Skip to main content

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

(a) To sketch: The direction field of the differential equation d y d x = y 2 − 3 y + 2 and guess the value of lim x → ∞ ϕ ( x ) from the graph. The given initial value problem is, d y d x = y 2 − 3 y + 2 , y ( 0 ) = 1.5 . Use online graphing calculator and sketch the direction field for the differential equation d y d x = y 2 − 3 y + 2 as shown below in Figure 1. From Figure 1, it is observed that as x → ∞ , the value of the slope field approaches 1. Let y = ϕ ( x ) denote the solution to the initial value problem. Then, using the slope field, it is concluded that lim x → ∞ ϕ ( x ) = 1 . (b) An approximation of ϕ ( x ) using Euler’s method of step size 0.1. An approximation of ϕ ( x ) using Euler’s method of step size 0.1 is 1.267 _ . Formula used: Consider the initial value problem y ′ = f ( x , y ) , y ( x 0 ) = y 0 . The summarized recursive formula for the Euler’s method is: x n + 1 = x n + h y n + 1 = y n + h f ( x n , y n ) , n = 0 , 1 , 2 , … Calculation: The given initial value problem is, d y d x = y 2 − 3 y + 2 , y ( 0 ) = 1.5 . Here, x 0 = 0 , y 0 = 1.5 , h = 0.1 and f ( x , y ) = y 2 − 3 y + 2 . The recursive formula will be, x n + 1 = x n + 0.1 y n + 1 = y n + h f ( x n , y n ) = y n + 0.1 ( y n 2 − 3 y n + 2 ) The aim is to find the approximation of ϕ ( x ) , where y = ϕ ( x ) denote the solution to the initial value problem. Then, with the step size 0.1, values of x will be x 0 = 0 , x 1 = 0.1 , x 2 = 0.2 , x 3 = 0.3 , x 4 = 0.4 , x 5 = 0.5 , x 6 = 0.6 , x 7 = 0.7 , x 8 = 0.8 , x 9 = 0.9 and x 10 = 1 . The first approximation y 1 for n = 0 will be, y 1 = y 0 + 0.1 ( y 0 2 − 3 y 0 + 2 ) = 1.5 + 0.1 ( ( 1.5 ) 2 − 3 ( 1.5 ) + 2 ) ( ∵ y 0 = 1.5 ) = 1.5 + 0.1 ( − 0.25 ) = 1.475 The second approximation y 2 for n = 1 will be, y 2 = y 1 + 0.1 ( y 1 2 − 3 y 1 + 2 ) = 1.475 + 0.1 ( ( 1.475 ) 2 − 3 ( 1.475 ) + 2 ) ( ∵ y 1 = 1.475 ) = 1.475 + 0.1 ( − 0.0249375 ) ≈ 1.450 The third approximation y 3 for n = 2 will be, y 3 = y 2 + 0.1 ( y 2 2 − 3 y 2 + 2 ) = 1.450 + 0.1 ( ( 1.450 ) 2 − 3 ( 1.450 ) + 2 ) ( ∵ y 2 = 1.450 ) = 1.450 + 0.1 ( − 0.2475 ) ≈ 1.425 The fourth approximation y 4 for n = 3 will be, y 4 = y 3 + 0.1 ( y 3 2 − 3 y 3 + 2 ) = 1.425 + 0.1 ( ( 1.425 ) 2 − 3 ( 1.425 ) + 2 ) ( ∵ y 3 = 1.425 ) = 1.425 + 0.1 ( − 0.244375 ) ≈ 1.401 The fifth approximation y 5 for n = 4 will be, y 5 = y 4 + 0.1 ( y 4 2 − 3 y 4 + 2 ) = 1.401 + 0.1 ( ( 1.401 ) 2 − 3 ( 1.401 ) + 2 ) ( ∵ y 4 = 1.401 ) = 1.401 + 0.1 ( − 0.240199 ) ≈ 1.377 The sixth approximation y 6 for n = 5 will be, y 6 = y 5 + 0.1 ( y 5 2 − 3 y 5 + 2 ) = 1.377 + 0.1 ( ( 1.377 ) 2 − 3 ( 1.377 ) + 2 ) ( ∵ y 5 = 1.377 ) = 1.377 + 0.1 ( − 0.234871 ) ≈ 1.353 The seventh approximation y 7 for n = 6 will be, y 7 = y 6 + 0.1 ( y 6 2 − 3 y 6 + 2 ) = 1.353 + 0.1 ( ( 1.353 ) 2 − 3 ( 1.353 ) + 2 ) ( ∵ y 6 = 1.353 ) = 1.353 + 0.1 ( − 0.228391 ) ≈ 1.331 The eighth approximation y 8 for n = 7 will be, y 8 = y 7 + 0.1 ( y 7 2 − 3 y 7 + 2 ) = 1.331 + 0.1 ( ( 1.331 ) 2 − 3 ( 1.331 ) + 2 ) ( ∵ y 7 = 1.331 ) = 1.331 + 0.1 ( − 0.221439 ) ≈ 1.308 The ninth approximation y 9 for n = 8 will be, y 9 = y 8 + 0.1 ( y 8 2 − 3 y 8 + 2 ) = 1.308 + 0.1 ( ( 1.308 ) 2 − 3 ( 1.308 ) + 2 ) ( ∵ y 8 = 1.308 ) = 1.308 + 0.1 ( − 0.213136 ) ≈ 1.287 The tenth approximation y 10 for n = 9 will be, y 10 = y 9 + 0.1 ( y 9 2 − 3 y 9 + 2 ) = 1.287 + 0.1 ( ( 1.287 ) 2 − 3 ( 1.287 ) + 2 ) ( ∵ y 9 = 1.287 ) = 1.287 + 0.1 ( − 0.204631 ) ≈ 1.267 The approximations are tabulated as follows: x n y n 0.1 1.475 0.2 1.450 0.3 1.425 0.4 1.401 0.5 1.377 0.6 1.353 0.7 1.331 0.8 1.308 0.9 1.287 1.0 1.267 Therefore, the approximation of ϕ ( x ) using Euler’s method of step size 0.1 is 1.267 _ . (c) The formula for ϕ ( x ) and sketch ϕ ( x ) on the direction field of the differential equation d y d x = y 2 − 3 y + 2 . The formula for ϕ ( x ) is ϕ ( x ) = 2 + e x 1 + e x _ . The given initial value problem is, d y d x = y 2 − 3 y + 2 , y ( 0 ) = 1.5 . Factorize y 2 − 3 y + 2 as, y 2 − 3 y + 2 = y 2 − 2 y − y + 2 = y ( y − 2 ) − 1 ( y − 2 ) = ( y − 2 ) ( y − 1 ) Thus, the differential equation d y d x = y 2 − 3 y + 2 becomes, d y d x = ( y − 2 ) ( y − 1 ) d y ( y − 2 ) ( y − 1 ) = d x Express 1 ( y − 2 ) ( y − 1 ) in the partial fraction form A y − 2 + B y − 1 and solve for values of A and B as follows: 1 ( y − 2 ) ( y − 1 ) = A y − 2 + B y − 1 ( y − 2 ) ( y − 1 ) ( y − 2 ) ( y − 1 ) = A y − 2 ( y − 2 ) ( y − 1 ) + B y − 1 ( y − 2 ) ( y − 1 ) ( multiply by ( y − 2 ) ( y − 1 ) on both sides ) 1 = A ( y − 1 ) + B ( y − 2 ) 1 = A y − A + B y − 2 B 1 = ( A + B ) y + ( − A − 2 B ) Comparing the coefficients of the equation 1 = ( A + B ) y + ( − A − 2 B ) , it is observed that A + B = 0 and − A − 2 B = 1 . From the equation A + B = 0 , the value of A will be A = − B . Substitute A = − B in − A − 2 B = 1 and simplify as, − ( − B ) − 2 B = 1 B − 2 B = 1 − B = 1 B = − 1 Thus, A = − B becomes A = 1 . The form A y − 2 + B y − 1 becomes 1 y − 2 + − 1 y − 1 . That is, the partial fraction form of 1 ( y − 2 ) ( y − 1 ) will be 1 y − 2 + − 1 y − 1 . Thus, the equation d y ( y − 2 ) ( y − 1 ) = d x becomes, ( 1 y − 2 + − 1 y − 1 ) d y = d x . Since the form ( 1 y − 2 + − 1 y − 1 ) d y = d x is variable separable, integrate on both sides as, ∫ ( 1 y − 2 + − 1 y − 1 ) d y = ∫ d x ln | y − 2 | − ln | y − 1 | = x + C ln | y − 2 y − 1 | = x + C ( ∵ ln a b = ln a − ln b ) Apply the initial condition y ( 0 ) = 1.5 to obtain the value of the constant C as follows: Substitute x = 0 and y ( 0 ) = 1.5 in ln | y − 2 y − 1 | = x + C , ln | 1.5 − 2 1.5 − 1 | = 0 + C ln | − 0.5 0.5 | = C ln | − 1 | = C ln 1 = C C = 0 Thus, ln | y − 2 y − 1 | = x + C becomes, ln | y − 2 y − 1 | = x | y − 2 y − 1 | = e x ( taking exponential on both sides ) y − 2 y − 1 = ± e x ( ∵ e ln x = x ) y − 2 = ± ( y − 1 ) e x On further simplifications, the required y value is obtained as, y − 2 = ± y e x ∓ e x y ∓ y e x = 2 ∓ e x y ( 1 ∓ e x ) = 2 ∓ e x y = 2 ∓ e x 1 ∓ e x Since the initial condition is y ( 0 ) = 1.5 , substituting x = 0 and y ( 0 ) = 1.5 gives, 1.5 = 2 ∓ e 0 1 ∓ e 0 1.5 = 2 ∓ 1 1 ∓ 1 The equation will be satisfied for 1.5 = 2 + 1 1 + 1 . Thus, y = 2 + e x 1 + e x . Assume that, y = ϕ ( x ) denote the solution to the initial value problem. Then, ϕ ( x ) = 2 + e x 1 + e x . Therefore, the formula for ϕ ( x ) is ϕ ( x ) = 2 + e x 1 + e x _ . Use online graphing calculator and sketch ϕ ( x ) = 2 + e x 1 + e x on the direction field for the differential equation d y d x = y 2 − 3 y + 2 as shown below in Figure 2. From Figure 2, it is observed that as x → ∞ , the value of ϕ ( x ) = 2 + e x 1 + e x approaches 1. (d) The exact value of ϕ ( 1 ) and compare with the approximation value from part (b). The exact value of ϕ ( 1 ) is 1.2689 _ . From part (c), the formula for ϕ ( x ) is ϕ ( x ) = 2 + e x 1 + e x . Substitute x = 1 to obtain the value of ϕ ( 1 ) as, ϕ ( 1 ) = 2 + e 1 1 + e 1 = 2 + e 1 + e = 4.718281828 4.718281828 ≈ 1.2689 Therefore, the exact value of ϕ ( 1 ) is 1.2689 _ . From part (b), the approximation of ϕ ( x ) using Euler’s method of step size 0.1 is 1.267 . Thus, the error will be 1.2689 − 1.267 ≈ 0.0019 . That is, Euler’s method of step size 0.1 gives an error of 0.0019 difference from the exact value. (e) The value of lim x → ∞ ϕ ( x ) using the exact solution and compare the initial guess for lim x → ∞ ϕ ( x ) . The value of lim x → ∞ ϕ ( x ) using the exact solution is 1 _ . From part (c), the formula for ϕ ( x ) is ϕ ( x ) = 2 + e x 1 + e x . Obtain the value of lim x → ∞ ϕ ( x ) as follows: lim x → ∞ ϕ ( x ) = lim x → ∞ 2 + e x 1 + e x = 2 + e ∞ 1 + e ∞ = ∞ ∞ Since the indeterminate form ∞ ∞ is obtained, L’Hopital’s rule is applied to find the limit value. Differentiate numerator and denominator of lim x → ∞ 2 + e x 1 + e x and apply the limits as, lim x → ∞ e x e x = lim x → ∞ 1 = 1 That is, as x → ∞ , the value of ϕ ( x ) tends to 1. From part (a), using the slope field, it is concluded that lim x → ∞ ϕ ( x ) = 1 . That is, the initial guess was correct.

