Question

Use undetermined coefficients to find the particular solution to y'' + 4y' – 3y = 4t² + 5t + 6 yp(t) = [ Preview

Answer 1

The corresponding homogeneous ODE is

`y''+4y'-3y=0`

with characteristic equation

`r^2+4r-3=0`

which has roots at
`r=-2\pm\sqrt7`, which gives two characteristic solutions

`y_c=C_1e^((-2+\sqrt7)t)+C_2e^((-2-\sqrt7)t)`

For the particular solution, let

`y_p=at^2+bt+c`

with derivatives

`{y_p}'=2at+b`

`{y_p}''=2a`

Substituting into the ODE gives

`2a+4(2at+b)-3(at^2+bt+c)=4t^2+5t+6`

`-3at^2+(8a-3b)t+(2a+4b-3c)=4t^2+5t+6`

`\implies a=-\frac43\implies b=-\frac{47}9\implies c=-(266)/(27)`

Then the ODE has general solution

`y(t)=C_1e^((-2+\sqrt7)t)+C_2e^((-2-\sqrt7)t)-\frac43t^2-\frac{47}9t-(266)/(27)`