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

(a) To show: That the general solution to d y d x + P ( x ) y = Q ( x ) has the form y ( x ) = C y h ( x ) + y p ( x ) where y h ( x ) ( ≡ 0 ) is a solution of that equation when Q ( x ) ≡ 0 and y p ( x ) = v ( x ) y h ( x ) . The given equation (20) is d y d x + P ( x ) y = Q ( x ) . Take μ ( x ) = e ∫ P ( x ) d x and differentiate it with respect to x . d μ d x = d d x ( e ∫ P ( x ) d x ) = d d x ( ∫ P ( x ) d x ) e ∫ P ( x ) d x = P ( x ) e ∫ P ( x ) d x = P ( x ) μ ( x ) Take y h ( x ) = μ − 1 ( x ) and differentiate it with respect to x . d ( y h ( x ) ) d x = − μ − 2 ( x ) d μ ( x ) d x d y h d x = − μ − 2 ( x ) d μ d x d y h d x = − μ − 2 ( x ) P ( x ) μ ( x ) ( ∵ d μ d x = P ( x ) μ ( x ) ) d y h d x = − μ − 1 ( x ) P ( x ) Simplify further, d y h d x = − y h P ( x ) ( ∵ μ − 1 ( x ) = y h ) d y h d x + y h P ( x ) = 0 Thus, the solution y h ( x ) = μ − 1 ( x ) satisfy the equation when Q ( x ) ≡ 0 . It is know that, the solution (8) to the linear equation d y d x + P ( x ) y = Q ( x ) is y ( x ) = 1 μ ( x ) [ ∫ μ ( x ) Q ( x ) d x + C ] . Assume that, the solution to the differential equation d y d x + P ( x ) y = Q ( x ) has the form y ( x ) = C y h ( x ) + y p ( x ) . Then it is enough to show that y p ( x ) = y h ( x ) v ( x ) . Thus, the solution becomes C y h ( x ) + y p ( x ) = 1 μ ( x ) ∫ μ ( x ) Q ( x ) d x + 1 μ ( x ) C C y h ( x ) + y p ( x ) = 1 μ ( x ) ∫ μ ( x ) Q ( x ) d x + C μ − 1 ( x ) C y h ( x ) + y p ( x ) = μ − 1 ( x ) ∫ μ ( x ) Q ( x ) d x + C μ − 1 ( x ) C y h ( x ) + y p ( x ) = y h ( x ) ∫ μ ( x ) Q ( x ) d x + C y h ( x ) ( ∵ μ − 1 ( x ) = y h ( x ) ) Simplify further, y p ( x ) = y h ( x ) ∫ μ ( x ) Q ( x ) d x y p ( x ) = y h ( x ) v ( x ) ( Take v = ∫ μ ( x ) Q ( x ) d x ) Thus, the assumption is true for by taking the function v = ∫ μ ( x ) Q ( x ) d x . Hence, it is showed that the solution to the differential equation d y d x + P ( x ) y = Q ( x ) has the form y ( x ) = C y h ( x ) + y p ( x ) where y h ( x ) ( ≡ 0 ) is a solution of that equation when Q ( x ) ≡ 0 and y p ( x ) = v ( x ) y h ( x ) . (b) To find: A nontrivial solution y h to the separable equation d y d x + 3 x y = 0 , x > 0 by using the procedure in part (a). The nontrivial solution is, y h ( x ) = 1 x 3 _ . The linear equation is d y d x + 3 x y = 0 , x > 0 . Here, P ( x ) = 3 x and Q ( x ) = 0 . From part (a), a nontrivial solution, y h ( x ) = μ − 1 ( x ) . Calculate the integrating factor μ ( x ) = e ∫ P ( x ) d x as follows. μ ( x ) = e ∫ 3 x d x = e 3 ln x = e ln x 3 = x 3 Thus, the nontrivial solution is, y h ( x ) = 1 x 3 _ . (c) To obtain: The derivative v ′ ( x ) = x 2 y h ( x ) by assuming equation (21), d y d x + 3 x y = x 2 , x > 0 has a solution of the form y p ( x ) = v ( x ) y h ( x ) and substitute it in equation (21) and then simplify. The value of v ′ ( x ) is x 5 _ . Consider the linear equation d y d x + 3 x y = x 2 , x > 0 . Here, P ( x ) = 3 x and Q ( x ) = x 2 . Assume that the solution of the differential equation d y d x + 3 x y = x 2 , x > 0 is y p ( x ) = v ( x ) y h ( x ) . Substitute y p ( x ) = v ( x ) y h ( x ) in d y d x + 3 x y = x 2 . d [ y p ( x ) ] d x + 3 x y p ( x ) = x 2 d [ v ( x ) y h ( x ) ] d x + 3 x ( v ( x ) y h ( x ) ) = x 2 y h ( x ) d v d x + v ( x ) d y h d x + 3 x [ v ( x ) y h ( x ) ] = x 2 v ′ ( x ) y h ( x ) + v ( x ) [ d y h d x + 3 x y h ( x ) ] = x 2 From part (b), d y h d x + 3 x y h ( x ) = 0 . Thus the above result becomes v ′ ( x ) y h ( x ) + v ( x ) ( 0 ) = x 2 v ′ ( x ) = x 2 y h ( x ) From part (b), y h ( x ) = 1 x 3 . Thus the above result becomes v ′ ( x ) = x 2 1 x 3 = x 2 ⋅ x 3 = x 5 Thus, the value of v ′ ( x ) is x 5 _ . (d) To integrate: The derivative v ′ ( x ) = x 5 The value of v ( x ) is x 6 6 _ . Consider the derivative v ′ ( x ) = x 5 . That is, d v d x = x 5 Separate the variables and rewrite the equation as follows, d v = x 5 d x Integrate both sides of the equation. ∫ d v = ∫ x 5 d x v = x 6 6 + C 1 v = x 6 6 ( Take C 1 = 0 ) Thus, the value of v ( x ) is x 6 6 _ . (e) To verify: That y ( x ) = C y h ( x ) + v ( x ) y h ( x ) is a general solution to the equation (21), d y d x + 3 x y = x 2 , x > 0 . The solution of the differential equation d y d x + 3 x y = x 2 , x > 0 is y = C x − 3 + x 3 6 _ . Consider the function y ( x ) = C y h ( x ) + v ( x ) y h ( x ) . From part (b), and (d), the value of y h ( x ) = 1 x 3 and v ( x ) = x 6 6 . Substitute the values y h ( x ) = 1 x 3 and v ( x ) = x 6 6 in the function y ( x ) = C y h ( x ) + v ( x ) y h ( x ) . y ( x ) = C x 3 + x 6 6 ⋅ 1 x 3 = C x − 3 + x 3 6 The differential equation (21) is, d y d x + 3 x y = x 2 , x > 0 . Now, check whether the function y = C x − 3 + x 3 6 satisfy a linear equation d y d x + 3 x y = x 2 , x > 0 by substituting y = C x − 3 + x 3 6 in the left side of the above linear equation. d y d x + 3 x y = d d x ( C x − 3 + x 3 6 ) + 3 x ( C x − 3 + x 3 6 ) = − 3 C x − 4 + 3 x 2 6 + 3 C x − 3 x + 3 x 3 6 x = − 3 C x − 4 + x 2 2 + 3 C x − 4 + x 2 2 = x 2 Thus, the solution y = C x − 3 + x 3 6 is true for the differential equation d y d x + 3 x y = x 2 , x > 0 . Therefore, the solution of the differential equation d y d x + 3 x y = x 2 , x > 0 is y = C x − 3 + x 3 6 _ .

