Differential Equations - Ordinary Differential Equations - First Order Linear Equations 1

For the nonhomogeneous case, suppose we have the particular solution y = y p so that y p′ + p(x)y p = g(x). If ya is any other solution of (I.2), then ya′ + p(x)ya = g(x). Setting z = ya − y p , we have z ′ + p(x)z = 0, that is, z is a solution of the homogeneous equation. The previous theorem shows that ya (x)− y p (x) = ce−P(x) ⇔ ya (x) = y p (x)+ce−P(x) .

Theorem I.2. The general solution of the first order linear o.d.e. y ′ + p(x)y = g(x) is y(x) = y p (x) + ce−P(x) ,

where y p (x) is any solution of the o.d.e., c is an arbitrary R constant, and P(x) = p(x) dx . OHP 16

Thus one method for finding the general solution of a first order o.d.e. is to find the solution of the homogeneous equation as well as a particular solution of the nonhomogeneous equation. We could ‘guess’ a particular solution y p .

Example 4.1. Consider a electrical circuit with current I (t) at time t, constant resistance R, constant inductance L, and voltage V (t) at time t.

R

L

V(t)

OHP 17

The current in this circuit satisfies the differential equation dI dI R V (t) L + R I = V (t) or + I = . dt dt L L

(I.4)

This last equation is of the form (I.2) with p(t) = R/L. Setting α = R/L, the general solution of the homogeneous equation is then I (t) = ce−αt . Suppose V (t) = V0 , a constant. We might expect a particular solution I p to be a constant, say A. Substitution of A into the first equation in (I.4) shows that R A = V0 , that is, I p (t) = A = V0 /R. So the solution of the problem is V0 I (t) = + ce−αt . R If we have the initial condition I (0) = 0, then c = −V0 /R. ⊠ OHP 18

Example 4.2. For the electrical circuit of Example 4.1, suppose V (t) = V0 t, that is, a straight line of slope V0 . If we try I p (t) = a + bt, then we require Lb + R(a + bt) = V0 t. This leads to the equations Lb + Ra = 0 and

Rb = V0 .

Solving these two simultaneous equations yields (with α = R/L) V0 b= R

Lb V0 and a = − =− . R αR

Thus the particular solution is   V0 t V0 V0 1 I p (t) = − = t− , R αR R α and so the general solution of the nonhomogeneous equation is   1 V0 t− + ce−αt . I (t) = R α OHP 19

If the current is 0 at t = 0, then −

V0 V0 +c =0⇔c = , Rα Rα

and hence
V0 −αt I (t) = αt − 1 + e . Rα

⊠

OHP 20