#$&* course Mth 174 8/6 12 017. `query 17 Cal 2
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Given Solution: Substitute y = cos(omega * t) into the equation. The goal is to find the value of omega that satisfies the equation: We first calculate y ‘’ y = cos(omega*t) so y' = -omega*sin(omega*t) and y"""" = -omega^2*cos(omega*t) Now substituting in y"""" + 9y = 0 we obtain -omega^2*cos(omega*t) + 9cos(omega*t) = 0 , which can be easily solved for omega: -omega^2*cos(omega*t) = -9cos(omega*t) omega^2 = 9 omega = +3, -3 Both solutions check in the original equation. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: Query problem 11.1.18 (4th edition 11.1.14 3d edition 11.1.13) (formerly 10.6.1) P = 1 / (1 + e^-t) satisfies dP/dt = P(1-P) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: multiplied the numerator and denominator and simplified and substituted. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: RESPONSE --> P= 1/(1+e^-t) ; multiplying numerator and denominator by e^t we have P = e^t/(e^t+1) , which is a form that makes the algebra a little easier. dP/dt = [ (e^t)’ ( e^t + 1) - (e^t + 1) ‘ e^t ] / (e^t + 1)^2, which simplifies to dP/dt = [ e^t ( e^t + 1) - e^t * e^t ] / (e^t+1)^2 = e^t / (e^t + 1)^2. Substituting: P(1-P) = e^t/(e^t+1) * [1 - e^t/(e^t+1)] . We simplify this in order to see if it is the same as our expression for dP/dt: e^t/(e^t+1) * [1 - e^t/(e^t+1)] = e^t / (e^t + 1) - (e^t)^2 / (e^t + 1)^2. Putting the first term into a form with common denominator e^t ( e^t + 1) / [ (e^t + 1) ( e^t + 1) ] - (e^t)^2 / (e^t + 1 ) ^ 2 = (e^(2 t) + e^t) / (e^t + 1)^2 - e^(2 t) / (e^t + 1) ^ 2 = (e^(2 t) + e^t - e^(2 t) ) / (e^t + 1)^2 = e^t / (e^t + 1)^2. This agrees with our expression for dP/dt and we see that dP/dt = P ( 1 - P). " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: Query problem 11.1.18 (4th edition 11.1.14 3d edition 11.1.13) (formerly 10.6.1) P = 1 / (1 + e^-t) satisfies dP/dt = P(1-P) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: multiplied the numerator and denominator and simplified and substituted. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: RESPONSE --> P= 1/(1+e^-t) ; multiplying numerator and denominator by e^t we have P = e^t/(e^t+1) , which is a form that makes the algebra a little easier. dP/dt = [ (e^t)’ ( e^t + 1) - (e^t + 1) ‘ e^t ] / (e^t + 1)^2, which simplifies to dP/dt = [ e^t ( e^t + 1) - e^t * e^t ] / (e^t+1)^2 = e^t / (e^t + 1)^2. Substituting: P(1-P) = e^t/(e^t+1) * [1 - e^t/(e^t+1)] . We simplify this in order to see if it is the same as our expression for dP/dt: e^t/(e^t+1) * [1 - e^t/(e^t+1)] = e^t / (e^t + 1) - (e^t)^2 / (e^t + 1)^2. Putting the first term into a form with common denominator e^t ( e^t + 1) / [ (e^t + 1) ( e^t + 1) ] - (e^t)^2 / (e^t + 1 ) ^ 2 = (e^(2 t) + e^t) / (e^t + 1)^2 - e^(2 t) / (e^t + 1) ^ 2 = (e^(2 t) + e^t - e^(2 t) ) / (e^t + 1)^2 = e^t / (e^t + 1)^2. This agrees with our expression for dP/dt and we see that dP/dt = P ( 1 - P). " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!