Question

Find a solution x = x(t) of the equation x′ + 2x = t2 + 4t + 7 in the form of a quadratic function of t, that is, of the form x(t) = at2 + bt + c, where a, b, and c are to be determined.

Answer 1

The particular quadratic solution to the ODE is found as follows:

$x=at^2+bt+c$
$x'=2at+b$

$(2at+b)+2(at^2+bt+c)=t^2+4t+7$
$2at^2+(2a+2b)t+(b+2c)=t^2+4t+7$

$\begin{cases}2a=1\\2(a+b)=4\\b+2c=7\end{cases}\implies a=\dfrac12,b=\dfrac32,c=\dfrac{11}4$

Note that there's also the fundamental solution to account for, which is obtained from the characteristic equation for the ODE:

$x'+2x=0\implies r+2=0\implies r=-2$

so that $x_c=Ce^{-2t}$ is a characteristic solution to the ODE, and the general solution would be

$x=Ce^{-2t}+\dfrac{t^2}2+\dfrac{3t}2+\dfrac{11}4$