University:
Whitman College
Course:
MATH-244 | Differential Equations
Academic year:

2024
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377

Pages:

3
Author:

gesaagdeq1j

2.1: Some Detailed Examples Summary: Given y 0 + p(t)y = f (t), we first find the integrating factor µ(t): µ(t) = e R p(t) dt Then multiply both sides of the DE by it: µ(t)y 0 + µ(t)p(t)y = µ(t)f (t) Since µ0 = µp, then this becomes µ(t)y 0 + µ0 (t)y = µ(t)f (t) (µ(t)y(t))0 = µ(t)f (t) ⇒ Then solve this by integrating both sides, then isolate y. Example 1 ty 0 + (t + 1)y = t y(ln(2)) = 1 SOLUTION: First get the DE in standard form, y 0 + p(t)y = f (t) by dividing by t: y0 + t+1 y=1 t Now compute the integrating factor µ: µ(t) = e R p(t) dt R =e 1+ 1t dt = et+ln(t) = et eln(t) = tet Multiply both sides by the integrating factor so that the LHS becomes a “perfect derivative”: 0 tet (y 0 + (1 + 1/t)y) = tet ⇒ tet y(t) = tet We integrate by parts using a table (see the Review sheet): Z tet dt + t et − 1 et + 0 et = tet − et + C so that tet y = tet − et + C and y(t) = 1 − 1 C −t + e t t To solve for C, put in t = ln(2) and y = 1: 1=1− 1 C − ln(2) + e ln(2) ln(2) ⇒ The solution is: y(t) = 1 − 1 1= 1 2 t + e t t C 2 ⇒ C=2 Example 2 (2.1, #4) Solve: y 0 + (1/t)y = 3 cos(2t), t > 0 SOLUTION: The DE is already in standard form, so we can compute µ(t) directly: R µ(t) = e 1/t dt = eln(t) = t Multiply both sides of the DE by t so that the left side of the DE can be written as: (y t)0 = 3t cos(2t) Integrate the right side of the DE by parts using a table (see the Review sheet): + t 3 cos(2t) − 1 32 sin(2t) + 0 − 34 cos(2t) Z ⇒ 3 3 3t cos(2t) dt = t sin(2t) + cos(2t) 2 4 Therefore, y(t) = 3 3 C sin(2t) + cos(2t) + 2 4t t NOTE: The following is incorrect: y(t) = 3 3 sin(2t) + cos(2t) + C 2 4t Example 3 (2.1, # 31) Solve the IVP below and describe how the initial value y0 changes the nature of the solution y(t). 3 y 0 − y = 3t + 2et , y(0) = y0 2 3 SOLUTION: The integrating factor can be computed quickly: µ(t) = e− 2 t so that 0 3 3 t e− 2 t y(t) = 3te− 2 t + 2e− 2 The first term is integrated by parts (use a table), and the second is done directly. The general solution is then 4 y(t) = −2t − − 4et + ce3t/2 3 16 Putting in the initial value, we see that c = 3 + y0 . How does this change the nature of the solution? (Also see the direction field below) • As t gets very large, and c 6= 0, the term e3t/2 will dominate the expression. • We see that if y0 > −16/3, then c > 0 and so y(t) will diverge to positive infinity. • If y0 < −16/3, then c < 0, and y(t) will diverge (to negative infinity). We still go to negative infinity if y0 = −16/3 as well. 2 Figure 1: Direction field with some solution curves, Exercise 31, 2.1 Example 4 (2.1, #37) Find a linear differential equation for which all solutions tend to y = 4 − t2 as t → ∞. SOLUTION: From our expression in linear DE’s, we might guess that: y(t) = 4 − t2 + Ce−t so that as t → ∞, y → 4 − t2 . Now we’ll see if y satisfies a linear DE. We’ll manipulate the expressions so that something of the form y 0 + ay gets rid of the arbitrary constant C. One way to do it: y 0 + y = (−2t − Ce−t ) + (4 − t2 + Ce−t ) = 4 − 2t − t2 Our ODE is therefore: y 0 + y = 4 − 2t − t2 3

Differential Equations 2.1 Some Detailed Examples

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