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Math EP FUND.OF DIFF.EQUATIONS-MYLAB (18 WK)

The continuous solution to the initial value problem d y d x + 2 y = Q ( x ) , y ( 0 ) = 0 , where Q ( x ) = { 2 , 0 ≤ x ≤ 3 , − 2 , x > 3. ; Sketch the graph of the solution from x = 2 to x = 7 . The continuous solution to the initial value problem d y d x + 2 y = Q ( x ) , y ( 0 ) = 0 , is y = { 1 − e − 2 x , if 0 ≤ x ≤ 3 − 1 + ( 2 e 6 − 1 ) e − 2 x , if x > 3 _ . Given: The initial value problem is, d y d x + 2 y = Q ( x ) , y ( 0 ) = 0 , where Q ( x ) = { 2 , 0 ≤ x ≤ 3 , − 2 , x > 3. Procedure used: “The procedure discussed in problem 31 to find the continuous solution to the initial value problem as given below. 1. Find the general solution for the interval 0 ≤ x ≤ a . 2. Choose the constant in the solution from step 1 so that the initial condition is satisfied. 3. Find the general solution for the interval x > a . 4. Now choose the constant in the general solution from step 3 so that the solution from step 2 and the solution from step 3 agree at x = a . 5. By patching the two solutions together, we can obtain a continuous function that satisfies the differential equation except at x = 3 , where its derivative is undefined.” Calculation: Use the above procedure to find the continuous solution to the initial value problem as follows. Step 1: Obtain the general solution for the interval 0 ≤ x ≤ 3 . The differential equation is, d y d x + 2 y = Q ( x ) ... ( 1 ) For the interval 0 ≤ x ≤ 3 , the value of Q ( x ) is equal to 2. Substitute Q ( x ) = 2 in equation (1), the differential equation becomes d y d x + 2 y = 2 Here, P ( x ) = 2 and Q ( x ) = 2 . Calculate the integrating factor μ ( x ) by the formula μ ( x ) = exp [ ∫ P ( x ) d x ] . μ ( x ) = exp [ ∫ ( 2 ) d x ] = e 2 x Multiply the equation (1) by μ ( x ) and obtain d d x ( μ ( x ) y ) = μ ( x ) Q ( x ) . e 2 x ( d y d x + 2 y ) = e 2 x 2 e 2 x d y d x + 2 y e 2 x = 2 e 2 x d d x ( e 2 x y ) = 2 e 2 x To obtain the general solution y ( x ) = 1 μ ( x ) [ ∫ μ ( x ) Q ( x ) d x + C ] , integrate both sides of the equation d d x ( e 2 x y ) = 2 e 2 x and solve for y ( x ) by dividing μ ( x ) . e 2 x y = ∫ 2 e 2 x d x e 2 x y = 2 e 2 x 2 + C y = e 2 x e 2 x + C e 2 x y = 1 + C e − 2 x Thus, the general solution for 0 ≤ x ≤ 3 is y = 1 + C e − 2 x . Step 2: From step 1, the general solution for 0 ≤ x ≤ 3 is y = 1 + C e − 2 x . The initial condition is y ( 0 ) = 0 . Apply the initial condition x = 0 and y = 0 in to y = 1 + C e − 2 x find the constant C . 