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To verify: The term x ( 1 − a ) 2 ln x is a solution of x 2 y ″ + a x y ′ + b y = 0 if it has double root and x m 1 ln x and x m 2 ln x are not solutions if x 2 y ″ + a x y ′ + b y = 0 has different roots m 1 and m 2 . Consider the Euler-Cauchy differential equation x 2 y ″ + a x y ′ + b y = 0 . The auxiliary equation is m 2 + ( a − 1 ) m + b = 0 If the equation has double root, then the discriminant value becomes zero. That is, ( a − 1 ) 2 − 4 b = 0 ⇒ b = ( a − 1 ) 2 4 . Now verify y = x ( 1 − a ) 2 ln x is a solution of x 2 y ″ + a x y ′ + b y = 0 as follows. x 2 y ″ + a x y ′ + b y = x 2 d d x 2 ( x ( 1 − a ) 2 ln x ) + a x d d x ( x ( 1 − a ) 2 ln x ) + b ( x ( 1 − a ) 2 ln x ) = { x 2 ( d d x ( ( 1 − a ) x ( − 1 − a ) 2 ln x 2 + x ( − 1 − a ) 2 ) ) + a x ( ( 1 − a ) x ( − 1 − a ) 2 ln x 2 + x ( − 1 − a ) 2 ) + b ( x ( 1 − a ) 2 ln x ) } = { x 2 ( ( a 2 − 1 ) x ( − a − 3 ) 2 ln x 4 + ( 1 − a ) x ( − 3 − a ) 2 2 + ( − 1 − a ) x ( − 3 − a ) 2 2 ) + a x ( ( 1 − a ) x ( − 1 − a ) 2 ln x 2 + x ( − 1 − a ) 2 ) + b ( x ( 1 − a ) 2 ln x ) } On further simplifications, the following is obtained. x 2 y ″ + a x y ′ + b y = { x 2 ( ( a 2 − 1 ) x ( − a − 3 ) 2 ln x 4 − a x ( − 3 − a ) 2 ) + a x ( ( 1 − a ) x ( − 1 − a ) 2 ln x 2 + x ( − 1 − a ) 2 ) + ( a − 1 ) 2 4 ( x ( 1 − a ) 2 ln x ) } = { ( a 2 − 1 ) x ( 1 − a ) 2 ln x 4 − a x ( 1 − a ) 2 + a ( 1 − a ) x ( 1 − a ) 2 ln x 2 + a x ( 1 − a ) 2 + ( a − 1 ) 2 4 ( x ( 1 − a ) 2 ln x ) } = ( a 2 − 1 ) x ( 1 − a ) 2 ln x 4 + a ( 1 − a ) x ( 1 − a ) 2 ln x 2 + ( a − 1 ) 2 4 ( x ( 1 − a ) 2 ln x ) = 0 Therefore, x ( 1 − a ) 2 ln x is a solution of x 2 y ″ + a x y ′ + b y = 0 if it has double root. Verify y = x m 1 ln x is solutions of x 2 y ″ + a x y ′ + b y = 0 as shown below. x 2 y ″ + a x y ′ + b y = { x 2 ( m 1 ( m 1 − 1 ) x m 1 − 2 ln x + m 1 x m 1 − 2 + ( m 1 − 1 ) x m 1 − 2 ) + a x ( m 1 x m 1 − 1 ln x + x m 1 − 1 ) + b ( x m 1 ln x ) } = { ( m 1 2 x m 1 ln x − m 1 x m 1 ln x + m 1 x m 1 + ( m 1 − 1 ) x m 1 ) + a m 1 x m 1 ln x + a x m 1 + b x m 1 ln x } ≠ 0 Similarly, x 2 y ″ + a x y ′ + b y ≠ 0 for y = x m 2 ln x . Therefore, x m 1 ln x and x m 2 ln x are not solutions of x 2 y ″ + a x y ′ + b y = 0 , if it has different roots m 1 and m 2 .

