A level differential equations help

Is that just H = e^(0.1sin(0.25t) + c) ?

Initial height is 5. I.e. H is 5, when t=0 but you need to have a workable equation to figure out what the constant of integration is first.

Yes, but you can split into a simpler form (which is something you should get used to doing).

$e^{0.1\sin(0.25t) + c} = e^{0.1\sin(0.25t)}\cdot e^c$

Now since $c$ is an arbitrary constant of integration, then so is $e^c$. We can denote it as $A = e^c$.

Thus the equation reduces to $H = Ae^{0.1\sin(0.25t)}$.

Now use the initial condition to determine $A$.

Oh no, it's okay I was trying to differentiate it
Now I get lnH = 0.1sin(0.25t) which makes a lot more sense, do I then just show H = e^(0.1sin(0.25t)) ? And then just put the 5 in as that's the initial value or what should I do about that bit?

Yes, but you can split into a simpler form (which is something you should get used to doing).

$e^{0.1\sin(0.25t) + c} = e^{0.1\sin(0.25t)}\cdot e^c$

Now since $c$ is an arbitrary constant of integration, then so is $e^c$. We can denote it as $A = e^c$.

Thus the equation reduces to $H = Ae^{0.1\sin(0.25t)}$.

Now use the initial condition to determine $A$.

You are giving FAR too much help here - this student needs to go back to the beginning of this topic.

Ok? Just let OP know that, no need to quote me lol...