0 = 1 + C e − 2 ( 0 ) 0 = 1 + C C = − 1 Substitute C = − 1 in the solution y = 1 + C e − 2 x and simplify further. y = 1 − e − 2 x Therefore, the solution is y = 1 − e − 2 x . Step 3: For the interval x > 3 , the value of Q ( x ) is equal to − 2 . Substitute P ( x ) = 3 in equation (1), the differential equation becomes d y d x + 2 y = − 2 Here, P ( x ) = 2 and Q ( x ) = − 2 . Calculate the integrating factor μ ( x ) by the formula μ ( x ) = exp [ ∫ P ( x ) d x ] . μ ( x ) = exp [ ∫ ( 2 ) d x ] = e 2 x Multiply the equation (1) by μ ( x ) and obtain d d x ( μ ( x ) y ) = μ ( x ) Q ( x ) . e 2 x ( d y d x + 2 y ) = e 2 x ( − 2 ) e 2 x d y d x + 2 y e 2 x = − 2 e 2 x d d x ( e 2 x y ) = − 2 e 2 x To obtain the general solution y ( x ) = 1 μ ( x ) [ ∫ μ ( x ) Q ( x ) d x + C ] , integrate both sides of the equation d d x ( e 2 x y ) = − 2 e 2 x and solve for y ( x ) by dividing μ ( x ) . e 2 x y = ∫ − 2 e 2 x d x e 2 x y = − 2 e 2 x 2 + C y = − e 2 x e 2 x + C e 2 x y = − 1 + C e − 2 x Thus, the general solution for x > 3 is y = − 1 + C e − 2 x . Step 4: The object is to find a constant C where the solution from step 2 and the solution from step 3 agree at x = 3 . From step 2, the solution for 0 ≤ x ≤ 3 is y = 1 − e − 2 x . From step 3, the general solution for x > 3 is y = − 1 + C e − 2 x . Equate the solution from step 2 and step 3 and substitute x = 3 . − 1 + C e − 2 x = 1 − e − 2 x − 1 + C e − 2 ( 3 ) = 1 − e − 2 ( 3 ) C e − 6 = 1 − e − 6 + 1 C e − 6 = 2 − e − 6 C = 2 e 6 − 1 Substitute C = 2 e 6 − 1 in the solution y = − 1 + C e − 2 x and simplify further. y = − 1 + ( 2 e 6 − 1 ) e − 2 x Thus, the solution is y = − 1 + ( 2 e 6 − 1 ) e − 2 x . Clearly, for x > 3 the derivative is undefined. Step 5: Combine the solutions for the intervals 0 ≤ x ≤ 3 and x > 3 to obtain a continuous function that satisfies the differential equation except at x = 3 . y = { 1 − e − 2 x , if 0 ≤ x ≤ 3 − 1 + ( 2 e 6 − 1 ) e − 2 x , if x > 3 Thus, the continuous solution to the initial value problem d y d x + 2 y = Q ( x ) , y ( 0 ) = 0 , is y = { 1 − e − 2 x , if 0 ≤ x ≤ 3 − 1 + ( 2 e 6 − 1 ) e − 2 x , if x > 3 _ . Use online graphing calculator to sketch the graph of the solution y = { 1 − e − 2 x , if 0 ≤ x ≤ 3 − 1 + ( 2 e 6 − 1 ) e − 2 x , if x > 3 from x = 2 to x = 7 as shown in Figure 1. From Figure 1 it is observed that the graph has a sharp point x = 3 . Thus, the function is not differential at x = 3 .