To verify: The term x ( 1 − a ) 2 ln x is a solution of x 2 y ″ + a x y ′ + b y = 0 if it has double root and x m 1 ln x and x m 2 ln x are not solutions if x 2 y ″ + a x y ′ + b y = 0 has different roots m 1 and m 2 . Consider the Euler-Cauchy differential equation x 2 y ″ + a x y ′ + b y = 0 . The auxiliary equation is m 2 + ( a − 1 ) m + b = 0 If the equation has double root, then the discriminant value becomes zero. That is, ( a − 1 ) 2 − 4 b = 0 ⇒ b = ( a − 1 ) 2 4 . Now verify y = x ( 1 − a ) 2 ln x is a solution of x 2 y ″ + a x y ′ + b y = 0 as follows. x 2 y ″ + a x y ′ + b y = x 2 d d x 2 ( x ( 1 − a ) 2 ln x ) + a x d d x ( x ( 1 − a ) 2 ln x ) + b ( x ( 1 − a ) 2 ln x ) = { x 2 ( d d x ( ( 1 − a ) x ( − 1 − a ) 2 ln x 2 + x ( − 1 − a ) 2 ) ) + a x ( ( 1 − a ) x ( − 1 − a ) 2 ln x 2 + x ( − 1 − a ) 2 ) + b ( x ( 1 − a ) 2 ln x ) } = { x 2 ( ( a 2 − 1 ) x ( − a − 3 ) 2 ln x 4 + ( 1 − a ) x ( − 3 − a ) 2 2 + ( − 1 − a ) x ( − 3 − a ) 2 2 ) + a x ( ( 1 − a ) x ( − 1 − a ) 2 ln x 2 + x ( − 1 − a ) 2 ) + b ( x ( 1 − a ) 2 ln x ) } On further simplifications, the following is obtained. x 2 y ″ + a x y ′ + b y = { x 2 ( ( a 2 − 1 ) x ( − a − 3 ) 2 ln x 4 − a x ( − 3 − a ) 2 ) + a x ( ( 1 − a ) x ( − 1 − a ) 2 ln x 2 + x ( − 1 − a ) 2 ) + ( a − 1 ) 2 4 ( x ( 1 − a ) 2 ln x ) } = { ( a 2 − 1 ) x ( 1 − a ) 2 ln x 4 − a x ( 1 − a ) 2 + a ( 1 − a ) x ( 1 − a ) 2 ln x 2 + a x ( 1 − a ) 2 + ( a − 1 ) 2 4 ( x ( 1 − a ) 2 ln x ) } = ( a 2 − 1 ) x ( 1 − a ) 2 ln x 4 + a ( 1 − a ) x ( 1 − a ) 2 ln x 2 + ( a − 1 ) 2 4 ( x ( 1 − a ) 2 ln x ) = 0 Therefore, x ( 1 − a ) 2 ln x is a solution of x 2 y ″ + a x y ′ + b y = 0 if it has double root. Verify y = x m 1 ln x is solutions of x 2 y ″ + a x y ′ + b y = 0 as shown below. x 2 y ″ + a x y ′ + b y = { x 2 ( m 1 ( m 1 − 1 ) x m 1 − 2 ln x + m 1 x m 1 − 2 + ( m 1 − 1 ) x m 1 − 2 ) + a x ( m 1 x m 1 − 1 ln x + x m 1 − 1 ) + b ( x m 1 ln x ) } = { ( m 1 2 x m 1 ln x − m 1 x m 1 ln x + m 1 x m 1 + ( m 1 − 1 ) x m 1 ) + a m 1 x m 1 ln x + a x m 1 + b x m 1 ln x } ≠ 0 Similarly, x 2 y ″ + a x y ′ + b y ≠ 0 for y = x m 2 ln x . Therefore, x m 1 ln x and x m 2 ln x are not solutions of x 2 y ″ + a x y ′ + b y = 0 , if it has different roots m 1 and m 2 .