(a) To sketch: The direction field of the differential equation d y d x = y 2 − 3 y + 2 and guess the value of lim x → ∞ ϕ ( x ) from the graph. The given initial value problem is, d y d x = y 2 − 3 y + 2 , y ( 0 ) = 1.5 . Use online graphing calculator and sketch the direction field for the differential equation d y d x = y 2 − 3 y + 2 as shown below in Figure 1. From Figure 1, it is observed that as x → ∞ , the value of the slope field approaches 1. Let y = ϕ ( x ) denote the solution to the initial value problem. Then, using the slope field, it is concluded that lim x → ∞ ϕ ( x ) = 1 . (b) An approximation of ϕ ( x ) using Euler’s method of step size 0.1. An approximation of ϕ ( x ) using Euler’s method of step size 0.1 is 1.267 _ . Formula used: Consider the initial value problem y ′ = f ( x , y ) , y ( x 0 ) = y 0 . The summarized recursive formula for the Euler’s method is: x n + 1 = x n + h y n + 1 = y n + h f ( x n , y n ) , n = 0 , 1 , 2 , … Calculation: The given initial value problem is, d y d x = y 2 − 3 y + 2 , y ( 0 ) = 1.5 . Here, x 0 = 0 , y 0 = 1.5 , h = 0.1 and f ( x , y ) = y 2 − 3 y + 2 . The recursive formula will be, x n + 1 = x n + 0.1 y n + 1 = y n + h f ( x n , y n ) = y n + 0.1 ( y n 2 − 3 y n + 2 ) The aim is to find the approximation of ϕ ( x ) , where y = ϕ ( x ) denote the solution to the initial value problem. Then, with the step size 0.1, values of x will be x 0 = 0 , x 1 = 0.1 , x 2 = 0.2 , x 3 = 0.3 , x 4 = 0.4 , x 5 = 0.5 , x 6 = 0.6 , x 7 = 0.7 , x 8 = 0.8 , x 9 = 0.9 and x 10 = 1 . The first approximation y 1 for n = 0 will be, y 1 = y 0 + 0.1 ( y 0 2 − 3 y 0 + 2 ) = 1.5 + 0.1 ( ( 1.5 ) 2 − 3 ( 1.5 ) + 2 ) ( ∵ y 0 = 1.5 ) = 1.5 + 0.1 ( − 0.25 ) = 1.475 The second approximation y 2 for n = 1 will be, y 2 = y 1 + 0.1 ( y 1 2 − 3 y 1 + 2 ) = 1.475 + 0.1 ( ( 1.475 ) 2 − 3 ( 1.475 ) + 2 ) ( ∵ y 1 = 1.475 ) = 1.475 + 0.1 ( − 0.0249375 ) ≈ 1.450 The third approximation y 3 for n = 2 will be, y 3 = y 2 + 0.1 ( y 2 2 − 3 y 2 + 2 ) = 1.450 + 0.1 ( ( 1.450 ) 2 − 3 ( 1.450 ) + 2 ) ( ∵ y 2 = 1.450 ) = 1.450 + 0.1 ( − 0.2475 ) ≈ 1.425 The fourth approximation y 4 for n = 3 will be, y 4 = y 3 + 0.1 ( y 3 2 − 3 y 3 + 2 ) = 1.425 + 0.1 ( ( 1.425 ) 2 − 3 ( 1.425 ) + 2 ) ( ∵ y 3 = 1.425 ) = 1.425 + 0.1 ( − 0.244375 ) ≈ 1.401 The fifth approximation y 5 for n = 4 will be, y 5 = y 4 + 0.1 ( y 4 2 − 3 y 4 + 2 ) = 1.401 + 0.1 ( ( 1.401 ) 2 − 3 ( 1.401 ) + 2 ) ( ∵ y 4 = 1.401 ) = 1.401 + 0.1 ( − 0.240199 ) ≈ 1.377 The sixth approximation y 6 for n = 5 will be, y 6 = y 5 + 0.1 ( y 5 2 − 3 y 5 + 2 ) = 1.377 + 0.1 ( ( 1.377 ) 2 − 3 ( 1.377 ) + 2 ) ( ∵ y 5 = 1.377 ) = 1.377 + 0.1 ( − 0.234871 ) ≈ 1.353 