The continuous solution to the initial value problem d y d x + 2 y = Q ( x ) , y ( 0 ) = 0 , where Q ( x ) = { 2 , 0 ≤ x ≤ 3 , − 2 , x > 3. ; Sketch the graph of the solution from x = 2 to x = 7 . The continuous solution to the initial value problem d y d x + 2 y = Q ( x ) , y ( 0 ) = 0 , is y = { 1 − e − 2 x , if 0 ≤ x ≤ 3 − 1 + ( 2 e 6 − 1 ) e − 2 x , if x > 3 _ . Given: The initial value problem is, d y d x + 2 y = Q ( x ) , y ( 0 ) = 0 , where Q ( x ) = { 2 , 0 ≤ x ≤ 3 , − 2 , x > 3. Procedure used: “The procedure discussed in problem 31 to find the continuous solution to the initial value problem as given below. 1. Find the general solution for the interval 0 ≤ x ≤ a . 2. Choose the constant in the solution from step 1 so that the initial condition is satisfied. 3. Find the general solution for the interval x > a . 4. Now choose the constant in the general solution from step 3 so that the solution from step 2 and the solution from step 3 agree at x = a . 5. By patching the two solutions together, we can obtain a continuous function that satisfies the differential equation except at x = 3 , where its derivative is undefined.” Calculation: Use the above procedure to find the continuous solution to the initial value problem as follows. Step 1: Obtain the general solution for the interval 0 ≤ x ≤ 3 . The differential equation is, d y d x + 2 y = Q ( x ) ... ( 1 ) For the interval 0 ≤ x ≤ 3 , the value of Q ( x ) is equal to 2. Substitute Q ( x ) = 2 in equation (1), the differential equation becomes d y d x + 2 y = 2 Here, P ( x ) = 2 and Q ( x ) = 2 . Calculate the integrating factor μ ( x ) by the formula μ ( x ) = exp [ ∫ P ( x ) d x ] . μ ( x ) = exp [ ∫ ( 2 ) d x ] = e 2 x Multiply the equation (1) by μ ( x ) and obtain d d x ( μ ( x ) y ) = μ ( x ) Q ( x ) . e 2 x ( d y d x + 2 y ) = e 2 x 2 e 2 x d y d x + 2 y e 2 x = 2 e 2 x d d x ( e 2 x y ) = 2 e 2 x To obtain the general solution y ( x ) = 1 μ ( x ) [ ∫ μ ( x ) Q ( x ) d x + C ] , integrate both sides of the equation d d x ( e 2 x y ) = 2 e 2 x and solve for y ( x ) by dividing μ ( x ) . e 2 x y = ∫ 2 e 2 x d x e 2 x y = 2 e 2 x 2 + C y = e 2 x e 2 x + C e 2 x y = 1 + C e − 2 x Thus, the general solution for 0 ≤ x ≤ 3 is y = 1 + C e − 2 x . Step 2: From step 1, the general solution for 0 ≤ x ≤ 3 is y = 1 + C e − 2 x . The initial condition is y ( 0 ) = 0 . Apply the initial condition x = 0 and y = 0 in to y = 1 + C e − 2 x find the constant C . 0 = 1 + C e − 2 ( 0 ) 0 = 1 + C C = − 1 Substitute C = − 1 in the solution y = 1 + C e − 2 x and simplify further. y = 1 − e − 2 x Therefore, the solution is y = 1 − e − 2 x . Step 3: For the interval x > 3 , the value of Q ( x ) is equal to − 2 . Substitute P ( x ) = 3 in equation (1), the differential equation becomes d y d x + 2 y = − 2 Here, P ( x ) = 2 and Q ( x ) = − 2 . Calculate the integrating factor μ ( x ) by the formula μ ( x ) = exp [ ∫ P ( x ) d x ] . μ ( x ) = exp [ ∫ ( 2 ) d x ] = e 2 x Multiply the equation (1) by μ ( x ) and obtain d d x ( μ ( x ) y ) = μ ( x ) Q ( x ) . e 2 x ( d y d x + 2 y ) = e 2 x ( − 2 ) e 2 x d y d x + 2 y e 2 x = − 2 e 2 x d d x ( e 2 x y ) = − 2 e 2 x To obtain the general solution y ( x ) = 1 μ ( x ) [ ∫ μ ( x ) Q ( x ) d x + C ] , integrate both sides of the equation d d x ( e 2 x y ) = − 2 e 2 x and solve for y ( x ) by dividing μ ( x ) . e 2 x y = ∫ − 2 e 2 x d x e 2 x y = − 2 e 2 x 2 + C y = − e 2 x e 2 x + C e 2 x y = − 1 + C e − 2 x Thus, the general solution for x > 3 is y = − 1 + C e − 2 x . Step 4: The object is to find a constant C where the solution from step 2 and the solution from step 3 agree at x = 3 . From step 2, the solution for 0 ≤ x ≤ 3 is y = 1 − e − 2 x . From step 3, the general solution for x > 3 is y = − 1 + C e − 2 x . Equate the solution from step 2 and step 3 and substitute x = 3 . − 1 + C e − 2 x = 1 − e − 2 x − 1 + C e − 2 ( 3 ) = 1 − e − 2 ( 3 ) C e − 6 = 1 − e − 6 + 1 C e − 6 = 2 − e − 6 C = 2 e 6 − 1 Substitute C = 2 e 6 − 1 in the solution y = − 1 + C e − 2 x and simplify further. y = − 1 + ( 2 e 6 − 1 ) e − 2 x Thus, the solution is y = − 1 + ( 2 e 6 − 1 ) e − 2 x . Clearly, for x > 3 the derivative is undefined. Step 5: Combine the solutions for the intervals 0 ≤ x ≤ 3 and x > 3 to obtain a continuous function that satisfies the differential equation except at x = 3 . y = { 1 − e − 2 x , if 0 ≤ x ≤ 3 − 1 + ( 2 e 6 − 1 ) e − 2 x , if x > 3 Thus, the continuous solution to the initial value problem d y d x + 2 y = Q ( x ) , y ( 0 ) = 0 , is y = { 1 − e − 2 x , if 0 ≤ x ≤ 3 − 1 + ( 2 e 6 − 1 ) e − 2 x , if x > 3 _ . Use online graphing calculator to sketch the graph of the solution y = { 1 − e − 2 x , if 0 ≤ x ≤ 3 − 1 + ( 2 e 6 − 1 ) e − 2 x , if x > 3 from x = 2 to x = 7 as shown in Figure 1. From Figure 1 it is observed that the graph has a sharp point x = 3 . Thus, the function is not differential at x = 3 .