Advanced Engineering Mathematics

Advanced Engineering Mathematics

10th Edition

ISBN: 9780470458365

Author: Erwin Kreyszig

Publisher: Wiley, John & Sons, Incorporated

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Chapter 2.4, Problem 20P

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Advanced Engineering Mathematics

Ch. 2.1 - Prob. 1P Ch. 2.1 - Reduction. Show that F(y, y′, y″) = 0 can be...Ch. 2.1 - 3–10 REDUCTION OF ORDER Reduce to first order and...Ch. 2.1 - 3–10 REDUCTION OF ORDER Reduce to first order and...Ch. 2.1 - 3–10 REDUCTION OF ORDER Reduce to first order and...Ch. 2.1 - 3–10 REDUCTION OF ORDER Reduce to first order and...Ch. 2.1 - 3–10 REDUCTION OF ORDER Reduce to first order and...Ch. 2.1 - 3–10 REDUCTION OF ORDER Reduce to first order and...Ch. 2.1 - 3–10 REDUCTION OF ORDER Reduce to first order and...Ch. 2.1 - 3–10 REDUCTION OF ORDER Reduce to first order and...

Ch. 2.1 - 11–14 APPLICATIONS OF REDUCIBLE...Ch. 2.1 - 11–14 APPLICATIONS OF REDUCIBLE ODEs 12. Hanging...Ch. 2.1 - APPLICATIONS OF REDUCIBLE ODEs 13. Motion. If, in...Ch. 2.1 - Motion. In a straight-line motion, let the...Ch. 2.1 - GENERAL SOLUTION. INITIAL VALUE PROBLEM...Ch. 2.1 - GENERAL SOLUTION. INITIAL VALUE PROBLEM...Ch. 2.1 - GENERAL SOLUTION. INITIAL VALUE PROBLEM...Ch. 2.1 - GENERAL SOLUTION. INITIAL VALUE PROBLEM...Ch. 2.1 - GENERAL SOLUTION. INITIAL VALUE PROBLEM...Ch. 2.2 - 1–15 GENERAL SOLUTION Find a general solution....Ch. 2.2 - 1–15 GENERAL SOLUTION Find a general solution....Ch. 2.2 - 1–15 GENERAL SOLUTION Find a general solution....Ch. 2.2 - Find a general solution. Check your answer by...Ch. 2.2 - Find a general solution. Check your answer by...Ch. 2.2 - Find a general solution. Check your answer by...Ch. 2.2 - 1–15 GENERAL SOLUTION Find a general solution....Ch. 2.2 - Find a general solution. Check your answer by...Ch. 2.2 - 1–15 GENERAL SOLUTION Find a general solution....Ch. 2.2 - GENERAL SOLUTION Find a general solution. Check...Ch. 2.2 - 1–15 GENERAL SOLUTION Find a general solution....Ch. 2.2 - 1–15 GENERAL SOLUTION Find a general solution....Ch. 2.2 - Find a general solution. Check your answer by...Ch. 2.2 - Find a general solution. Check your answer by...Ch. 2.2 - Find a general solution. Check your answer by...Ch. 2.2 - 16–20 FIND AN ODE for the given basis. 16. Ch. 2.2 - 16–20 FIND AN ODE for the given basis. 17. Ch. 2.2 - 16–20 FIND AN ODE for the given basis. 18. Ch. 2.2 - 16–20 FIND AN ODE for the given basis. 19. Ch. 2.2 - 16–20 FIND AN ODE for the given basis. 20. Ch. 2.2 - Solve the IVP. Check that your answer satisfies...Ch. 2.2 - Solve the IVP. Check that your answer satisfies...Ch. 2.2 - Solve the IVP. Check that your answer satisfies...Ch. 2.2 - Solve the IVP. Check that your answer satisfies...Ch. 2.2 - Solve the IVP. Check that your answer satisfies...Ch. 2.2 - Solve the IVP. Check that your answer satisfies...Ch. 2.2 - Solve the IVP. Check that your answer satisfies...Ch. 2.2 - Solve the IVP. Check that your answer satisfies...Ch. 