To find a, b, c for the solution:
Let's start by writing down the expression for the function x(t) and its derivative:
We have:
x(t) = at² + bt + c
and
x'(t) = 2at + b
Using x' and x into the differential equation x′ + 2x = t² + 4t + 7 gives us:
2at + b + 2*(at² + bt + c) = t² + 4t + 7
Expanding this gives:
2at² + 2bt + b + 4at + 2c = t² + 4t + 7
By equating the coefficients of equivalent powers of t on both sides, we get three equations:
For t² :
2a = 1
So, a = 1/2
For t:
2b + 4a = 4
Substitute a = 1/2 into the equation gives b = 1 - 2 = -1
For the constant term:
b + 2c = 7
Substituting b = -1 gives c = 4.
So the solution is a = 1/2, b = -1, c = 4.
So the specific solution of this differential equation is given by x(t) = (1/2)t² - t + 4.