Solution Summary: The author describes the procedure used to find the continuous solution to the initial value problem.

EP FUND.OF DIFF.EQUATIONS-MYLAB (18 WK)

EP FUND.OF DIFF.EQUATIONS-MYLAB (18 WK)

9th Edition

ISBN: 9780135963777

Author: Nagle

Publisher: PEARSON CO

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EP FUND.OF DIFF.EQUATIONS-MYLAB (18 WK)

Ch. 2.2 - In Problems 16, determine whether the given...Ch. 2.2 - In Problems 16, determine whether the given...Ch. 2.2 - In Problems 16, determine whether the given...Ch. 2.2 - In Problems 16, determine whether the given...Ch. 2.2 - In Problems 16, determine whether the given...Ch. 2.2 - In Problems 16, determine whether the given...Ch. 2.2 - In Problems 716, solve the equation. 7. xdydx=1y3 Ch. 2.2 - In Problems 716, solve the equation. 8. dxdt=3xt2 Ch. 2.2 - In Problems 716, solve the equation. 9....Ch. 2.2 - In Problems 716, solve the equation. 10....

Ch. 2.2 - In Problems 716, solve the equation. 11....Ch. 2.2 - In Problems 716, solve the equation. 12....Ch. 2.2 - In Problems 716, solve the equation. 13....Ch. 2.2 - In Problems 716, solve the equation. 14. dxdtx3=x Ch. 2.2 - In Problems 716, solve the equation. 15....Ch. 2.2 - In Problems 716, solve the equation. 16. y1 dy +...Ch. 2.2 - In Problems 1726, solve the initial value problem....Ch. 2.2 - In Problems 1726, solve the initial value problem....Ch. 2.2 - In Problems 1726, solve the initial value problem....Ch. 2.2 - In Problems 1726, solve the initial value problem....Ch. 2.2 - In Problems 1726, solve the initial value problem....Ch. 2.2 - In Problems 1726, solve the initial value problem....Ch. 2.2 - Prob. 23E Ch. 2.2 - In Problems 1726, solve the initial value problem....Ch. 2.2 - In Problems 1726, solve the initial value problem....Ch. 2.2 - In Problems 1726, solve the initial value problem....Ch. 2.2 - Prob. 27E Ch. 2.2 - Sketch the solution to the initial value problem...Ch. 2.2 - Uniqueness Questions. In Chapter 1 we indicated...Ch. 2.2 - As stated in this section, the separation of...Ch. 2.2 - Prob. 31E Ch. 2.2 - Prob. 32E Ch. 2.2 - Mixing. Suppose a brine containing 0.3 kilogram...Ch. 2.2 - Newtons Law of Cooling. According to Newtons law...Ch. 2.2 - Prob. 35E Ch. 2.2 - Prob. 36E Ch. 2.2 - Compound Interest. If P(t) is the amount of...Ch. 2.2 - Free Fall. In Section 2.1, we discussed a model...Ch. 2.2 - Grand Prix Race. Driver A had been leading...Ch. 2.2 - Prob. 40E Ch. 2.3 - In Problems 16, determine whether the given...Ch. 2.3 - In Problems 16, determine whether the given...Ch. 2.3 - In Problems 16, determine whether the given...Ch. 2.3 - In Problems 16, determine whether the given...Ch. 