Question

[10 pts] b) y"+12y'+36 y = 0

Answer 1

The general solution of the differential equation is:

`y=c_1e^(-6t)+c_2te^(-6t)`

Explanation:

We have a second order homogeneous differential equation
`y''+12y'+36 y = 0`

We need to find the characteristic polynomial

`x^2+12x-36=0`

Next, we find the roots as follows:

`\mathrm{Solve\:by\:factoring}\\\\\mathrm{Rewrite\:}x^2+12x+36\mathrm{\:as\:}x^2+2x\cdot \:6+6^2\\\\\mathrm{Apply\:Perfect\:Square\:Formula}:\quad \left(a+b\right)^2=a^2+2ab+b^2\\\\\left(x+6\right)^2=0\\\\\mathrm{Solve\:}\:x+6=0:\quad x=-6`

The roots of characteristic polynomial are
`r=-6` and
`s=-6`

When the roots are real and equal the general solution of the differential equation is:

`y=c_1e^(-6t)+c_2te^(-6t)`