# Question #2ef29

Feb 12, 2017

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

#### Explanation:

This is homogeneous so we only need the complementary solution.

The most common approach which is to go straight to the "characteristic equation", which here is:

${\lambda}^{2} + 4 \lambda + 5 = 0$

And which has complex roots:

${\lambda}_{1 , 2} = \frac{- 4 \pm \sqrt{{4}^{2} - 4 \left(5\right)}}{2} = - 2 \pm i$

The solution to this equation is therefore:

$y = A {e}^{\left(- 2 + i\right) x} + B {e}^{\left(- 2 - i\right) x}$

$= {e}^{- 2 x} \left(A {e}^{i x} + B {e}^{- i} x\right)$

Using Euler's Formula, ${e}^{i \theta} = \cos \theta + i \sin \theta$, we can expand this to:

$= {e}^{- 2 x} \left(A \left(\cos x + i \sin x\right) + B \left(\cos x - i \sin x\right)\right)$

$= {e}^{- 2 x} \left(\left(A + B\right) \cos x + i \left(A - B\right) \sin x\right)$

And then simplify with new integration constants:

$= {e}^{- 2 x} \left(C \cos x + D \sin x\right)$

We can now apply the IV's:

$y \left(0\right) = 2 \implies 2 = \left(1\right) \left(C + D \left(0\right)\right) \implies C = 2$

We then have:

$y = {e}^{- 2 x} \left(2 \cos x + D \sin x\right)$

So by the product rule:

$y ' = - 2 {e}^{- 2 x} \left(2 \cos x + D \sin x\right) + {e}^{- 2 x} \left(- 2 \sin x + D \cos x\right)$

$= {e}^{- 2 x} \left(\left(- 4 + D\right) \cos x - \left(2 D + 2\right) \sin x\right)$

Applying the other IV:

$y ' \left(0\right) = - 7 \implies - 7 = \left(1\right) \left(\left(- 4 + D\right) - \left(2 D + 2\right) \left(0\right)\right)$

$\implies D = - 3$

$\implies y = {e}^{- 2 x} \left(2 \cos x - 3 \sin x\right)$