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DIFFERENTIAL EQUATIONS

9. DIFFERENTIAL EQUATIONS. DIFFERENTIAL EQUATIONS. 9.5 Linear Equations. In this section, we will learn: How to solve linear equations using an integrating factor. LINEAR EQUATIONS. Equation 1. A first-order linear differential equation is one that can be put into the form

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DIFFERENTIAL EQUATIONS

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  1. 9 DIFFERENTIAL EQUATIONS

  2. DIFFERENTIAL EQUATIONS 9.5Linear Equations • In this section, we will learn: • How to solve linear equations • using an integrating factor.

  3. LINEAR EQUATIONS Equation 1 • A first-order linear differential equation is one that can be put into the form • where P and Q are continuous functions on a given interval. • This type of equation occurs frequently in various sciences, as we will see.

  4. LINEAR EQUATIONS Equation 2 • An example of a linear equation is • xy’ + y = 2x • because, for x≠ 0, it can be written in the form

  5. LINEAR EQUATIONS • Notice that this differential equation is not separable. • It’s impossible to factor the expression for y’as a function of x times a function of y.

  6. LINEAR EQUATIONS • However, we can still solve the equation by noticing, by the Product Rule, that xy’ + y = (xy)’ • So, we can rewrite the equation as: (xy)’ = 2x

  7. LINEAR EQUATIONS • If we now integrate both sides, we get: xy = x2 + C or y = x + C/x • If the differential equation had been in the form of Equation 2, we would have had to initially multiply each side of the equation by x.

  8. INTEGRATING FACTOR • It turns out that every first-order linear differential equation can be solved in a similar fashion by multiplying both sides of Equation 1 by a suitable function I(x). • This is called an integrating factor.

  9. LINEAR EQUATIONS Equation 3 • We try to find I so that the left side of Equation 1, when multiplied by I(x), becomes the derivative of the product I(x)y: I(x)(y’ + P(x)y) = (I(x)y)’

  10. LINEAR EQUATIONS • If we can find such a function I, then Equation 1 becomes: • (I(x)y)’ = I(x)Q(x) • Integrating both sides, we would have: • I(x)y = ∫I(x)Q(x) dx + C

  11. LINEAR EQUATIONS Equation 4 • So, the solution would be:

  12. LINEAR EQUATIONS • To find such an I, we expand Equation 3 and cancel terms:I(x)y’ + I(x)P(x)y = (I(x)y)’ = I’(x)y + I(x)y’ • I(x)P(x) = I’(x)

  13. SEPERABLE DIFFERENTIAL EQUATIONS • This is a separable differential equation for I, which we solve as follows: • where A = ±eC.

  14. LINEAR EQUATIONS Equation 5 • We are looking for a particular integrating factor—not the most general one. • So, we take A = 1 and use:

  15. LINEAR EQUATIONS • Thus, a formula for the general solution to Equation 1 is provided by Equation 4, where I is given by Equation 5. • However, instead of memorizing the formula, we just remember the form of the integrating factor—as follows.

  16. LINEAR EQUATIONS • To solve the linear differential equation • y’ + P(x)y = Q(x) • multiply both sides by the integrating factor • and integrate both sides.

  17. LINEAR EQUATIONS Example 1 • Solve the differential equation • The given equation is linear as it has the form of Equation 1: P(x) = 3x2 and Q(x) = 6x2 • An integrating factor is:

  18. LINEAR EQUATIONS Example 1 • Multiplying both sides of the equation by ex3, we get or

  19. LINEAR EQUATIONS Example 1 • Integrating both sides, we have:

  20. LINEAR EQUATIONS Example 1 • The figure shows the graphs of several members of the family of solutions in Example 1. • Notice that they all approach 2 as x → ∞.

