Question

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

Answer 1

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

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

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