#### 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:
- a
^{d²y}/_{dx²}+ b^{dy}/_{dx}+ cy = 0

- a
- 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:
- a
^{d²y}/_{dx²}+ b^{dy}/_{dx}+ cy = f(x)

- a
- 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 $(10
^{8}y + 10^{9}). The variables y and t are related by the differential equation:

^{d²y}/_{dt²}+ 5^{dy}/_{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}+ 5^{dy}/_{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}

- y = Ae
- 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

- y = Ae
- 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

- y = -7e

- Form the auxiliary equation by assuming f(x) = 0: