# Thread: Find the solution in the form of a quadratic function of t, where a,b, and c tbd

1. ## Find the solution in the form of a quadratic function of t, where a,b, and c tbd

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.

At first I thought you would solve this by finding the integrating factor, but after speaking with the teacher he said to plug in the form x(t)=at2+bt+c into the Diff. Eq. and use the resulting expression to solve for a,b, and c by comparing the expression on the left side to the expression on the right side.

So now I get this, but I am not sure what to do from here.

x'(t)=2t+4
2t2+10t+18 = t2+4t+7

2. ## Re: Find the solution in the form of a quadratic function of t, where a,b, and c tbd

$x(t)=at^2+bt+c$
$x'(t)=2at+b$

Plugging in:

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

Equate terms:
$2at^2=t^2$
$2 (a+b)t=4t$
$b+2c=7$

This gives:
$a=\dfrac {1}{2},b=\dfrac {3}{2}, c=\dfrac {11}{4}$

thank you.