Explanation Given information: The differential equation y ″ + a y ′ + b y = 0 such that a and b are positive. Formula used: If m 1 ≠ m 2 are distinct real zeros of the characteristics equation, then the general solution is y = K 1 e m 1 x + K 2 e m 2 x . Proof: The characteristic equation for the differential equation y ″ + a y ′ + b y = 0 is m 2 + a m + b = 0 . The zeros of m 2 + a m + b = 0 can be calculated as: m 2 + a m + b = 0 m = − a ± a 2 − 4 b 2 Thus, the general solution of the given differential equationis y = K 1 e − a + a 2 − 4 b 2 x + K 2 e − a − a 2 − 4 b 2 x . The value of − a 2 ± ( a 2 ) 2 − b is always negative as a and b are positive and the value under the square root is always less than a 2

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Author: Larson

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Chapter 16, Problem 6PS

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To prove: That limx→∞y(x)=0 where y(x) is the solution of the differential equation y″+ay′+by=0 such that a and b are positive.

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Chapter 16, Problem 5PS

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To prove: That lim x → ∞ y ( x ) = 0 where y ( x ) is the solution of the differential equation y ″ + a y ′ + b y = 0 such that a and b are positive.

Solution Summary: The author proves that undersetxto 'inftymathrmlimy(x)=0 is the solution of the differential equation.

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To prove: That limx→∞y(x)=0 where y(x) is the solution of the differential equation y″+ay′+by=0 such that a and b are positive.

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