2.2 - INITIAL VALUES PROBLEMS Solve the IVP. Check that...Ch. 2.2 - Solve the IVP. Check that your answer satisfies...Ch. 2.2 - LINEAR INDEPENDENCE is of basic importance, in...Ch. 2.2 - LINEAR INDEPENDENCE is of basic importance, in...Ch. 2.2 - Prob. 33P Ch. 2.2 - Prob. 34P Ch. 2.2 - Prob. 35P Ch. 2.2 - LINEAR INDEPENDENCE is of basic importance, in...Ch. 2.2 - Instability. Solve y″ − y = 0 for the initial...Ch. 2.3 - Apply the given operator to the given functions....Ch. 2.3 - Apply the given operator to the given functions....Ch. 2.3 - Apply the given operator to the given functions....Ch. 2.3 - Prob. 4P Ch. 2.3 - Apply the given operator to the given functions....Ch. 2.3 - Factor as in the text and solve. (D2 + 4.00D +...Ch. 2.3 - Factor as in the text and solve. (4D2 − I)y = 0 Ch. 2.3 - Factor as in the text and solve. (D2 + 3I)y = 0 Ch. 2.3 - Factor as in the text and solve. (D2 − 4.20D +...Ch. 2.3 - Factor as in the text and solve. (D2 + 4.80D +...Ch. 2.3 - Factor as in the text and solve. (D2 − 4.00D +...Ch. 2.3 - Prob. 12P Ch. 2.3 - Linear operator. Illustrate the linearity of L in...Ch. 2.3 - Double root. If D2 + aD + bI has distinct roots μ...Ch. 2.3 - Definition of linearity. Show that the definition...Ch. 2.4 - Initial value problem. Find the harmonic motion...Ch. 2.4 - Frequency. If a weight of 20 nt (about 4.5 lb)...Ch. 2.4 - Frequency. How does the frequency of the harmonic...Ch. 2.4 - Initial velocity. Could you make a harmonic...Ch. 2.4 - Springs in parallel. What are the frequencies of...Ch. 2.4 - Spring in series. If a body hangs on a spring s1...Ch. 2.4 - Prob. 7P Ch. 2.4 - Prob. 8P Ch. 2.4 - HARMONIC OSCILLATIONS (UNDAMPED MOTION) 9....Ch. 2.4 - Prob. 11P Ch. 2.4 - DAMPED MOTION 12. Overdamping. Show that in the...Ch. 2.4 - DAMPED MOTION 13. Initial value problem. Find the...Ch. 2.4 - DAMPED MOTION 14. Shock absorber. What is the...Ch. 2.4 - DAMPED MOTION 15. Frequency. Find an approximation...Ch. 2.4 - DAMPED MOTION 16. Maxima. Show that the maxima of...Ch. 2.4 - DAMPED MOTION 17. Underdamping. Determine the...Ch. 2.4 - DAMPED MOTION 18. Logarithmic decrement. Show that...Ch. 2.4 - DAMPED MOTION 19. Damping constant. Consider an...Ch. 2.5 - Prob. 1P Ch. 2.5 - Find a real general solution. Show the details of...Ch. 2.5 - Find a real general solution. Show the details of...Ch. 2.5 - Find a real general solution. Show the details of...Ch. 2.5 - Find a real general solution. Show the details of...Ch. 2.5 - Find a real general solution. Show the details of...Ch. 2.5 - Find a real general solution. Show the details of...Ch. 2.5 - Find a real general solution. Show the details of...Ch. 2.5 - Prob. 9P Ch. 2.5 - Find a real general solution. Show the details of...Ch. 2.5 - Find a real general solution. Show the details of...Ch. 2.5 - INITIAL VALUE PROBLEM Solve and graph the...Ch. 2.5 - INITIAL VALUE PROBLEM Solve and graph the...Ch. 2.5 - INITIAL VALUE PROBLEM Solve and graph the...Ch. 2.5 - INITIAL VALUE PROBLEM Solve and graph the...Ch. 2.5 - INITIAL