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

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.

Answer 2

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=\frac12,b=\frac32,c=\frac{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)+\frac{t^2}2+\frac{3t}2+\frac{11}4`