Sorry to be asking so many questions during exam time, but how would I integrate the equation dy/dx=yx. My first thoughts are to find what y=, but I have been unable to be able to do this without invoking hours of trial and error.

Brad

dy/y=dx

ò(1/y)dy=òdx

log(y)=x+C (some constant C)

y=A e^x (some constant A > 0)

Which is of course, complete nonsense (but it works). Similarly for this problem, you write

dy/y=x dx

ò(1/y) dy=òx dx

log(y)=0.5x²+C (some constant C)

y=Ae^0.5x2 (some constant A > 0)

See if you can use this method to integrate:

dy/dx=y²cos(x)

I see how you obtained the result log(y)=x+C (I had not done it this way before). I am not quite sure how you received Ae^x though. I did this a bit differently (I didn't get this from a book though so it may be wrong) and didn't get the A: First, make x a function of y:

dx/dy=1/y.

Therefore,
x=lny (or ln of the modulus of y, but I didn't really consider that).

So

y=e^x .

I do see how you worked through the second problem, but I do not see why it works to integrate on both sides of the equation- my book does not say anything concerning this. Anyway, using this method

dy/dx=y² cos(x)

integrates to

ò(y^-2) dy=ò(cos(x)) dx
So,
-y^-1 =sin(x),(neglecting the constants to appear on both sides).

From this,

y=-1/sin(x)+C.

I will bet that the A constant should be there, but I am not sure.

Thanks,

Brad

If you get to log(y)=x+C, take exponentials of both sides to y=e^x+C =e^x e^C , if C is a real number then e^C > 0, let A=e^C to get y=Ae^x .

With the second problem, I think you might have it right, but I'm not sure. It should read y=-1/(sin(x)+C) (note the brackets). The reason you get these constants is that when you take an indefinite integral (one without specific limits) you can add any constant value to it.

I'm not sure why this method works at the moment (it's quite late and my head is full of Quantum Mechanics which I'm revising for exams, yuch!) but I'll get back to you on this soon if you're interested (unless someone else replies first).