#$&* course MTH 279 Part I: The equation m x '' = - k x*********************************************
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Given Solution: If x = cos(t) then x ' = - sin(t) and x '' = - cos(t). Substituting the expressions for x and x '' into the equation we obtain m * (-cos(t)) = - k * cos(t). Dividing both sides by cos(t) we obtain m = k. If m = k, then the equation is satisfied. If m is not equal to k, it is not. If x = sin(sqrt(k/m) * t) then x ' = sqrt(k / m) cos(sqrt(k/m) * t) and x '' = -k / m sin(sqrt(k/m) * t). Substituting this into the equation we have m * (-k/m sin(sqrt(k/m) * t) ) = -k sin(sqrt(k/m) * t Simplifying both sides we see that the equation is true. The same procedure can and should be used to show that the third equation is true, while the fourth is not. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: `q002. An incorrect integration of the equation x ' = 2 x + t yields x = x^2 + t^2 / 2. After all the integral of x is x^2 / 2 and the integral of t is t^2 / 2. Show that substituting x^2 + t^2 / 2 (or, if you prefer to include an integration constant, x^2 + t^2 / 2 + c) for x in the equation x ' = 2 x + t does not lead to equality. Explain what is wrong with the reasoning given above. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If x' = 2 x + t, and x = x^2 + t^2 / 2. Then x' also is equal to x'= 2(x^2 + t^2 / 2) + t. Therefore 2 x + t would be equal to 2(x^2 + t^2 / 2) + t because both are equalto x'. If you solve the equation, 2 x + t = 2(x^2 + t^2 / 2) + t you get the following, 2 x + t = 2(x^2 + t^2 / 2) + t 2 x + t = 2 x^2 + t^2 + t 2 x = 2 x^2 + t^2 , which is impossible. The original integration is incorrect because the terms were integrated with respect to different variables. the first with respect to x, and the second with respect to t. once this is done, the two cannot be the recombined into a single expression for an integral with respect to x. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: The given function is a solution to the equation, provided its derivative x ' satisfies x ' = 2 x + t. It would be tempting to say that the derivative of x^2 is 2 x, and the derivative of t^2 / 2 is t. The problem with this is that the derivative of x^2 was taken with respect to x and the derivative of t^2 / 2 with respect to t. We have to take both derivatives with respect to the same variable. Similarly we can't integrate the expression 2 x + t by integrating the first term with respect to x and the second with respect to t. Since in this context x ' represent the derivative of our solution function x with respect to t, the variable of integration therefore must be t. We will soon see a method for solving this equation, but at this point we simply cannot integrate our as-yet-unknown x(t) function with respect to t. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: `q003. The general solution to the equation m x '' = - k x is of the form x(t) = A cos(omega * t + theta_0), where A, omega and theta_0 are constants. (There are reasons for using the symbols omega and theta_0, but for right now just treat these symbols as you would any other constant like b or c). Find the general solution to the equation 5 x'' = - 2000 x: Substitute A cos(omega * t + theta_0) for x in the given equation. The value of one of the three constants A, omega and theta_0 is dictated by the numbers in the equation. Which is it and what is its value? One of the unspecified constants is theta_0. Suppose for example that theta_0 = 0. What is the remaining unspecified constant? Still assuming that theta_0 = 0, describe the graph of the solution function x(t). Repeat, this time assuming that theta_0 = 3 pi / 2. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x = A cos(omega * t + theta_0) x ' = - omega A sin(omega * t + theta_0) x '' = -omega^2 A cos(omega * t + theta_0) m * (-omega^2 A cos(omega * t + theta_0) ) = - k A cos(omega * t + theta_0) -m omega^2 = - k ;omega = sqrt(k/m) x(t) = A cos(20 t + theta_0) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 2
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Given Solution: If x = A cos(omega * t + theta_0) then x ' = - omega A sin(omega * t + theta_0) and x '' = -omega^2 A cos(omega * t + theta_0). Our equation therefore becomes m * (-omega^2 A cos(omega * t + theta_0) ) = - k A cos(omega * t + theta_0). Rearranging we obtain -m omega^2 A cos(omega * t + theta_0) = -k A cos(omega * t + theta_0) so that -m omega^2 = - k and omega = sqrt(k/m). Thus the constant omega is determined by the equation. The constants A and theta_0 are not determined by the equation and can therefore take any values. No matter what values we choose for A and theta_0, the equation will be satisfied as long as omega = sqrt(k / m). Our second-order equation m x '' = - k x therefore has a general solution containing two arbitrary constants. In the present equation m = 5 and k = 2000, so that omega = sqrt(k / m) = sqrt(2000 / 5) = sqrt(400) = 20. Our solution x(t) = A cos(omega * t + theta_0) therefore becomes x(t) = A cos(20 t + theta_0). If theta_0 = 0 the function becomes x(t) = A cos( 20 t ). The graph of this function will be a 'cosine wave' with a 'peak' at the origin, and a period of pi / 10. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I didn't take the time to find the particulars of the cosine curve. ------------------------------------------------ Self-critique rating: 2"