Question

Find the particular solution of the initial value problem
d²y/dx² + 4 dy/dx + 13y = 0
y(0)=1
y¹(0)=4

Answer 1

$\frac{d^2 y}{dx^2} + 4 \frac{dy}{dx} + 13y = 0; y(0)=1, y'(0)=4$

The given equation can be written as

$(D^2 + 4D + 13)y = 0; y(0)=1, y'(0)=4.$

The auxiliary equation is,

$m^2 + 4m + 13 = 0$

Solving for m,

$m = \frac{-4 \pm \sqrt{16 - 4*13}}{2}\\ ~~~~~=\frac{-4 \pm \sqrt{-36}}{2} \\~~~~~ = -2 \pm 3i.\\$

The solution is,

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

where A and B are constants.

To find A and B, we use the given initial conditions

y(0)=1, y'(0)=4.

Now,

$y'(x) = e^{-2x}(-3A \sin 3x + 3B \cos 3x)\\~~~~~~~~~~~~~~~~~ - 2e^-2x(A \cos 3x + B \sin 3x)\\ ~~~~~= e^{-2x}(-3A \sin 3x + 3B \cos 3x) - 2y(x).$

Using the condition

y(0) = 1, in y(x) we get A = 1.

Using the condition

y(0) = 1, y'(0) = 4 and A = 1 in y'(x), we get

y'(0) = (0 + 3B) - 2y(0)

4 = 3B - 2

B = 2.

Thus, the solution is

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