Find the general solution to y′′+6y′+13y=0. Give your answer as y=.... In your answer, use c1 and c2 to denote arbitrary constants and x the independent variable. Enter c1 as c1 and c2 as c2.

In order to find the general solution of a homogeneous second order differential equation, we need to solve the characteristic equation. This is basically as easy as solving a quadratic.

For a second order differential equation of type:

[tex]ay''+by'+cy=0[/tex]

Has characteristic equation:

[tex]a r^{2} +br+c=0[/tex]

Whose solutions [tex]r_1 , r_2 ,.., r_n[/tex] are the roots from which the general solution can be formed. There are three cases:

Real roots:

[tex]y(x)=c_1e^{r_1 x} +c_2e^{r_2 x}[/tex]

Repeated roots:

[tex]y(x)=c_1e^{r x} +c_1 xe^{r x}[/tex]

Complex roots:

[tex]y(x)=c_1e^{\lambda x}cos(\mu x) +c_2e^{\lambda x}sin(\mu x)\\\\Where:\\\\r_1_,_2=\lambda \pm \mu i[/tex]

Therefore:

The characteristic equation for:

[tex]y''+6y'+13y=0[/tex]

Is:

[tex]r^{2} +6r+13=0[/tex]

Solving for [tex]r[/tex] :

[tex]r_1_,_2= -3 \pm 2i[/tex]

So:

[tex]\mu = 2\\\\and\\\\\lambda=-3[/tex]

Hence, the general solution of the differential equation will be given by:

[tex]y(x)=c_1e^{-3x} cos(2x)+c_2e^{-3x} sin(2x)[/tex]