Answer:
[tex]y(x)=e^{-2x}[3cos(\sqrt{6}x)+\frac{2\sqrt{6}}{3}sin(\sqrt{6}x)][/tex] (See attached graph)
Step-by-step explanation:
To solve a second-order homogeneous differential equation, we need to substitute each term with the auxiliary equation [tex]am^2+bm+c=0[/tex] where the values of [tex]m[/tex] are the roots:
[tex]y''+4y'+10y=0\\\\m^2+4m+10=0\\\\m^2+4m+10-6=0-6\\\\m^2+4m+4=-6\\\\(m+2)^2=-6\\\\m+2=\pm\sqrt{6}i\\\\m=-2\pm\sqrt{6}i[/tex]
Since the values of [tex]m[/tex] are complex conjugate roots, then the general solution is [tex]y(x)=e^{\alpha x}[C_1cos(\beta x)+C_2sin(\beta x)][/tex] where [tex]m=\alpha\pm\beta i[/tex].
Thus, the general solution for our given differential equation is [tex]y(x)=e^{-2x}[C_1cos(\sqrt{6}x)+C_2sin(\sqrt{6}x)][/tex].
To account for both initial conditions, take the derivative of [tex]y(x)[/tex], thus, [tex]y'(x)=-2e^{-2x}[C_1cos(\sqrt{6}x+C_2sin(\sqrt{6}x)]+e^{-2x}[-C_1\sqrt{6}sin(\sqrt{6}x)+C_2\sqrt{6}cos(\sqrt{6}x)][/tex]
Now, we can create our system of equations given our initial conditions:
[tex]y(x)=e^{-2x}[C_1cos(\sqrt{6}x)+C_2sin(\sqrt{6}x)]\\\\y(0)=e^{-2(0)}[C_1cos(\sqrt{6}(0))+C_2sin(\sqrt{6}(0))]=3\\\\C_1=3[/tex]
[tex]y'(x)=-2e^{-2x}[C_1cos(\sqrt{6}x+C_2sin(\sqrt{6}x)]+e^{-2x}[-C_1\sqrt{6}sin(\sqrt{6}x)+C_2\sqrt{6}cos(\sqrt{6}x)]\\\\y'(0)=-2e^{-2(0)}[C_1cos(\sqrt{6}(0))+C_2sin(\sqrt{6}(0))]+e^{-2(0)}[-C_1\sqrt{6}sin(\sqrt{6}(0))+C_2\sqrt{6}cos(\sqrt{6}(0))]=-2\\\\-2C_1+\sqrt{6}C_2=-2[/tex]
We then solve the system of equations, which becomes easy since we already know that [tex]C_1=3[/tex]:
[tex]-2C_1+\sqrt{6}C_2=-2\\\\-2(3)+\sqrt{6}C_2=-2\\\\-6+\sqrt{6}C_2=-2\\\\\sqrt{6}C_2=4\\\\C_2=\frac{4}{\sqrt{6}}\\ \\C_2=\frac{4\sqrt{6}}{6}\\ \\C_2=\frac{2\sqrt{6}}{3}[/tex]
Thus, our final solution is:
[tex]y(x)=e^{-2x}[C_1cos(\sqrt{6}x)+C_2sin(\sqrt{6}x)]\\\\y(x)=e^{-2x}[3cos(\sqrt{6}x)+\frac{2\sqrt{6}}{3}sin(\sqrt{6}x)][/tex]
