Question

2nd order differential equations

Answer 1

I mainly need help on part (a). I think I'll be ok with the rest.

Answer 2

Cool, so we need a coupled system that's the technical term for it, it's in the form \[\frac{ dx }{ dt } = ax+by\] and \[\frac{ dy }{ dt } = cx+dy\] you can use elimination here to make it into a second order dif. eqn. It's like a systems of equations but with differentials, so solving for x or y and yadi yadi yada.

Answer 3

that's what i was thinking but one has dx/dt and the other has dy/dt. Thats where i'm getting stuck.

Answer 4

You shouldn't once you make the substitution so lets say \[\frac{ dy }{ dt } = -4x-3y \] is your equation 1 what do we get if we solve for x?

Answer 5

\[x=(-3y+\frac{ dy }{ dt })*\frac{ 1 }{ 4 }\]

Answer 6

sorry its -dy/dt

Answer 7

\[\frac{ dy }{ dt } + 3y = -4x \implies x = -\frac{ 1 }{ 4 }\left( 3y+\frac{ dy }{ dt } \right)\] right good :)

Answer 8

Now our second equation will be \[\frac{ dx}{ dt } = 2x+y\] so we can now make the substitution in this equation, notice \[\frac{ d }{ dt } \left\{ -\frac{ 1 }{ 4 }\left( 3y+\frac{ dy }{ dt } \right) \right\} = 2\left\{ -\frac{ 1 }{ 4 }\left( 3y+\frac{ dy }{ dt } \right) \right\}+y\]

Answer 9

yh I understand that

Answer 10

Ok now simplify

Answer 11

I'm trying but I'm not sure how.

Answer 12

Distribute as you would normally

Answer 13

I got:
\[-\frac{ 1 }{ 4 }\frac{ dy }{ dt }-\frac{ d^2y }{ dt^2 }=-\frac{ y }{ 2 }\]

Answer 14

sorry my connections a bit weak

Answer 15

I got \[\frac{ d^2y }{ dt^2 }+\frac{ dy }{ dt }-2y=0\] but you can check the math, notice it's a homogeneous equation now so you can solve it now

Answer 16

But notice I did this example on purpose as they want you to do it for x(t) :)

Answer 17

yes its right i just forgot to put -1/4 in.

Answer 18

Thanks for doing that. I like to make sure I can do it myself.