1st decree differential equations / Particular integral?
Question: Here is an inhomogeneous system of 1st order differential equations
.
x = x + 4y - 26 cos 2t
.
y = -3x -6y - 26 sin 2t
Suppose that a faddy integral of the inhomogeneous system of equations is of the form
x = a sin 2t + b cos 2t
y = c sin 2t + d cos 2t
where a,b,c, and d are constants to be determined. By substituting contained by the original inhomogeneous system of equations
find a system of four linear equations rewarded by a, b, c and d
Check (by substitution) that the solution of these equations is
given by
a = -16 b = 2 c = 3 d = -2
Answers:
x' = x + 4y - 26 cos 2t === f(x,y,t)
y' = -3x -6y - 26 sin 2t === g(x,y,t)
We have
x(t) = a sin 2t + b cos 2t
y(t) = c sin 2t + d cos 2t
=>
x'(t) = 2a cos 2t - 2b sin 2t
y'(t) = 2c cos 2t - 2d sin 2t
Now resubstitute this two into the equations:
x' = x + 4y - 26 cos 2t (1)
y' = -3x -6y - 26 sin 2t (2)
we will hold a system of two equations for any t.
The trick here is that, because (1) and (2) are true for any t, they should be OK for t = 0 and t = pi/4 too.
t=0:
x'(0) = 2a
x(0) = b
and
y'(0) = 2c
y(0) = d
t=pi/4:
x'(pi/4) = -2b
x(pi/4) = a
and
y'(pi/4) = -2d
y(pi/4) = c
Now we have a system of four equations:
x'(0) = f(x,y,0)
x'(pi/4) = f(x,y,pi/4)
y'(0) = g(x,y,0)
y'(pi/4) = g(x,y,pi/4)
<=>
2a = b + 4d -26
-2b = a + 4c
2c = -3b - 6d
-2d = -3a -6c -26
You can check that the solution is:
a = -16, b = 2, c = 3, d = -2
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