Week 9 4. Method of variation of parameters Consider a 2nd-order linear non-homogeneous ODE, (1) If p and q are constants and r = eax, sin ωx, etc., (1) can be solved using the method of undetermined coefficients. The method of variation of parameters enables us to solve (1) for any p, q and r – but only if we know the general solution of the corresponding homogeneous equation, (2)

Assume that you do know two independent solutions of the homogeneous equation, y1,2 – hence, the general solution is Let’s seek the solution of the non-homogeneous equation (1) in the form (3) where u(x) are v(x) undetermined functions. To determine u are v, let’s substitute (3) into ODE (1) and try to satisfy it – but, to do so, we need y'(x) and y''(x), so...

Let’s calculate and require that (4) Hence, Next, calculate Substituting the expressions for y' and y'' into the original ODE (1), we obtain...

y" y' y = 0 = 0 Rearrange this equation in the form Recall that y1,2(x) satisfy the homogeneous equation – hence, the above equation reduces to (5) Let’s write equations (4) and (5) next to each other...

Since y1,2 are assumed to be known, these are algebraic equations for u' and v'. To solve these equations, express u' from the first equation, (6) and substitute into the second equation, which yields hence...

(7) hence, (8) Substituting (7) into (6), we find u' and then obtain (9) Summarising (3) and (8)-(9), we obtain...

u v where ۞ W is called the Wronskian. Comment: A summary of the MoVoP will be included at the back of the exam paper.

Example 1: Solve the following ODE: Hint: The soln: Step 1: Find y1,2, i.e. solve the corresponding homogeneous equation: hence, hence...

cot x Step 2: Calculate the Wronskian, Step 3: Calculate u and v:

hence, = 1 – sin2 x hence, hence, using the hint provided and your knowledge of the basics,

The answer: Warning: Don’t lose C1,2, otherwise you’ll find a particular solution instead of the general one.

yp yh Comment: Rearrange the answer for Example 1 in the form In terms of Theorem 8.2, the two terms in this expression are yp and yh.

5. Reduction of order In some cases, one can solve a 2nd-order ODE by reducing its order. ۞ In an ODE with the standard notation, x and y are called the independent and dependent variables, respectively. Example 2: ODEs with the dependent variable missing Solve the following ODE: (10) Soln: Look at equation (10)... anything catches your eye? Observe that the dependent variable y doesn’t appear in this ODE (only y' and y")...

In such cases, change the dependent variable y → z, where z is such that Then it follows that and equation (10) becomes This is a linear 1st-order ODE for z(x), the solution is To return to the original dependent variable, recall that z = y'...

hence, the answer is Example 3: ODEs with the independent variable missing Solve the following ODE: Soln: Observe that the independent variable x doesn’t appear in this ODE...

x y y z = y' Let’s change both variables : variable old new independent x y dependent y z = y' Warning: Since we are going to change x (i.e. the variable with respect to which we differentiate), it’s a good idea to rewrite the original ODE in the form (11)

We’ll also need an expression to substitute for y" – hence, consider z hence, hence, using the chain rule, z hence,

y" Now, equation (11) becomes hence, This is a linear 1st-order ODE for z(y), the solution is Now, go back to the original variables...

This is a separable ODE, the solution is Comment: In Examples 2 and 3, reduction of order resulted in linear ODEs. Generally, this isn’t necessarily the case.

Reduction of order (summary) (1) ODEs with y missing The change of variables: The derivatives: The inverse change: once you’ve found z = f(x), write

(2) ODEs with x missing The change of variables: The derivatives: The inverse change: once you’ve found z = f(y), write and solve by separation of variables.