C4 calculus help?

Now I'm not quite sure what to write...Do I find C? Did I even need to include C?

I know I should probably bring up the -1/2 and make it into a power...

How do I show that M = Kt/sqr(1+t^2) like I'm asked to?

Where does the K come from? Is it the integration constant C?

Please help...

lnM = lnt- 1/2ln(1+t^2) + c

From here you can already see $\displaystyle \ln M = \ln t - \ln \sqrt{1+t^2} + c \Rightarrow \ln M = \ln \frac{t}{\sqrt{1+t^2}} + c$

Now if you take the exponential of both sides: $\displaystyle M = \frac{e^c t}{\sqrt{1+t^2}}$, now just let $K = e^c$.

Oh thank you for the help! I didn't realise you could take the exponential of both sides?

You can usually do whatever you'd like to both sides of an equation, for example, if you have the equation $x^3 = 27$, when you're solving this, you're implicitly taking the cube root of both sides to get $x = 27^{1/3} = 3$.

If you want another way of looking at it, though, we have:

$\displaystyle \ln M = \ln \frac{t}{\sqrt{1+t^2}} + c = \ln \frac{t}{\sqrt{1+t^2}} + \ln e^c \Rightarrow \ln M = \ln \frac{e^c t}{\sqrt{1+t^2}}$

Now the logs 'cancel' to get you $\displaystyle M = \frac{e^c t}{\sqrt{1+t^2}}$, if you're more comfortable working this way.

Yes, that makes complete sense now! Thank you very much for the guidance!