  21. LINEAR EQUATIONS Example 2 • Find the solution of the initial-value problem x2y’ + xy = 1 x > 0 y(1) = 2

  22. LINEAR EQUATIONS E. g. 2—Equation 6 • First, we must divide both sides by the coefficient of y’ to put the differential equation into standard form:

  23. LINEAR EQUATIONS Example 2 • The integrating factor is: • I(x) = e ∫ (1/x) dx = eln x = x • Multiplication of Equation 6 by x gives:

  24. LINEAR EQUATIONS Example 2 • Hence, • Thus, • Since y(1) = 2, we have:

  25. LINEAR EQUATIONS Example 2 • Thus, the solution to the initial-value problem is:

  26. LINEAR EQUATIONS Example 3 • Solve y’ +2xy = 1 • The equation is in the standard form for a linear equation. • Multiplying by the integrating factor e ∫ 2x dx = ex2, weget ex2y’ + 2xex2y = ex2or (ex2y)’ = ex2

  27. LINEAR EQUATIONS Example 3 • Therefore, ex2y = ∫ ex2dx + C • Recall from Section 7.5 that ∫ex2dxcan’t be expressed in terms of elementary functions. • Nonetheless, it’s a perfectly good function and we can leave the answer as: y = e-x2 ∫ex2 dx + Ce-x2

  28. LINEAR EQUATIONS Example 3 • Another way of writing the solution is: • Any number can be chosen for the lower limit of integration.

  29. LINEAR EQUATIONS • Though the solutions of the differential equation in Example 3 are expressed in terms of an integral, they can still be graphed by a computer algebra system (CAS).

  30. APPLICATION TO ELECTRIC CIRCUITS • In Section 9.2, we considered the simple electric circuit shown here.

  31. APPLICATION TO ELECTRIC CIRCUITS • An electromotive force (usually a battery or generator) produces: • A voltage of E(t) volts (V) • A current of I(t) amperes (A) at time t

  32. APPLICATION TO ELECTRIC CIRCUITS • The circuit also contains: • A resistor with a resistance of R ohms (Ω) • An inductor with an inductance of L henries (H)

  33. ELECTRIC CIRCUITS • Ohm’s Law gives the drop in voltage due to the resistor as RI. • The drop due to the inductor is L(dI/dt).

  34. ELECTRIC CIRCUITS • One of Kirchhoff’s laws says that the sum of the voltage drops is equal to the supplied voltage E(t).

  35. ELECTRIC CIRCUITS Equation 7 • Thus, we have • which is a first-order linear differential equation. • The solution gives the current I at time t.

  36. ELECTRIC CIRCUITS Example 4 • In this simple circuit, suppose: • The resistance is 12 Ω • The inductance is 4 H

  37. ELECTRIC CIRCUITS Example 4 • Then, a battery gives a constant voltage of 60 V and the switch is closed when t = 0, so the current starts with I(0) = 0.

  38. ELECTRIC CIRCUITS Example 4 • Now, find: • I(t) • The current after 1 s • The limiting value of the current

  39. ELECTRIC CIRCUITS Example 4 a • If we put L = 4, R = 12, and E(t) = 60 in Equation 7, we obtain the initial-value problem • or

  40. ELECTRIC CIRCUITS Example 4 a • Multiplying by the integrating factor e ∫ 3 dt = e3t, we get:

  41. ELECTRIC CIRCUITS Example 4 a • Since I(0) = 0, we have: 5 + C = 0 • Thus, C = –5 and I(t) = 5(1 – e-3t)

  42. ELECTRIC CIRCUITS Example 4 b • After 1 second, the current is: I(1) = 5(1 – e-3) ≈ 4.75 A

  43. ELECTRIC CIRCUITS Example 4 c • The limiting value of the current is given by:

  44. ELECTRIC CIRCUITS • The figure shows how the current in Example 4 approaches its limiting value.

  45. ELECTRIC CIRCUITS • The differential equation in Example 4 is both linear and separable. • So, an alternative method is to solve it as a separable equation (Example 4 in Section 9.3).

  46. ELECTRIC CIRCUITS • However, if we replace the battery by a generator, we get an equation that is linear but not separable—as in the next example.

  47. ELECTRIC CIRCUITS Example 5 • Suppose the resistance and inductance remain as in Example 4 but—instead of the battery—we use a generator that produces a variable voltage of: • E(t) = 60 sin 30t volts • Find I(t).

  48. ELECTRIC CIRCUITS Example 5 • This time, the differential equation becomesor

  49. ELECTRIC CIRCUITS Example 5 • The same integrating factor e3tgives:

  50. ELECTRIC CIRCUITS Example 5 • Using Formula 98 in the Table of Integrals, we have:

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