Firstly we find the auxiliary equation and its roots. Write down the general solution. Substitute the initial conditions and work out the arbitrary constants.

What is an auxiliary equation?

Letting and substituting this in the problem gives . As e^{mx} cannot be 0, we see that is a solution of the given differential equation if .

is the Auxiliary Equation.

The following table shows the general solution of the differential equation for different values of the discriminant.

Roots of

Auxiliary Equation

2 real and distinct roots m, n

2 real and equal roots

m

complex roots p + iq, p - iq

p = real part &

q = imaginary part

General solution

of ODE

For example:

Find the solution of with initial conditions y(0) = 1 and y'(0) = 0.

Step 1: First we find the auxiliary equation

Step 2: The roots of this equation are -1, -3.

Step 3: Hence the general solution is .

Step 4:Substituting the initial conditions in the general solution gives A + B = 1 and -A - 3B = 0. Solving these equations gives and .

First we solve the homogeneous problem. Then we find a particular solution for the inhomogeneous part. Substitute the initial conditions and work out the arbitrary constants.

How do we find the particular solution?

If f(x) is a polynomial then we try a solution of the same degree, e.g. if f(x) = 2x + 3 then try a particular solution of the form y = ax + b. Similarly if we have exponential or trigonometric terms as f(x) we try e^{ax} or asin(x) + bcos(x).

For example:

How do we solvewith initial conditions y(0) = 1, y'(0) = 0?

Step1:First we find the auxiliary equation .

Step2: The roots of this equation are -1, -3.

Step3:Hence the solution to the homogeneous problem is .

Step4: Now try a solution in the form y = ax + b. , . Subtituting these in the differential equation gives

Step5: Comparing the coefficients of x and the constant term gives . Hence the general solution is .

Step6:Use the initial conditions to find A and B. The solution is .