The seventh approximation y 7 for n = 6 will be, y 7 = y 6 + 0.1 ( y 6 2 − 3 y 6 + 2 ) = 1.353 + 0.1 ( ( 1.353 ) 2 − 3 ( 1.353 ) + 2 ) ( ∵ y 6 = 1.353 ) = 1.353 + 0.1 ( − 0.228391 ) ≈ 1.331 The eighth approximation y 8 for n = 7 will be, y 8 = y 7 + 0.1 ( y 7 2 − 3 y 7 + 2 ) = 1.331 + 0.1 ( ( 1.331 ) 2 − 3 ( 1.331 ) + 2 ) ( ∵ y 7 = 1.331 ) = 1.331 + 0.1 ( − 0.221439 ) ≈ 1.308 The ninth approximation y 9 for n = 8 will be, y 9 = y 8 + 0.1 ( y 8 2 − 3 y 8 + 2 ) = 1.308 + 0.1 ( ( 1.308 ) 2 − 3 ( 1.308 ) + 2 ) ( ∵ y 8 = 1.308 ) = 1.308 + 0.1 ( − 0.213136 ) ≈ 1.287 The tenth approximation y 10 for n = 9 will be, y 10 = y 9 + 0.1 ( y 9 2 − 3 y 9 + 2 ) = 1.287 + 0.1 ( ( 1.287 ) 2 − 3 ( 1.287 ) + 2 ) ( ∵ y 9 = 1.287 ) = 1.287 + 0.1 ( − 0.204631 ) ≈ 1.267 The approximations are tabulated as follows: x n y n 0.1 1.475 0.2 1.450 0.3 1.425 0.4 1.401 0.5 1.377 0.6 1.353 0.7 1.331 0.8 1.308 0.9 1.287 1.0 1.267 Therefore, the approximation of ϕ ( x ) using Euler’s method of step size 0.1 is 1.267 _ . (c) The formula for ϕ ( x ) and sketch ϕ ( x ) on the direction field of the differential equation d y d x = y 2 − 3 y + 2 . The formula for ϕ ( x ) is ϕ ( x ) = 2 + e x 1 + e x _ . The given initial value problem is, d y d x = y 2 − 3 y + 2 , y ( 0 ) = 1.5 . Factorize y 2 − 3 y + 2 as, y 2 − 3 y + 2 = y 2 − 2 y − y + 2 = y ( y − 2 ) − 1 ( y − 2 ) = ( y − 2 ) ( y − 1 ) Thus, the differential equation d y d x = y 2 − 3 y + 2 becomes, d y d x = ( y − 2 ) ( y − 1 ) d y ( y − 2 ) ( y − 1 ) = d x Express 1 ( y − 2 ) ( y − 1 ) in the partial fraction form A y − 2 + B y − 1 and solve for values of A and B as follows: 1 ( y − 2 ) ( y − 1 ) = A y − 2 + B y − 1 ( y − 2 ) ( y − 1 ) ( y − 2 ) ( y − 1 ) = A y − 2 ( y − 2 ) ( y − 1 ) + B y − 1 ( y − 2 ) ( y − 1 ) ( multiply by ( y − 2 ) ( y − 1 ) on both sides ) 1 = A ( y − 1 ) + B ( y − 2 ) 1 = A y − A + B y − 2 B 1 = ( A + B ) y + ( − A − 2 B ) Comparing the coefficients of the equation 1 = ( A + B ) y + ( − A − 2 B ) , it is observed that A + B = 0 and − A − 2 B = 1 . From the equation A + B = 0 , the value of A will be A = − B . Substitute A = − B in − A − 2 B = 1 and simplify as, − ( − B ) − 2 B = 1 B − 2 B = 1 − B = 1 B = − 1 Thus, A = − B becomes A = 1 . The form A y − 2 + B y − 1 becomes 1 y − 2 + − 1 y − 1 . That is, the partial fraction form of 1 ( y − 2 ) ( y − 1 ) will be 1 y − 2 + − 1 y − 1 . Thus, the equation d y ( y − 2 ) ( y − 1 ) = d x becomes, ( 1 y − 2 + − 1 y − 1 ) d y = d x . Since the form ( 1 y − 2 + − 1 y − 1 ) d y = d x is variable separable, integrate on both sides as, ∫ ( 1 y − 2 + − 1 y − 1 ) d y = ∫ d x ln | y − 2 | − ln | y − 1 | = x + C ln | y − 2 y − 1 | = x + C ( ∵ ln a b = ln a − ln b ) Apply the initial condition y ( 0 ) = 1.5 to obtain the value of the constant C as follows: Substitute x = 0 and y ( 0 ) = 1.5 in ln | y − 2 y − 1 | = x + C , ln | 1.5 − 2 1.5 − 1 | = 0 + C ln | − 0.5 0.5 | = C ln | − 1 | = C ln 1 = C C = 0 Thus, ln | y − 2 y − 1 | = x + C becomes, ln | y − 2 y − 1 | = x | y − 2 y − 1 | = e x ( taking exponential on both sides ) y − 2 y − 1 = ± e x ( ∵ e ln x = x ) y − 2 = ± ( y − 1 ) e x On further simplifications, the required y value is obtained as, y − 2 = ± y e x ∓ e x y ∓ y e x = 2 ∓ e x y ( 1 ∓ e x ) = 2 ∓ e x y = 2 ∓ e x 1 ∓ e x Since the initial condition is y ( 0 ) = 1.5 , substituting x = 0 and y ( 0 ) = 1.5 gives, 1.5 = 2 ∓ e 0 1 ∓ e 0 1.5 = 2 ∓ 1 1 ∓ 1 The equation will be satisfied for 1.5 = 2 + 1 1 + 1 . Thus, y = 2 + e x 1 + e x . Assume that, y = ϕ ( x ) denote the solution to the initial value problem. Then, ϕ ( x ) = 2 + e x 1 + e x . Therefore, the formula for ϕ ( x ) is ϕ ( x ) = 2 + e x 1 + e x _ . Use online graphing calculator and sketch ϕ ( x ) = 2 + e x 1 + e x on the direction field for the differential equation d y d x = y 2 − 3 y + 2 as shown below in Figure 2. From Figure 2, it is observed that as x → ∞ , the value of ϕ ( x ) = 2 + e x 1 + e x approaches 1. (d) The exact value of ϕ ( 1 ) and compare with the approximation value from part (b). The exact value of ϕ ( 1 ) is 1.2689 _ . From part (c), the formula for ϕ ( x ) is ϕ ( x ) = 2 + e x 1 + e x . Substitute x = 1 to obtain the value of ϕ ( 1 ) as, ϕ ( 1 ) = 2 + e 1 1 + e 1 = 2 + e 1 + e = 4.718281828 4.718281828 ≈ 1.2689 Therefore, the exact value of ϕ ( 1 ) is 1.2689 _ . From part (b), the approximation of ϕ ( x ) using Euler’s method of step size 0.1 is 1.267 . Thus, the error will be 1.2689 − 1.267 ≈ 0.0019 . That is, Euler’s method of step size 0.1 gives an error of 0.0019 difference from the exact value. (e) The value of lim x → ∞ ϕ ( x ) using the exact solution and compare the initial guess for lim x → ∞ ϕ ( x ) . The value of lim x → ∞ ϕ ( x ) using the exact solution is 1 _ . From part (c), the formula for ϕ ( x ) is ϕ ( x ) = 2 + e x 1 + e x . Obtain the value of lim x → ∞ ϕ ( x ) as follows: lim x → ∞ ϕ ( x ) = lim x → ∞ 2 + e x 1 + e x = 2 + e ∞ 1 + e ∞ = ∞ ∞ Since the indeterminate form ∞ ∞ is obtained, L’Hopital’s rule is applied to find the limit value. Differentiate numerator and denominator of lim x → ∞ 2 + e x 1 + e x and apply the limits as, lim x → ∞ e x e x = lim x → ∞ 1 = 1 That is, as x → ∞ , the value of ϕ ( x ) tends to 1. From part (a), using the slope field, it is concluded that lim x → ∞ ϕ ( x ) = 1 . That is, the initial guess was correct.

