Answer:
[tex]y(x)=c_1e^{2ix}+c_2e^{-2ix}[/tex]
Step-by-step explanation:
You have the following differential equation:
[tex]3y''+12y=0[/tex] (1)
In order to find the solution to the equation, you can use the method of the characteristic polynomial.
The characteristic polynomial of the given differential equation is:
[tex]3m^2+12=0\\\\m^2=-\frac{12}{3}=-4\\\\m_{1,2}=\pm2\sqrt{-1}=\pm2i[/tex]
The solution of the differential equation is:
[tex]y(x)=c_1e^{m_1x}+c_2e^{m_2x}[/tex] (2)
where m1 and m2 are the roots of the characteristic polynomial.
You replace the values obtained for m1 and m2 in the equation (2). Then, the solution to the differential equation is:
[tex]y(x)=c_1e^{2ix}+c_2e^{-2ix}[/tex]
