What Are the Values of k That Satisfy This Differential Equation?

sristi
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I missed class last Thursday and I am completely confused about the homework. I tried reading the text but there aren't any example problems like this one. Anyone know any free resources for James Stewart Single Variable Calculus?

Well here is the question:

For what values of k does the function y=cos kt satisfy the differential equation 4y"=-25y.
For those values of k, verify that every member of the family of functions y=A sin kt + B cos Kt.

I tried solving for y" and setting the equations equal to each other. To tell the truth I don't understand what the question is saying.

Thanks for any help!
 
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Have you tried to solve the equation directly

y'' = -k^2*cos(kt)
y = cos(kt)

than you get two values of k +/-.
 
For a problem where you are asked "show that A is a solution to problem B", actually solving the problem is overkill. Suppose you were asked to show that x= 1 is a solution to the equation x5- 3x4+ 2x3- 4x2+ 3x- 1= 0. Would you solve the equation or just substitute 1 for x and see if the resulting equation is true?

sristi, substute cos(kt) into 4y"= 5y and see what happens. For what values of k is the resulting equation true?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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