First Order Differential Equations

Transcription

1 First Order Differential Equations Linear Equations Philippe B. Laval KSU Philippe B. Laval (KSU) 1st Order Linear Equations 1 / 11

2 Introduction We are still looking at 1st order equations. In today s lecture we will discuss linear equations that is equations which can be expressed in the form a 1 (x) dy + a 0 (x) y = b (x) (1) Note that for this to be a first order equation, we must have a 1 (x) 0. In these slides, we assume that the dependent variable is y and the independent variable is x. When we say y, we really mean "the dependent variable" and when we say x we really mean "the independent variable". The reader needs to be able to adapt this if different variables are used in an equation. Philippe B. Laval (KSU) 1st Order Linear Equations 2 / 11

3 Examples Example Determine if the differential equations below are linear or not and explain why. 1 x 2 sin x (cos x) y = sin x dy 2 sin x dy + ex y = cos x 3 sin x dy + ex y = cos y 4 y dy + (sin x) y 3 = e x Philippe B. Laval (KSU) 1st Order Linear Equations 3 / 11

4 Case when the Coeffi cient of y is 0 Equation 1 reduces to a 1 (x) dy = b (x). Philippe B. Laval (KSU) 1st Order Linear Equations 4 / 11

5 Case when the Coeffi cient of y is 0 Equation 1 reduces to a 1 (x) dy = b (x). The solution is b (x) y (x) = + C (2) a 1 (x) Philippe B. Laval (KSU) 1st Order Linear Equations 4 / 11

6 Case when the Coeffi cient of y is 0 Equation 1 reduces to a 1 (x) dy = b (x). The solution is b (x) y (x) = + C (2) a 1 (x) Solve 1 x dy ex = 0 Philippe B. Laval (KSU) 1st Order Linear Equations 4 / 11

7 Case when the Coeffi cient of y is the Derivative of the coeffi cient of y 1 Equation 1 reduces to a 1 (x) dy + a 1 (x) y = b (x). Philippe B. Laval (KSU) 1st Order Linear Equations 5 / 11

8 Case when the Coeffi cient of y is the Derivative of the coeffi cient of y 1 Equation 1 reduces to a 1 (x) dy + a 1 (x) y = b (x). 2 The solution is y (x) = 1 [ ] b (x) + C a 1 (x) (3) Philippe B. Laval (KSU) 1st Order Linear Equations 5 / 11

9 Case when the Coeffi cient of y is the Derivative of the coeffi cient of y 1 Equation 1 reduces to a 1 (x) dy + a 1 (x) y = b (x). 2 The solution is y (x) = 1 [ ] b (x) + C a 1 (x) (3) 3 Solve x 2 dy + 2xy = ex Philippe B. Laval (KSU) 1st Order Linear Equations 5 / 11

10 General Case Write our equation in standard form that is dy + P (x) y = Q (x) (4) Next, we try to rewrite equation 4 in such a way that its left-hand side is a derivative. For this, we multiply both sides by a quantity we call µ (x). We obtain µ (x) dy + µ (x) P (x) y = µ (x) Q (x). We want to find µ (x) such that the left-hand side is the derivative of µ (x) y that is µ (x) dy d (µ (x) y) + µ (x) P (x) y = dµ (x) = y + µ (x) dy So, we see that we must have µ (x) P (x) = dµ Philippe B. Laval (KSU) 1st Order Linear Equations 6 / 11

11 General Case This is a separable equation which we can write as dµ = P (x). µ Integrating on each side gives µ (x) = e P(x). µ (x) is called the integrating factor. With this choice for µ, our equation becomes d (µ (x) y (x)) = µ (x) Q (x) We can now easily solve this. The solution is y (x) = 1 [ ] µ (x) Q (x) + C with µ (x) = P(x) e (5) µ (x) Philippe B. Laval (KSU) 1st Order Linear Equations 7 / 11

12 General Case Theorem To solve a linear differential equation of the first order, follow the steps below: 1 Write the equation in the form dy + P (x) y = Q (x). 2 Find the integrating factor µ (x) = e P(x) 3 Multiply the equation in standard form by µ (x) to obtain d (µ (x) y (x)) = µ (x) Q (x) 4 Integrate both sides and solve for y to obtain ] µ (x) Q (x) + C y (x) = 1 [ µ (x) This is called the general solution to equation 4. with µ (x) = e P(x) Philippe B. Laval (KSU) 1st Order Linear Equations 8 / 11

13 Examples Example Find the general solution of 1 x Example dy 2y x 2 = x cos x For the initial value problem y + y = 1 + cos 2 x, y (1) = 4, find y (2). Note: You will need to approximate the integral you obtain as it is hard to evaluate. Example A rock contains 2 isotopes, RA1 and RA2, that belong to the same radioactive series; that is RA1 decays to RA2 which then decays into stable atoms. Assume the rate at which RA1 decays into RA2 is 50e 10t kg/s and the rate at which RA2 decays is proportional to y, the mass of RA2 and is given by 2y. Find y = y (t) for t 0 if y (0) = 40kg. Philippe B. Laval (KSU) 1st Order Linear Equations 9 / 11

14 Existence and Uniqueness Theorem Revisited The existence and uniqueness theorem we studied in the previous chapter applies here since it was for first order differential equations. For first order linear differential equations, we can state a slightly stronger result. Theorem Suppose that P (x) and Q (x) are continuous on an interval (a, b) that contains the point x 0. Then, for any choice of initial value y 0, there exists a unique solution y (x) on (a, b) to the initial value problem dy + P (x) y = Q (x), y (x 0) = y 0 In fact the solution is given by equation 5 for a suitable value of C. Philippe B. Laval (KSU) 1st Order Linear Equations 10 / 11

15 Assignment Do the problems at the end of section 2.2 in my notes on linear equations. Philippe B. Laval (KSU) 1st Order Linear Equations 11 / 11