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Math Advanced Math Fundamentals of Differential Equations (9th Edition)

(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=

Fundamentals of Differential Equations (9th Edition)

Fundamentals of Differential Equations (9th Edition)

9th Edition

ISBN: 9780321977069

Author: R. Kent Nagle, Edward B. Saff, Arthur David Snider

Publisher: PEARSON

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Fundamentals of Differential Equations (9th Edition)

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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