# differential equations problem

I haven't dealt with differential equations in a while so I completely forget how to do this problem.

Given dx/dt=y and dy/dt=x

Verify that f(t)=Ce^t+De^-t and g(t)=Ce^t - De^-t give a solution to the system of differential equations, where C and D are arbitrary constants.

Thanks!

### Re: differential equations problem

Thanks perucho!

Yea, you don't have to find C and D they are just arbitrary constants.
Although could you explain where to go from y''-y=0 using the chain rule? I'm a little hazy on that...

### Re: differential equations problem

I see cr but it's strange for me because I think it's evident. Anyway, you posted the ODEs
dx/dt = y, dy/dt = x,
so insert the second one into the first one, that is
d(dy/dt) = y.
Do you get it now?

### Re: differential equations problem

No, excuse my typo. Correct is
d/dt(dy/dt) = y.
Sorry.

### Re: differential equations problem

By the way,
y'' = dy'/dt = dy'/dy.dy/dt = y'.dy'/dy.

### Re: differential equations problem

thanks darkgently!

also how does one find the values of the constants C and D if you have x_0=3 and y_0=1? I'm confused as to where to plug in x_0 and y_0....

### Re: differential equations problem

> also how does one find the values of the constants C and D
> if you have x_0=3 and y_0=1? I'm confused as to where to
> plug in x_0 and y_0....

That means when t=0, x=3 and y=1. Just substitute that information into the solutions to get two equations in C and D, which you can easily solve.

### Re: differential equations problem

i'm sorry that i'm still stuck on this problem

i know i'm probably missing a really simple step but I can't seem to differentiate f(t)=Ce^t+De^-t and g(t)=Ce^t-De^-t to give me y and x. I'm confused as to which substitutions I need to make..

### Re: differential equations problem

Hi crazziqt,
darkgently realizes your problem was more elementary than I thought. And f(t) and g(t) confused me as they also are confusing you. It's not necessary call f(t) and g(t) to the solutions x=x(t) and y=y(t). So let's back to your original problem.
dx/dt = y and dy/dt = x (1)
You only must verify that the functions
x(t) = f(t) = Ce^t + De^-t
and
y(t) = g(t) = Ce^t - De^{-t}
are general solutions of the system of ODEs (1). Therefore if you find the derivative
dx/dt = Ce^t - De^{-t} (d/dt(e^{-t} = - e^{-t})
you see this is precisely y(t), and analogously by calculating the derivative
dy/dt = Ce^t + De{-t}
it's exactly x(t). Like you see it's a trivial problem.
perucho

### Re: differential equations problem

What is that crazzi?
f(t) & g(t) don't make sense, as you must find x(t) & y(t), and +D or -D is irrelevant because C & D depend on the initial conditions for the independent variable t.
From your ODEs you will get
y''-y=0,
and by the chain rule
y'dy'=ydy.
Not known initial conditions, not possible to find C&D.
perucho

### Re: differential equations problem

It seems to me that the question is only asking you to *verify* that those functions f and g (which I assume mean x and y) are solutions, not to derive those solutions from scratch. So you'd just have to substitute them into the equations, i.e. check that one is the derivative of the other, which is easy!