Explanation Given: 4 y ′′ − 4 y ′ + 10 y = 0 Approach: The differential equation 4 y ′′ − 4 y ′ + 10 y = 0 (1) is homogeneous. The characteristic equation to (1) is given by a r 2 + b r + c = 0 a is a coefficient of y ′′ , b is a coefficient of y ′ and, c is a coefficient of y . Linearly independent solutions of (1) are y 1 ( t ) = e λ t cos ( μ t ) y 2 ( t ) = e λ t sin ( μ t ) ( 2 ) The general solution to (1) is given by y ( t ) = c 1 y 1 ( t ) + c 2 y 2 ( t ) (3) c 1 and c 2 are constants. Calculation: From (1) substitute 4 for a , − 4 for b and 10 for c in the characteristic equation, a r 2 + b r + c = 0 4 r 2 − 4 r + 10 = 0 r = 1 2 ( 1 ± 3 i ) ( 4 ) From (4) substitute 1 2 for λ and 3 2 for μ in in (2)

Fundamentals of Differential Equations and Boundary Value Problems

7th Edition

ISBN: 9780321977106

Author: Nagle, R. Kent

Publisher: Pearson Education, Limited

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One hundred students were surveyed about their preference between dogs and cats. The following two-way table displays data for the sample of students who responded to the survey. Preference Male Female TOTAL Prefers dogs \[36\] \[20\] \[56\] Prefers cats \[10\] \[26\] \[36\] No preference \[2\] \[6\] \[8\] TOTAL \[48\] \[52\] \[100\] problem 1 Find the probability that a randomly selected student prefers dogs.Enter your answer as a fraction or decimal. \[P\left(\text{prefers dogs}\right)=\] Incorrect Check Hide explanation Preference Male Female TOTAL Prefers dogs \[\blueD{36}\] \[\blueD{20}\] \[\blueE{56}\] Prefers cats \[10\] \[26\] \[36\] No preference \[2\] \[6\] \[8\] TOTAL \[48\] \[52\] \[100\] There were \[\blueE{56}\] students in the sample who preferred dogs out of \[100\] total students.

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Fundamentals of Differential Equations and Boundary Value Problems

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To Find:

A general solution to the given differential equation.