Solve dy/dx= (x√(x^2+3))/e^2y given that y=0 when x=1, giving your answer in the form y = f(x)

The first thing we notice is that this differential equation is seperable, meaning we can get all of our y's on the left with a dy and all of our x's on the right with a dx. Doing this by multiplying both sides by e^2y and dx gives:

e^2y dy = x√(x^2+3) dx

Now we need to integrate on both sides. Integrating the left with respect to y won't be too bad, we'll get 

e^2y/2     Whenever we integrate by inspection, it's a good idea to check that we have the right answer by                        differentiating and making sure we get back what we put in to start with. This works in this case, as                    differentiating by the chain rule with u = 2y gets us 2(e^2y/2) = e^2y, which is what's on the left.

So far so good, but a tricky one now. On the right we have an x-term multiplied by a function of x, which is usually a signal telling us to try integrating by parts. This would work, but we're still going to have to integrate √(x^2+3), which is tough. Seems like integration by parts is a bit of a dead end, so we're going to have to think of something else to try. Using the substitution u = x^2+3 (when we have a square root it's often really useful to take u='whatever's in the square root') will get the right-hand side to be

= x√u dx   

That's good, but we don't really want dx in there (we're integrating with respect to u now), so let's get du in terms of dx. Differentiating the expression u = x^2+3 will get us:

du/dx = 2x  =>  dx = du/2x      Substituting that into our last equation gets us:

RHS = (x√u)(du/2x)         Great news! Our x's cancel, just leaving us with

the integral of √u/2 du which we can do by inspection: it's (1/2)u^3/2 divided by 3/2 

Dividing by 3/2 is the same as multiplying by 2/3, so our 2's cancel and we're just left with:

e^2y/2 = 1/3(u^3/2)     Oops, don't want u's in there, we're working in terms of y and x. Also we forgot to add c

e^2y/2 = 1/3(x^2+3)^3/2 + c   where c is our constant of integration. That's better. 

Wow; Long question! But it's a lot of marks so let's stick at it. Now to find c using our intitial values. Plugging in y = 0 when x = 1 will give us:

(e^0)/2 = 1/3(1+3)3/2 + c     =>       1/2 = 1/3*8 + c     =>     c = 1/2-3/8 = 3/6-16/6 = -13/6  

We're almost there. but the question wanted y = f(x) so we've got a bit of rearranging to do. Multiply by 2 first:

         e^2y/2 = 1/3(x^2+3)^3/2 -13/6

=>     e^2y = 2/3(x^2+3)^3/2 - 13/3      Now we're going to have to take logs to get rid of that e:

=>    ln(e^2y) = ln(2/3(x^2+3)^3/2  - 13/3)           ln(e^a) = a by definition, so on the left we're just left with 2y.

=>    2y =  ln(2/3(x^2+3)^3/2  - 13/3)           and finally we just need to divide by 2!

=>    y = 1/2 ln(2/3(x^2+3)^3/2 - 13/3)

Well that question required a lot of perseverance but nothing in it was too tricky. It's so important that we can do things like integrations quickly, picking up lots of marks on questions like this and giving ourselves plenty of time for the really fiendish quesitons right at the end of the paper.