[tex]x''+4x'+53x=15[/tex]
has characteristic equation
[tex]r^2+4r+53=0[/tex]
with roots at [tex]r=-2\pm7i[/tex]. Then the characteristic solution is
[tex]x_c=C_1e^{(-2+7i)t}+C_2e^{(-2-7i)t}=e^{-2t}\left(C_1\cos(7t)+C_2\sin(7t)\right)[/tex]
For the particular solution, consider the ansatz [tex]x_p=a_0[/tex], whose first and second derivatives vanish. Substitute [tex]x_p[/tex] and its derivatives into the equation:
[tex]53a_0=15\implies a_0=\dfrac{15}{53}[/tex]
Then the general solution to the equation is
[tex]x=e^{-2t}\left(C_1\cos(7t)+C_2\sin(7t)\right)+\dfrac{15}{53}[/tex]
With [tex]x(0)=8[/tex], we have
[tex]8=C_1+\dfrac{15}{53}\implies C_1=\dfrac{409}{53}[/tex]
and with [tex]x'(0)=-19[/tex],
[tex]-19=-2C_1+7C_2\implies C_2=-\dfrac{27}{53}[/tex]
Then the particular solution to the equation is
[tex]\boxed{x(t)=\dfrac1{53}e^{-2t}(409cos(7t)-27\sin(7t)+15)}[/tex]
