The type of differential we solve are linear, second order, constant coefficient differential equations.
There are two types:
Homogeneous
Non-homogeneous
Homogeneous Equations
This type of differential equations will be of the form:
ad²y/dx² + bdy/dx + cy = 0
First step to solving them is to find the complementary function by forming an auxiliary equation:
aλ² + bλ + c = 0
The complementary function formed depends on the discriminant of the auxiliary function
Solve the quadratic equation and use the following cases to work out the complementary function
The complementary function is also the general solution and we need to substitute values, find A and B to get the particular solution
Non-homogeneous Equations
This type of differential equations will be of the form:
ad²y/dx² + bdy/dx + cy = f(x)
We find a complementary function in the same way as above, by making f(x) = 0
After finding a complementary function, we have to add a particular integral
The particular integral we add is dependent upon f(x) and the trial solution used:
Once we have the trial solution, we equate it to y and differentiate it to find y’ and differentiate again to find y” in terms of P, Q , and R
Then substitute these into the initial differential equation and solve simultaneously for P, Q, and R
The trial solution with substituted values of P, Q, and R becomes the particular integral
Addition of complementary function and particular integral gives use the particular solution
Method Summary
Examples
The value of the assets of a large commercial organization at time t, measured in years, is $(108y + 109). The variables y and t are related by the differential equation: d²y/dt² + 5dy/dt + 6y = 15 cos 3t – 3 sin 3t
Find y in terms of t, given that y = 3 and dy/dt = 2 when t = 0
Form the auxiliary equation by assuming f(x) = 0:
d²y/dt² + 5dy/dt + 6y = 0
∴ λ² + 5λ + 6 = 0
Solve the quadratic equation:
(λ + 3)(λ + 2) = 0
∴ λ = 3 or 2
The answers are two real solutions, so the complementary function is the following:
y = Ae-3t + Be-2t
Now we have to find the particular integral using a trial solution:
f(x) = 15 cost 3t – 3 sin 3t
Thus, our trial solution and its derivatives are:
y = P cos 3t + Q sin 3t
y’ = -3P sin 3t + 3Q cos 3t
y” = -9P cos 3t – 9Q sin 3t
Substitute into the original equation and simplify to get:
(-3P + 15Q) cos 3t + (-3Q + 15P) sin 3t
Take the coefficient from both sides to form the simultaneous equations:
-3P + 15Q = 15
-3Q + 15P = -3
Solve these to get P and Q
P = 0 ad Q = 1
Thus, the particular integral is y = sin 3t
Our general solution is the addition of the complementary function and the articular integral:
y = Ae-3t + Be-2t + sin 3t
First step to finding the particular solution is to differentiate the general solution:
dy/dt = -3Ae-3t – 2Be-2t + cos 3t
Substitute the given information into the equations above to form the simultaneous equations:
