- #1
DaxInvader
Hello Guys, We haven't yet covered on how to solve 2nd order equation in class however we have this assignment given to us. Any tips would be appreciated for these 2 little problems.
1. Homework Statement
We have this initial Equation: d2y/dt2−7dy/dt+ky=0, and we need to find the values of k in which the solution y=e3t applies and the general solution.
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In this case, I simply started to find k but substituting y into the equations.
y(t) = e3t, y'(t) = 3*e3t, y''(t) = 9*e3t
We get: 9e3t - 7*(3*e3t) + k*e3t = 0
=> 9*e3t-21*e3t + k*e3t = 0
=>e3t * (k-12) = 0.
I find the value of K in which y = e3t is a solution to be 12.
Where I am lost is to find the general solution? Do I already have the necessary information?
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Find all values of k for which the function y=sin(kt) satisfies the differential equation y′′+11y=0. Hint: There are more than 2 values of k
We know that y is a solution of the DE.
y'(t) = sin(kt), y'(t) = kcos(kt), y''(t)= -k2sin(kt)
By substitution:
-k2sin(kt) + 11sin(kt) = 0
=> sin(kt)(11-k2)=0
I find that √11 for k is a solution, 0 is a solution, and any multiple of π is a solution as well. I tried entering the following:
√11, 0, nπ
And It is not correct. Any tips to see if I did something wrong?Thanks for your help!
1. Homework Statement
We have this initial Equation: d2y/dt2−7dy/dt+ky=0, and we need to find the values of k in which the solution y=e3t applies and the general solution.
The Attempt at a Solution
[/B]
In this case, I simply started to find k but substituting y into the equations.
y(t) = e3t, y'(t) = 3*e3t, y''(t) = 9*e3t
We get: 9e3t - 7*(3*e3t) + k*e3t = 0
=> 9*e3t-21*e3t + k*e3t = 0
=>e3t * (k-12) = 0.
I find the value of K in which y = e3t is a solution to be 12.
Where I am lost is to find the general solution? Do I already have the necessary information?
Homework Statement
[/B]
Find all values of k for which the function y=sin(kt) satisfies the differential equation y′′+11y=0. Hint: There are more than 2 values of k
The Attempt at a Solution
We know that y is a solution of the DE.
y'(t) = sin(kt), y'(t) = kcos(kt), y''(t)= -k2sin(kt)
By substitution:
-k2sin(kt) + 11sin(kt) = 0
=> sin(kt)(11-k2)=0
I find that √11 for k is a solution, 0 is a solution, and any multiple of π is a solution as well. I tried entering the following:
√11, 0, nπ
And It is not correct. Any tips to see if I did something wrong?Thanks for your help!