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

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

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

$\implies a=-\dfrac43\implies b=-\dfrac{47}9\implies c=-\dfrac{266}{27}$

Then the ODE has general solution

$y(t)=C_1e^{(-2+\sqrt7)t}+C_2e^{(-2-\sqrt7)t}-\dfrac43t^2-\dfrac{47}9t-\dfrac{266}{27}$