2.3 - In Problems 16, determine whether the given...Ch. 2.3 - In Problems 16, determine whether the given...Ch. 2.3 - In Problems 716, obtain the general solution to...Ch. 2.3 - In Problems 716, obtain the general solution to...Ch. 2.3 - In Problems 716, obtain the general solution to...Ch. 2.3 - In Problems 716, obtain the general solution to...Ch. 2.3 - In Problems 716, obtain the general solution to...Ch. 2.3 - In Problems 716, obtain the general solution to...Ch. 2.3 - In Problems 716, obtain the general solution to...Ch. 2.3 - In Problems 716, obtain the general solution to...Ch. 2.3 - Prob. 15E Ch. 2.3 - Prob. 17E Ch. 2.3 - Prob. 18E Ch. 2.3 - Prob. 19E Ch. 2.3 - Prob. 20E Ch. 2.3 - In Problems 1722, solve the initial value problem....Ch. 2.3 - In Problems 1722, solve the initial value problem....Ch. 2.3 - Radioactive Decay. In Example 2 assume that the...Ch. 2.3 - Prob. 24E Ch. 2.3 - (a) Using definite integration, show that the...Ch. 2.3 - Prob. 26E Ch. 2.3 - Constant Multiples of Solutions. (a) Show that y =...Ch. 2.3 - Prob. 29E Ch. 2.3 - Bernoulli Equations. The equation (18) dydx+2y=xy2...Ch. 2.3 - Prob. 31E Ch. 2.3 - Prob. 32E Ch. 2.3 - Prob. 33E Ch. 2.3 - Prob. 34E Ch. 2.3 - Prob. 35E Ch. 2.3 - Prob. 36E Ch. 2.3 - Prob. 37E Ch. 2.3 - Prob. 38E Ch. 2.3 - Prob. 39E Ch. 2.4 - In Problems 18, classify the equation as...Ch. 2.4 - In Problems 18, classify the equation as...Ch. 2.4 - Prob. 3E Ch. 2.4 - Prob. 4E Ch. 2.4 - In Problems 18, classify the equation as...Ch. 2.4 - In Problems 18, classify the equation as...Ch. 2.4 - In Problems 18, classify the equation as...Ch. 2.4 - In Problems 18, classify the equation as...Ch. 2.4 - Prob. 9E Ch. 2.4 - In Problems 920, determine whether the equation is...Ch. 2.4 - Prob. 11E Ch. 2.4 - Prob. 12E Ch. 2.4 - Prob. 13E Ch. 2.4 - In Problems 920, determine whether the equation is...Ch. 2.4 - Prob. 15E Ch. 2.4 - In Problems 920, determine whether the equation is...Ch. 2.4 - Prob. 17E Ch. 2.4 - In Problems 920, determine whether the equation is...Ch. 2.4 - Prob. 19E Ch. 2.4 - Prob. 20E Ch. 2.4 - In Problems 2126, solve the initial value problem....Ch. 2.4 - Prob. 22E Ch. 2.4 - Prob. 23E Ch. 2.4 - In Problems 2126, solve the initial value problem....Ch. 2.4 - Prob. 25E Ch. 2.4 - In Problems 2126, solve the initial value problem....Ch. 2.4 - Prob. 27E Ch. 2.4 - For each of the following equations, find the most...Ch. 2.4 - Prob. 29E Ch. 2.4 - Prob. 30E Ch. 2.4 - Prob. 31E Ch. 2.4 - Orthogonal Trajectories. A geometric problem...Ch. 2.4 - Prob. 33E Ch. 2.4 - Prob. 34E Ch. 2.4 - Prob. 35E Ch. 2.4 - Prob. 36E Ch. 2.5 - Prob. 1E Ch. 2.5 - In Problems 16, identify the equation as...Ch. 