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Unformatted text preview: Linear Differential Equations A firstorder linear differential equation is one that can be put into the form where and are continuous functions on a given interval. This type of equation occurs frequently in various sciences, as we will see. An example of a linear equation is because, for , it can be written in the form Notice that this differential equation is not separable because it’s impossible to factor the expression for as a function of x times a function of y . But we can still solve the equa tion by noticing, by the Product Rule, that and so we can rewrite the equation as If we now integrate both sides of this equation, we get or If we had been given the differential equation in the form of Equation 2, we would have had to take the preliminary step of multiplying each side of the equation by x . It turns out that every firstorder linear differential equation can be solved in a similar fashion by multiplying both sides of Equation 1 by a suitable function called an integrating factor. We try to find so that the left side of Equation 1, when multiplied by , becomes the derivative of the product : If we can find such a function , then Equation 1 becomes Integrating both sides, we would have so the solution would be To find such an , we expand Equation 3 and cancel terms: I x P x I x I x y I x P x y I x y I x y I x y I y x 1 I x y I x Q x dx C 4 I x y y I x Q x dx C I x y I x Q x I I x y P x y I x y 3 I x y I x I I x y x C x xy x 2 C xy 2 x xy y xy y y 1 x y 2 2 x xy y 2 x Q P dy dx P x y Q x 1 1 This is a separable differential equation for , which we solve as follows: where . We are looking for a particular integrating factor, not the most general one, so we take A 1 and use Thus, a formula for the general solution to Equation 1 is provided by Equation 4, where is given by Equation 5. Instead of memorizing this formula, however, we just remember the form of the integrating factor. To solve the linear differential equation , multiply both sides by the integrating factor and integrate both sides. EXAMPLE 1 Solve the differential equation . SOLUTION The given equation is linear since it has the form of Equation 1 with and . An integrating factor is Multiplying both sides of the differential equation by , we get or Integrating both sides, we have EXAMPLE 2 Find the solution of the initialvalue problem SOLUTION We must first divide both sides by the coefficient of to put the differential equation into standard form: The integrating factor is I x e x 1 x dx e ln x x x y 1 x y 1 x 2 6 y y 1 2 x x 2 y xy 1 y 2 Ce x 3 e x 3 y y 6 x 2 e x 3 dx 2 e x 3 C d dx e x 3 y 6 x 2 e x 3 e x 3 dy dx 3 x 2 e x 3 y 6 x 2 e x 3 e x 3 I x e x 3 x 2 dx e x 3 Q x 6 x 2 P x 3 x 2 dy dx 3 x 2 y 6 x 2 I x e x P x dx y P x y Q x I I x e x P x dx 5 A e C I Ae x P x dx ln I y P x dx y dI I y P x dx I 2 ■ L I N E A R D I F F E R E N T I A L E Q U AT I O N S FIGURE 1 6 _3 _1.5 1.8 C=2 C=1 C=_2 C=_1 C=0 ■ ■ Figure 1 shows the graphs of several mem bers of the family of solutions in Example 1.bers of the family of solutions in Example 1....
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 Spring '09
 quach
 Differential Equations, Calculus, Equations, Derivative, Linear Differential Equations

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