{-2(0)}(C_1\sin{3(0)} + C_2\cos{3(0)})$

Since cos(0) = 1, we have that:

$C_2 = 4$

Then:

$y = e^{-2x}(C_1\sin{3x} + 4\cos{3x})$

The derivative is:

$y^\prime(x) = -2e^{-2x}(C_1\sin{3x} + 4\cos{3x}) + e^{-2x}(3C_1\cos{3x} - 4\sin{3x})$

Since y'(0) = -17, then:

$-17 = -8 + 3C_1$

$3C_1 = -9$

$C_1 = -3$

Then the specific solution is:

$y = e^{-2x}(-3\sin{3x} + 4\cos{3x})$

More can be learned about differential equations at brainly.com/question/8427134

#SPJ1