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A first order linear equation has the following form.

(20) [Graphics:2txgr137.gif]

Note that [Graphics:2txgr138.gif] is a function of x alone. If [Graphics:2txgr139.gif], then the equation is homogeneous. If we rearange (20) as follows

(21) [Graphics:2txgr140.gif]

Comparing this form to (1) shows that the first order linear is very similar to the exact type differential equation. From this we see that

(22) [Graphics:2txgr141.gif]


(23) [Graphics:2txgr142.gif]

Applying the partial derivative test to this gives [Graphics:2txgr143.gif] and [Graphics:2txgr144.gif], as verified below. Unless [Graphics:2txgr145.gif] the linear first order will not pass the exactness test. This suggests we try the integrating factor method suggested above. Equation (18) is what we need. It is reproduced below.


Now looking at equation (18),

(24) [Graphics:2txgr148.gif]

we see that [Graphics:2txgr149.gif], [Graphics:2txgr150.gif], and [Graphics:2txgr151.gif]. This gives the following result for the integrating factor for the linear first-order equation.

(25) [Graphics:2txgr152.gif]

Multiplying (21) by this results gives the following:

(26) [Graphics:2txgr153.gif]

Notice that the left side is [Graphics:2txgr154.gif] so if we rewrite (26) and then integrate with respect to x we get

(27) [Graphics:2txgr155.gif].

The left side of this result is integrated by inspection to give

(28) [Graphics:2txgr156.gif]

or solving for y we have the following:

(29) [Graphics:2txgr157.gif]

Once a differential equation has been analyzed as first-order linear, then all that must be done is to identify the [Graphics:2txgr158.gif] and the [Graphics:2txgr159.gif] then substitute into (29) and do the integration. This result also constitutes a proof that a solution exists for the first order linear differential equation.Let us look at an example.

Example 3: [Graphics:2txgr160.gif]

In this case we also give an initial condition, that is [Graphics:2txgr161.gif]. First we rearange the equation to the form recognizable as first-order linear.


From this we see that [Graphics:2txgr163.gif] and [Graphics:2txgr164.gif]. We will try (29) on this problem. Because of similar names for several of the variables, I will turn off Mathematica's spell checker.



Next we must satisfy the initial condition in order to evaluate the constant of integration, CC3.



We will plot our solution near the initial condition point (1,1).




For comparison, we will look at the DSolve results, which is the same as above.



EXERCISES: Solve and plot when initial conditions are given.

1 [Graphics:2txgr178.gif] 2 [Graphics:2txgr179.gif] 3 [Graphics:2txgr180.gif] 4 [Graphics:2txgr181.gif]


1 [Graphics:2txgr182.gif]





                [Graphics:2txgr189.gif] 2` [Graphics:2txgr190.gif]


3 [Graphics:2txgr194.gif]





                [Graphics:2txgr201.gif] 4 [Graphics:2txgr202.gif]






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