VALUE PROBLEM Solve and graph the...Ch. 2.5 - INITIAL VALUE PROBLEM Solve and graph the...Ch. 2.5 - INITIAL VALUE PROBLEM Solve and graph the...Ch. 2.5 - INITIAL VALUE PROBLEM Solve and graph the...Ch. 2.6 - Derive (6*) from (6). Ch. 2.6 - BASIS OF SOLUTIONS. WRONSKIAN Find the Wronskian....Ch. 2.6 - BASIS OF SOLUTIONS. WRONSKIAN Find the Wronskian....Ch. 2.6 - BASIS OF SOLUTIONS. WRONSKIAN Find the Wronskian....Ch. 2.6 - BASIS OF SOLUTIONS. WRONSKIAN Find the Wronskian....Ch. 2.6 - BASIS OF SOLUTIONS. WRONSKIAN Find the Wronskian....Ch. 2.6 - BASIS OF SOLUTIONS. WRONSKIAN Find the Wronskian....Ch. 2.6 - BASIS OF SOLUTIONS. WRONSKIAN Find the Wronskian....Ch. 2.6 - Prob. 9P Ch. 2.6 - ODE FOR GIVEN BASIS. WRONSKIAN. IVP (a) Find a...Ch. 2.6 - ODE FOR GIVEN BASIS. WRONSKIAN. IVP (a) Find a...Ch. 2.6 - ODE FOR GIVEN BASIS. WRONSKIAN. IVP (a) Find a...Ch. 2.6 - ODE FOR GIVEN BASIS. WRONSKIAN. IVP (a) Find a...Ch. 2.6 - ODE FOR GIVEN BASIS. WRONSKIAN. IVP (a) Find a...Ch. 2.6 - ODE FOR GIVEN BASIS. WRONSKIAN. IVP (a) Find a...Ch. 2.7 - NONHOMOGENEOUS LINEAR ODEs: GENERAL SOLUTION Find...Ch. 2.7 - NONHOMOGENEOUS LINEAR ODEs: GENERAL SOLUTION Find...Ch. 2.7 - NONHOMOGENEOUS LINEAR ODEs: GENERAL SOLUTION Find...Ch. 2.7 - NONHOMOGENEOUS LINEAR ODEs: GENERAL SOLUTION Find...Ch. 2.7 - NONHOMOGENEOUS LINEAR ODEs: GENERAL SOLUTION Find...Ch. 2.7 - NONHOMOGENEOUS LINEAR ODEs: GENERAL SOLUTION Find...Ch. 2.7 - NONHOMOGENEOUS LINEAR ODEs: GENERAL SOLUTION Find...Ch. 2.7 - NONHOMOGENEOUS LINEAR ODEs: GENERAL SOLUTION Find...Ch. 2.7 - NONHOMOGENEOUS LINEAR ODEs: GENERAL SOLUTION Find...Ch. 2.7 - NONHOMOGENEOUS LINEAR ODEs: GENERAL SOLUTION Find...Ch. 2.7 - NONHOMOGENEOUS LINEAR ODEs: IVPs Solve the initial...Ch. 2.7 - NONHOMOGENEOUS LINEAR ODEs: IVPs Solve the initial...Ch. 2.7 - NONHOMOGENEOUS LINEAR ODEs: IVPs Solve the initial...Ch. 2.7 - NONHOMOGENEOUS LINEAR ODEs: IVPs Solve the initial...Ch. 2.7 - NONHOMOGENEOUS LINEAR ODEs: IVPs Solve the initial...Ch. 2.7 - NONHOMOGENEOUS LINEAR ODEs: IVPs Solve the initial...Ch. 2.7 - NONHOMOGENEOUS LINEAR ODEs: IVPs Solve the initial...Ch. 2.7 - NONHOMOGENEOUS LINEAR ODEs: IVPs Solve the initial...Ch. 2.7 - CAS PROJECT. Structure of Solutions of Initial...Ch. 2.8 - Prob. 2P Ch. 2.8 - Find the steady-state motion of the mass–spring...Ch. 2.8 - Find the steady-state motion of the mass–spring...Ch. 2.8 - Find the steady-state motion of the mass–spring...Ch. 2.8 - Prob. 6P Ch. 2.8 - Prob. 7P Ch. 2.8 - TRANSIENT SOLUTIONS Find the transient motion of...Ch. 2.8 - TRANSIENT SOLUTIONS Find the transient motion of...Ch. 2.8 - TRANSIENT SOLUTIONS Find the transient motion of...Ch. 2.8 - TRANSIENT SOLUTIONS Find the transient motion of...Ch. 2.8 - Prob. 12P Ch. 2.8 - Prob. 13P Ch. 2.8 - TRANSIENT SOLUTIONS Find the transient motion of...Ch. 2.8 - TRANSIENT SOLUTIONS Find the transient motion of...Ch. 2.8 - INITIAL VALUE PROBLEMS Find the motion of the...Ch. 2.8 - Prob. 17P Ch. 2.8 - INITIAL VALUE PROBLEMS Find the motion of the...Ch. 2.8 - Prob. 19P