2.5 - Prob. 3E Ch. 2.5 - Prob. 4E Ch. 2.5 - In Problems 16, identify the equation as...Ch. 2.5 - Prob. 6E Ch. 2.5 - Prob. 7E Ch. 2.5 - Prob. 8E Ch. 2.5 - Prob. 9E Ch. 2.5 - Prob. 10E Ch. 2.5 - Prob. 11E Ch. 2.5 - Prob. 12E Ch. 2.5 - Prob. 13E Ch. 2.5 - Prob. 14E Ch. 2.5 - Prob. 15E Ch. 2.5 - Prob. 16E Ch. 2.5 - Prob. 17E Ch. 2.5 - Prob. 18E Ch. 2.5 - Prob. 19E Ch. 2.5 - Verify that when the linear differential equation...Ch. 2.6 - In Problems 18, identify (do not solve) the...Ch. 2.6 - Prob. 2E Ch. 2.6 - Prob. 3E Ch. 2.6 - Prob. 4E Ch. 2.6 - Prob. 5E Ch. 2.6 - Prob. 6E Ch. 2.6 - In Problems 18, identify (do not solve) the...Ch. 2.6 - Prob. 8E Ch. 2.6 - Use the method discussed under Homogeneous...Ch. 2.6 - Prob. 10E Ch. 2.6 - Prob. 11E Ch. 2.6 - Prob. 12E Ch. 2.6 - Prob. 13E Ch. 2.6 - Prob. 14E Ch. 2.6 - Prob. 15E Ch. 2.6 - Prob. 16E Ch. 2.6 - Prob. 17E Ch. 2.6 - Prob. 18E Ch. 2.6 - Prob. 19E Ch. 2.6 - Prob. 20E Ch. 2.6 - Prob. 21E Ch. 2.6 - Prob. 22E Ch. 2.6 - Use the method discussed under Bernoulli Equations...Ch. 2.6 - Prob. 24E Ch. 2.6 - Prob. 25E Ch. 2.6 - Prob. 26E Ch. 2.6 - Prob. 27E Ch. 2.6 - Prob. 28E Ch. 2.6 - Use the method discussed under Equations with...Ch. 2.6 - Prob. 30E Ch. 2.6 - Prob. 31E Ch. 2.6 - Prob. 32E Ch. 2.6 - Prob. 33E Ch. 2.6 - Prob. 34E Ch. 2.6 - Prob. 35E Ch. 2.6 - In Problems 3340, solve the equation given in: 36....Ch. 2.6 - Prob. 37E Ch. 2.6 - Prob. 38E Ch. 2.6 - Prob. 39E Ch. 2.6 - Prob. 40E Ch. 2.6 - Prob. 41E Ch. 2.6 - Prob. 42E Ch. 2.6 - Prob. 43E Ch. 2.6 - Show that equation (13) reduces to an equation of...Ch. 2.6 - Prob. 45E Ch. 2.6 - Prob. 46E Ch. 2.6 - Prob. 47E Ch. 2.6 - Prob. 48E Ch. 2 - In Problems 130, solve the equation. 1....Ch. 2 - Prob. 2RP Ch. 2 - Prob. 3RP Ch. 2 - Prob. 4RP Ch. 2 - Prob. 5RP Ch. 2 - In Problems 130, solve the equation. 6. 2xy3 dx ...Ch. 2 - In Problems 130, solve the equation. 7. t3y2 dt +...Ch. 2 - Prob. 8RP Ch. 2 - In Problems 130, solve the equation. 9. (x2 + y2)...Ch. 2 - Prob. 10RP Ch. 2 - Prob. 11RP Ch. 2 - Prob. 12RP Ch. 2 - Prob. 13RP Ch. 2 - Prob. 14RP Ch. 2 - Prob. 15RP Ch. 2 - Prob. 16RP Ch. 2 - Prob. 17RP Ch. 2 - Prob. 18RP Ch. 2 - Prob. 19RP Ch. 2 - Prob. 20RP Ch. 2 - Prob. 21RP Ch. 2 - Prob. 22RP Ch. 2 - Prob. 23RP Ch. 2 - In Problems 130, solve the equation. 24. (y/x +...Ch. 2 - Prob. 25RP Ch. 2 - Prob. 26RP Ch. 2 - Prob. 27RP Ch. 2 - Prob. 28RP Ch. 2 - Prob. 29RP Ch. 2 - Prob. 30RP Ch. 2 - Prob. 31RP Ch. 2 - Prob. 32RP Ch. 2 - Prob. 33RP Ch. 2 - Prob. 34RP Ch. 2 - Prob. 35RP Ch. 2 - Prob. 36RP Ch. 2 - Prob. 37RP Ch. 2 - Prob. 38RP Ch. 2 - Prob. 39RP Ch. 2 - Prob. 40RP Ch. 2 - Prob. 41RP

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