Week 4 4. Linear first-order ODEs ۞ The general form of a linear 1st-order ODE is (1) where p(x) and p(x) are given functions. ۞ If r(x) = 0, the ODE is called homogeneous. If r(x) ≠ 0, the ODE is called non-homogeneous. Linear homogeneous ODEs of the 1st order are separable:
hence, hence, where C = ± exp (– C1). To solve the non-homogeneous ODE (1), introduce the so-called “integrating factor”, (3)
Observe that hence, (2) Now, multiply Eq. (1) by I(x), and use (2) to obtain hence, hence...
...hence, integrating and dividing by I(x), obtain (4) This is the general solution of the original linear non-homogeneous ODE, i.e. Eq. (1). Q: The solution of any 1st-order ODE should involve one arbitrary constant. Why doesn’t solution (4) involve one? Q: Recall that the integrating factor involves indefinite integration too [see formula (3)], which seems to imply that (4) involves an extra arbitrary constant... or does it?
Linear 1st-order ODEs (summary) 1. A homogeneous linear ODE has the form and it’s solution is 2. A non-homogeneous linear ODE has the form and it’s solution is where
Example 1: Solve the initial-value problem, (5) (6) Solution: To identify p(x) and r(x), rearrange (5) into the standard form: hence, Step 1: Find the integrating factor...
hence, integrating and omitting the constant of integration, (7) hence, (8) Observe the omitted modulus in Eq. (7). In this particular case, the omission does not give rise to complex numbers. You can also keep the modulus (will get the same answer – more details to follow).
1/I(x) I(x) r(x) Step 2: Find the solution (9) hence, hence, using the initial condition (6), Hence, the answer is
Remarks: Observe that the logarithm disappears when we transform (7) into (8) – that’s why the omission of the modulus in the former doesn’t give rise to complex numbers. Look at formula (9) and explain why the answer doesn’t depend on whether or not the modulus is kept in formula (7). Keep the constant of integration in formula (7) and show that it doesn’t affect the answer.