Ch. 2.8 - Prob. 20P Ch. 2.8 - Prob. 21P Ch. 2.8 - Prob. 22P Ch. 2.8 - Prob. 24P Ch. 2.9 - RC-Circuit. Model the RC-circuit in Fig. 64. Find...Ch. 2.9 - RC-Circuit. Solve Prob. 1 when E = E0 sin ωt and...Ch. 2.9 - RL-Circuit. Model the RL-circuit in Fig. 66. Find...Ch. 2.9 - RL-Circuit. Solve Prob. 3 when E = E0 sin ωt and...Ch. 2.9 - LC-Circuit. This is an RLC-circuit with negligibly...Ch. 2.9 - LC-Circuit. Find the current when L = 0.5 H, C =...Ch. 2.9 - Prob. 7P Ch. 2.9 - 8–14 Find the steady-state current in the...Ch. 2.9 - 8–14 Find the steady-state current in the...Ch. 2.9 - 8–14 Find the steady-state current in the...Ch. 2.9 - 8–14 Find the steady-state current in the...Ch. 2.9 - Find the steady-state current in the RLC-circuit...Ch. 2.9 - Find the steady-state current in the RLC-circuit...Ch. 2.9 - Prob. 14P Ch. 2.9 - Prob. 15P Ch. 2.9 - Solve the initial value problem for the...Ch. 2.9 - Prob. 17P Ch. 2.9 - Prob. 18P Ch. 2.9 - Complex Solution Method. Solve , by substituting...Ch. 2.10 - Solve the given nonhomogeneous linear ODE by...Ch. 2.10 - Solve the given nonhomogeneous linear ODE by...Ch. 2.10 - Solve the given nonhomogeneous linear ODE by...Ch. 2.10 - Solve the given nonhomogeneous linear ODE by...Ch. 2.10 - Prob. 5P Ch. 2.10 - Solve the given nonhomogeneous linear ODE by...Ch. 2.10 - Solve the given nonhomogeneous linear ODE by...Ch. 2.10 - Solve the given nonhomogeneous linear ODE by...Ch. 2.10 - Solve the given nonhomogeneous linear ODE by...Ch. 2.10 - Solve the given nonhomogeneous linear ODE by...Ch. 2.10 - Solve the given nonhomogeneous linear ODE by...Ch. 2.10 - Prob. 12P Ch. 2.10 - Solve the given nonhomogeneous linear ODE by...Ch. 2 - Prob. 1RQ Ch. 2 - Prob. 2RQ Ch. 2 - By what methods can you get a general solution of...Ch. 2 - Prob. 4RQ Ch. 2 - Prob. 5RQ Ch. 2 - Prob. 6RQ Ch. 2 - Find a general solution. Show the details of your...Ch. 2 - Find a general solution. Show the details of your...Ch. 2 - Find a general solution. Show the details of your...Ch. 2 - Find a general solution. Show the details of your...Ch. 2 - Find a general solution. Show the details of your...Ch. 2 - Find a general solution. Show the details of your...Ch. 2 - Find a general solution. Show the details of your...Ch. 2 - Find a general solution. Show the details of your...Ch. 2 - Prob. 15RQ Ch. 2 - Prob. 16RQ Ch. 2 - Prob. 17RQ Ch. 2 - Find a general solution. Show the details of your...Ch. 2 - Solve the problem, showing the details of your...Ch. 2 - Solve the problem, showing the details of your...Ch. 2 - Solve the problem, showing the details of your...Ch. 2 - Solve the problem, showing the details of your...Ch. 2 - Find the steady-state current in the RLC-circuit...Ch. 2 - Find a general solution of the homogeneous linear...Ch. 2 - Find the steady-state current in the RLC-circuit...Ch. 2 - Find the current in the RLC-circuit in Fig. 71...Ch. 2 - Prob. 27RQ Ch. 2 - Prob. 28RQ Ch. 2 - Prob. 29RQ Ch. 2 - Prob. 30RQ

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