(a) To show: That the general solution to d y d x + P ( x ) y = Q ( x ) has the form y ( x ) = C y h ( x ) + y p ( x ) where y h ( x ) ( ≡ 0 ) is a solution of that equation when Q ( x ) ≡ 0 and y p ( x ) = v ( x ) y h ( x ) . The given equation (20) is d y d x + P ( x ) y = Q ( x ) . Take μ ( x ) = e ∫ P ( x ) d x and differentiate it with respect to x . d μ d x = d d x ( e ∫ P ( x ) d x ) = d d x ( ∫ P ( x ) d x ) e ∫ P ( x ) d x = P ( x ) e ∫ P ( x ) d x = P ( x ) μ ( x ) Take y h ( x ) = μ − 1 ( x ) and differentiate it with respect to x . d ( y h ( x ) ) d x = − μ − 2 ( x ) d μ ( x ) d x d y h d x = − μ − 2 ( x ) d μ d x d y h d x = − μ − 2 ( x ) P ( x ) μ ( x ) ( ∵ d μ d x = P ( x ) μ ( x ) ) d y h d x = − μ − 1 ( x ) P ( x ) Simplify further, d y h d x = − y h P ( x ) ( ∵ μ − 1 ( x ) = y h ) d y h d x + y h P ( x ) = 0 Thus, the solution y h ( x ) = μ − 1 ( x ) satisfy the equation when Q ( x ) ≡ 0 . It is know that, the solution (8) to the linear equation d y d x + P ( x ) y = Q ( x ) is y ( x ) = 1 μ ( x ) [ ∫ μ ( x ) Q ( x ) d x + C ] . Assume that, the solution to the differential equation d y d x + P ( x ) y = Q ( x ) has the form y ( x ) = C y h ( x ) + y p ( x ) . Then it is enough to show that y p ( x ) = y h ( x ) v ( x ) . Thus, the solution becomes C y h ( x ) + y p ( x ) = 1 μ ( x ) ∫ μ ( x ) Q ( x ) d x + 1 μ ( x ) C C y h ( x ) + y p ( x ) = 1 μ ( x ) ∫ μ ( x ) Q ( x ) d x + C μ − 1 ( x ) C y h ( x ) + y p ( x ) = μ − 1 ( x ) ∫ μ ( x ) Q ( x ) d x + C μ − 1 ( x ) C y h ( x ) + y p ( x ) = y h ( x ) ∫ μ ( x ) Q ( x ) d x + C y h ( x ) ( ∵ μ − 1 ( x ) = y h ( x ) ) Simplify further, y p ( x ) = y h ( x ) ∫ μ ( x ) Q ( x ) d x y p ( x ) = y h ( x ) v ( x ) ( Take v = ∫ μ ( x ) Q ( x ) d x ) Thus, the assumption is true for by taking the function v = ∫ μ ( x ) Q ( x ) d x . Hence, it is showed that the solution to the differential equation d y d x + P ( x ) y = Q ( x ) has the form y ( x ) = C y h ( x ) + y p ( x ) where y h ( x ) ( ≡ 0 ) is a solution of that equation when Q ( x ) ≡ 0 and y p ( x ) = v ( x ) y h ( x ) . (b) To find: A nontrivial solution y h to the separable equation d y d x + 3 x y = 0 , x > 0 by using the procedure in part (a). The nontrivial solution is, y h ( x ) = 1 x 3 _ . The linear equation is d y d x + 3 x y = 0 , x > 0 . Here, P ( x ) = 3 x and Q ( x ) = 0 . From part (a), a nontrivial solution, y h ( x ) = μ − 1 ( x ) . Calculate the integrating factor μ ( x ) = e ∫ P ( x ) d x as follows. μ ( x ) = e ∫ 3 x d x = e 3 ln x = e ln x 3 = x 3 Thus, the nontrivial solution is, y h ( x ) = 1 x 3 _ . (c) To obtain: The derivative v ′ ( x ) = x 2 y h ( x ) by assuming equation (21), d y d x + 3 x y = x 2 , x > 0 has a solution of the form y p ( x ) = v ( x ) y h ( x ) and substitute it in equation (21) and then simplify. The value of v ′ ( x ) is x 5 _ . Consider the linear equation d y d x + 3 x y = x 2 , x > 0 . Here, P ( x ) = 3 x and Q ( x ) = x 2 . Assume that the solution of the differential equation d y d x + 3 x y = x 2 , x > 0 is y p ( x ) = v ( x ) y h ( x ) . Substitute y p ( x ) = v ( x ) y h ( x ) in d y d x + 3 x y = x 2 . d [ y p ( x ) ] d x + 3 x y p ( x ) = x 2 d [ v ( x ) y h ( x ) ] d x + 3 x ( v ( x ) y h ( x ) ) = x 2 y h ( x ) d v d x + v ( x ) d y h d x + 3 x [ v ( x ) y h ( x ) ] = x 2 v ′ ( x ) y h ( x ) + v ( x ) [ d y h d x + 3 x y h ( x ) ] = x 2 From part (b), d y h d x + 3 x y h ( x ) = 0 . Thus the above result becomes v ′ ( x ) y h ( x ) + v ( x ) ( 0 ) = x 2 v ′ ( x ) = x 2 y h ( x ) From part (b), y h ( x ) = 1 x 3 . Thus the above result becomes v ′ ( x ) = x 2 1 x 3 = x 2 ⋅ x 3 = x 5 Thus, the value of v ′ ( x ) is x 5 _ . (d) To integrate: The derivative v ′ ( x ) = x 5 The value of v ( x ) is x 6 6 _ . Consider the derivative v ′ ( x ) = x 5 . That is, d v d x = x 5 Separate the variables and rewrite the equation as follows, d v = x 5 d x Integrate both sides of the equation. ∫ d v = ∫ x 5 d x v = x 6 6 + C 1 v = x 6 6 ( Take C 1 = 0 ) Thus, the value of v ( x ) is x 6 6 _ . (e) To verify: That y ( x ) = C y h ( x ) + v ( x ) y h ( x ) is a general solution to the equation (21), d y d x + 3 x y = x 2 , x > 0 . The solution of the differential equation d y d x + 3 x y = x 2 , x > 0 is y = C x − 3 + x 3 6 _ . Consider the function y ( x ) = C y h ( x ) + v ( x ) y h ( x ) . From part (b), and (d), the value of y h ( x ) = 1 x 3 and v ( x ) = x 6 6 . Substitute the values y h ( x ) = 1 x 3 and v ( x ) = x 6 6 in the function y ( x ) = C y h ( x ) + v ( x ) y h ( x ) . y ( x ) = C x 3 + x 6 6 ⋅ 1 x 3 = C x − 3 + x 3 6 The differential equation (21) is, d y d x + 3 x y = x 2 , x > 0 . Now, check whether the function y = C x − 3 + x 3 6 satisfy a linear equation d y d x + 3 x y = x 2 , x > 0 by substituting y = C x − 3 + x 3 6 in the left side of the above linear equation. d y d x + 3 x y = d d x ( C x − 3 + x 3 6 ) + 3 x ( C x − 3 + x 3 6 ) = − 3 C x − 4 + 3 x 2 6 + 3 C x − 3 x + 3 x 3 6 x = − 3 C x − 4 + x 2 2 + 3 C x − 4 + x 2 2 = x 2 Thus, the solution y = C x − 3 + x 3 6 is true for the differential equation d y d x + 3 x y = x 2 , x > 0 . Therefore, the solution of the differential equation d y d x + 3 x y = x 2 , x > 0 is y = C x − 3 + x 3 6 _ .

Solution Summary: The author explains that the general solution to dy dx+P(x))y=Q left (x

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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)5. The operator T on a vector space V over Q has as its sequence of invariant factors the sequence (x-1)3(x-2)5(x-3)2, (x-1)(x-2)(x-3)2, (x-3). Find the Jordan normal form of T.

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Chapter 2 Solutions

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. 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