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QveST [7]
3 years ago
13

Solve the following equation with the initial conditions. x¨ + 4 ˙x + 53x = 15 , x(0) = 8, x˙ = −19

Mathematics
1 answer:
Katen [24]3 years ago
8 0

x''+4x'+53x=15

has characteristic equation

r^2+4r+53=0

with roots at r=-2\pm7i. Then the characteristic solution is

x_c=C_1e^{(-2+7i)t}+C_2e^{(-2-7i)t}=e^{-2t}\left(C_1\cos(7t)+C_2\sin(7t)\right)

For the particular solution, consider the ansatz x_p=a_0, whose first and second derivatives vanish. Substitute x_p and its derivatives into the equation:

53a_0=15\implies a_0=\dfrac{15}{53}

Then the general solution to the equation is

x=e^{-2t}\left(C_1\cos(7t)+C_2\sin(7t)\right)+\dfrac{15}{53}

With x(0)=8, we have

8=C_1+\dfrac{15}{53}\implies C_1=\dfrac{409}{53}

and with x'(0)=-19,

-19=-2C_1+7C_2\implies C_2=-\dfrac{27}{53}

Then the particular solution to the equation is

\boxed{x(t)=\dfrac1{53}e^{-2t}(409cos(7t)-27\sin(7t)+15)}

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