Math Help Please!! (the problem is attached)

Question

anonymous · Answer

attachment

anonymous · Answer

I will try to help you with the first steps. 
let $$y=e^{rx}$$ then
$$y''=r^2e^{rx}$$so the equation becomes$$r^2e^{rx}+49e^{rx}=0$$ after dividing with $$e^{rx}$$ you get an equation that is squared for r and has complex solution

anonymous · Answer

$$r=\pm i 7$$ with this you should be able to solve your problem.

anonymous · Answer

The characteristic equation for complex roots is: $$y(x)=c_1e^{\lambda x}\cos( \mu x) + c_2e^{\lambda x } \sin ( \mu x)$$
in our case, since our root is a complex number without any real part, only imaginary we can set: $$\mu = r$$
and $$\lambda = 0$$

anonymous · Answer

what I ended up doing was 8=C1cos(7x)+C2sin(7x)
I replaced the x's with zeros and got C1=8
but idk how to solve for c2

anonymous · Answer

im kind of confused to what u are doing?

anonymous · Answer

!*

anonymous · Answer

Ok, it doesn't really matter if you don't understand what I did above, I was just using the ordinary way how to solve this sort of equation. Nevertheless, what you did seems to be right. So you have to IVP, the initial conditions you can use now, you know that: $$y(0)=8$$
for the equation (we) you have found that means:$$y(0)=c_1\cos(7 * 0 ) + c_2 \sin (7 * 0 ) = c_1 = 8 $$ so that is indeed correct.

Next thing I would do is set the entire derivative of the equation to zero and see if we can match another IVP.

$$y'(x)=-c_1\sin(7x)*7 + c_2\cos(7x)*7 $$

anonymous · Answer

so if you use that you will get $$y'(0)=-c_1\sin(7*0)*7 + c_2 \cos (7 *0 ) 7 = -7 \rightarrow c_2=-1$$

anonymous · Answer

Thanks so much!!!!

anonymous · Answer

very welcome, solution is: $$y(x)=8\cos(x)-\sin(7x)$$

anonymous · Answer

:)