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Linear Differential Equation of Second Order 
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#1
May412, 03:28 PM

P: 614

1. The problem statement, all variables and given/known data
I have the equation mx'' + cx' + kx = 0 where m = 2, c = 12, k = 50, x_{0} = 0, x'(0) = 8. x is a function of t, and primes denote derivatives w.r.t. t. 2. Relevant equations 3. The attempt at a solution so the equation is 2x'' + 12x' + 50x = 0, which I simplify to x'' + 6x' + 25x = 0. Characteristic equation is r^{2} + 6r + 25 = 0 using the quadratic formula, I find roots r = 3 ± 4i therefore the general solution is x(t) = e^{3t}(c_{1}cos(4t) + c_{2}sin(4t)) I then solve for the unknown constants, c_{1} and c_{2}: x(0) = 0 = c_{1} so x(t) = e^{3t}(c_{2}sin(4t)) x'(t) = 3e^{3t}(c_{2}sin(4t) + e^{3t}(4c_{2}cos(4t)) x'(0) = 8 = 4c_{2} so c_{2} = 2 ∴ x(t) = e^{3t}(2sin(4t)) but... the back of the book gives x(t) = 2e^{3t}cos(4t  3[itex]\pi[/itex]/2) as the answer... so I must be doing something horribly wrong, but I have no idea what 


#2
May412, 03:38 PM

P: 614

oh whoops, I just realized that, for the set of problems that I'm working on, it asks me to write the function in the form
C_{1}e^{pt}cos(ω_{1}t  [itex]\alpha[/itex]_{1}) so I guess it's just a matter of converting it 


#4
May412, 04:18 PM

PF Gold
P: 836

Linear Differential Equation of Second Order
[tex]x(t)=e^{3t} (2\sin 4t)=2e^{3t} (\sin 4t)[/tex]
Now, you try to obtain the above from the given answer, to check for equivalence: [tex]x(t)=2e^{3t} (\cos (4t  \frac{3\pi}{2}))[/tex] Comparing the two answers above, what you need is to prove that: [tex]\cos (4t  \frac{3\pi}{2})=\sin 4t[/tex] Use the cosine subtraction formula: [tex]\cos (4t  \frac{3\pi}{2})=(\cos 4t)(\cos \frac{3\pi}{2})+ (\sin 4t)(\sin \frac{3\pi}{2})=\sin 4t[/tex]Proved! 


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