Solution Summary: The recursive formula for the Euler's method is: lx_n+1=

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

See similar textbooks

Concept explainers

Differential Equations

Have you ever observed the movement of the satellites and rockets or how the planets go around the sun in their orbits? How will you determine their motion? Definitely they follow some scientific laws and these kinds of problems are framed as differentia…

Videos

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

Blurred answer

Chapter 2.2, Problem 31E

Chapter 2.2, Problem 33E

Students have asked these similar questions

The statistics problem is on page 95. Compute the mean (average) of the given dataset. Link: [https://drive.google.com/file/d/1RQ2OZK-LSxpRyejKEMg 1t2q15dbpVLCS/view? usp=sharing] Show your calculations clearly.

Refer to page 118 for the optimization problem. Solve the constrained optimization problem using the Lagrangian method. Link: [https://drive.google.com/file/d/1RQ2OZk-LSxpRyejKEMg1t2q15dbpVLCS/view? usp=sharing] Provide all derivations and interpretations of your solution.

Refer to page 113 of the document for the vector calculus problem. Verify the divergence theorem for the given vector field and surface. Link: [https://drive.google.com/file/d/1RQ2OZK-LSxpRyejKEMg1t2q15dbpVLCS/view? usp=sharing] Show the calculation for both sides of the theorem.

Chapter 2 Solutions

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

Knowledge Booster

Background pattern image

Math

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, subject and related others by exploring similar questions and additional content below.

Similar questions

SEE MORE QUESTIONS

Recommended textbooks for you

Advanced Engineering Mathematics
Advanced Math
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Wiley, John & Sons, Incorporated
Numerical Methods for Engineers
Advanced Math
ISBN:9780073397924
Author:Steven C. Chapra Dr., Raymond P. Canale
Publisher:McGraw-Hill Education
Introductory Mathematics for Engineering Applicat...
Advanced Math
ISBN:9781118141809
Author:Nathan Klingbeil
Publisher:WILEY
Mathematics For Machine Technology
Advanced Math
ISBN:9781337798310
Author:Peterson, John.
Publisher:Cengage Learning,
Basic Technical Mathematics
Advanced Math
ISBN:9780134437705
Author:Washington
Publisher:PEARSON
Topology
Advanced Math
ISBN:9780134689517
Author:Munkres, James R.
Publisher:Pearson,

Text book image

Advanced Engineering Mathematics

Advanced Math

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Wiley, John & Sons, Incorporated

Text book image

Numerical Methods for Engineers

Advanced Math

ISBN:9780073397924

Author:Steven C. Chapra Dr., Raymond P. Canale

Publisher:McGraw-Hill Education

Text book image

Introductory Mathematics for Engineering Applicat...

Advanced Math

ISBN:9781118141809

Author:Nathan Klingbeil

Publisher:WILEY

Text book image

Mathematics For Machine Technology

Advanced Math

ISBN:9781337798310

Author:Peterson, John.

Publisher:Cengage Learning,

Text book image

Basic Technical Mathematics

Advanced Math

ISBN:9780134437705

Author:Washington

Publisher:PEARSON

Text book image

Advanced Math

ISBN:9780134689517

Author:Munkres, James R.

Publisher:Pearson,

SEE MORE TEXTBOOKS

01 - What Is A Differential Equation in Calculus? Learn to Solve Ordinary Differential Equations.; Author: Math and Science;https://www.youtube.com/watch?v=K80YEHQpx9g;License: Standard YouTube License, CC-BY

Higher Order Differential Equation with constant coefficient (GATE) (Part 1) l GATE 2018; Author: GATE Lectures by Dishank;https://www.youtube.com/watch?v=ODxP7BbqAjA;License: Standard YouTube License, CC-BY

Solution of Differential Equations and Initial Value Problems; Author: Jefril Amboy;https://www.youtube.com/watch?v=Q68sk7XS-dc;License: Standard YouTube License, CC-BY