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#### Differential Equations

• 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

###### Examples
1. 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
• (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:
• From y: 3 = A + B
• From dy/dt: 3A + 2B = -1
• Solve the simultaneous equations to find A and B
• A = -7 and B = 10
• Thus, our particular solution is:
• y = -7e-3t + 10e-2t + sin 3t
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