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Question: exercise 021 check that the y given is really a...

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Exercise 0.2.1: Check that the y given is really a solution to the equation Next, take the second order differential equation dx for some constant k > 0. The general solution for this equation is y(x)-C, cos(kx) + C, sin(kr). Note that because we have a second order differential equation, we have two constants in our general solution Exercise 0.2.2: Check that the y given is really a solution to the equation. And finally, take the second order differential equation d. for some constant k > 0. The general solution for this equation is x)- CeCe or y(x) Di cosh(kx) + D2 sinh(kx) For those that do not know, cosh and sinh are defined by sinh x . These functions are sometimes easier to work with than exponentials. They have some nice familiar properties such as cosh 0-1, sinh 0 0, and芸cosh x sinh x (no that is not a typo) and dsinh cosh x Exercise 0.2.3: Check that both forms of the y given are really solutions to the equation An interesting note about cosh: The graph of cosh is the exact shape a hanging chain will make. This shape is called a catenary. Contrary to popular belief this is not a parabola. If you invert the graph of cosh it is also the ideal arch for supporting its own weight. For example, the gateway arch in Saint Louis is an inverted graph of cosh if it were just a parabola it might fall down. The formula used in the design is inscribed inside the arch: y -127.7A . cosh(x/127.7 ft) + 757.7 ft.
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