dy dt 2tk y = 0 2t k dt

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1 LESSON 7: DIFFERENTIAL EQUATIONS SEPARATION OF VARIABLES (I) MATH 6020 FALL 208 ELLEN WELD Example. Find y(t) such that where y(0) = and y() = e 2/7.. Solutions to In-Class Examples 2tk y = 0 Solution: We separate variables in our equation: 2tk y = 0 = 2tk y y = 2tk Next, we set up our integral. y = 2t k. Now, to integrate the RHS, we need to determine whether k = or k. Because we have y(0) =, we know that y(t) exists at the value t = 0. But if k < 0, then t k does not exist at t = 0. Thus, we may conclude that, in particular, k. We integrate: y = 2t k ln y = 2 k + tk+ + C We assume that y > 0 and so we have Since y(0) =, we have y = e 2 k+ tk+ +C = Ce 2 k+ tk+ y = Ce 2 k+ tk+. = Ce 2 k+ (0)k+ = C.

2 2 ELLEN WELD Thus, y = e 2 k+ tk+. Further, y() = e 2/7 which means Therefore, e 2/7 = e 2 k+ ()k+ = e 2 k+ ln e 2/7 = ln e 2 k+ 2 7 = 2 k + 7 = k + k + = 7 k = 6 y = e (2/7)t7. Example 2. A clay mug is 500 F when it is removed from a kiln and placed in a room with a constant temperature of 70 F. After 2 hours, the mug is 200 F. What is the temperature of the mug after 5 hours? Round your answer to the nearest degree. Solution: We use Newton s Cooling Formula: = k(t S) where T (t) is the temperature at time t and S is the surrounding temperature. Let T (t) be the temperature of the mug after t hours. Since the room is 70 F, we have = k(t 70). We separate the variables and integrate: = k(t 70) T 70 = k T 70 = k ln T 70 = kt + C T 70 = e kt+c = Ce kt We may assume that the temperature of the mug never dips below 70 F and so we don t need the absolute values. We have T 70 = Ce kt T = Ce kt + 70.

3 AN UNOFFICIAL GUIDE TO MATH 6020 FALL We know T (0) = 500. Thus, we have 500 = Ce k = C = C We conclude T = 430e kt + 70 and since T (2) = 200, we may write This means Finally, we compute T (5): 200 = 430e k = 430e 2k = e2k ( ) 3 ln = 2k 43 ( ) 3 2 ln = k. 43 T = 430e 2 T (5) = 430e 5 2 ln( 3 43)t ln( 3 43) F. Example 3. Suppose the volume of a balloon being inflated satisfies dv = 0 5 V 2 where t is time in seconds after the balloon begins to inflate. If the balloon pops when it reaches a volume of 400 cm 3, after how many seconds will the balloon pop? Round your answer to 3 decimal places. Solution: We are given dv = 0 5 V 2 = 0(V 2 ) /5 = 0V 2/5. So, dv = 0V 2/5 dv = 0 V 2/5 V 2/5 dv = 0 V 2/5 dv = 0 ( ) V 2/5+ = 0t + C 2/5 +

4 4 ELLEN WELD ( ) V 3/5 = 0t + C 3/5 Since V (0) = 0, Our equation is then 5 3 V 3/5 = 0t + C. 5 3 (0)3/5 = 0(0) + C C = V = 0t. We could solve for V, but observe that our ultimate goal is to find the t such that V (t) = 400. Hence, we need only write and solve for t. But this is just 5 3 ( }{{} 400 ) 3/5 = 0t V (t) t = 6 (400)3/ seconds. Example 4. A wet towel hung on a clothesline to dry outside loses moisture at a rate proportional to its moisture content. After hour, the towel has lost 5% of its original moisture content. After how long will the towel have lost 80% of its original moisture content? Round your answer to the nearest hundreh. Solution: Let M(t) be the percentage of the original moisture content the towel has after t hours. Then M (t) = km(t) for k the proportionality constant. We know M(0) = 00 because we assume at time t that the towel has not lost any moisture content and M() = 00 5 = 85 because after hour the towel is less 5% of its moisture. Our goal is to find the time t such that M(t) = = 20. The general solution to a differential equation of the form M (t) = km(t) is So, we need to solve for k. Since M() = 85, which implies M(t) = M(0)e kt. M(t) = 00e kt 85 = 00e k() = 00e k,.85 = e k ln(.85) = k. We want to find the t such that 20 = M(t) = 00e ln(.85)t.

5 Write AN UNOFFICIAL GUIDE TO MATH 6020 FALL = e ln(.85)t, which, after applying ln to both sides, becomes ln(.2) = ln(.85)t t = ln(.2) ln(.85) 9.9 hours. 2. Additional Examples Examples.. Find the general solution to + 50y = 0. Solution: Since we are finding a general solution, we are not asked to solve for all the unknowns. Write + 50y = 0 = 50y = 50 y y = ( 50) ln y = 50t + C e ln y = e 50t+C y = e 50t+C. Here, we have a choice for y: either y = e 50t+C or y = e 50t+C. For this class, we re always going to want our functions to be non-negative, so we drop the absolute values and write Hence, y = e 50t+C = e 50t e C }{{} C y = Ce 50t. 2. Find the particular solution to the equation da = Ce 50t. = (20 A) such that A(0) = 00, A < 20 for all t.

6 6 ELLEN WELD Solution: The fact that A < 20 for all t tells us that 20 A > 0. So, da = 20 A da = 20 A 20 A da = The LHS is a u-sub problem. If u = 20 A, then du = da. So 20 A da = du = ln u = ln 20 A. u Thus, 20 A da = ln 20 A = t + C ln 20 A = t C }{{} +C We have assumed that 20 A > 0, so we may drop the absolute values. Hence, ln 20 A = t + C ln(20 A) = t + C e ln(20 A) = e t+c 20 A = e t+c 20 e t+c = A So our solution is of the form 20 }{{} e C e t = A. C A = 20 Ce t. We were told that A(0) = 00, which means Therefore, 00 = 20 Ce 0 = 20 C C = 20. A = 20 20e t.