I reached a dead end with differential equations - a level maths

Where did I go wrong?

Helps to post a question/working.

sorry for some reason my threat keeps being reviewed

At the start of the third line you attach the -xe^y to the dy/dx, but this does not follow from line 2. In the mark scheme, they rearrange to get something like
f(y) dy/dx = g(x)
then integrate wrt x ("multiply through by dx"). You had something like
h(y) + f(y) dy/dx = g(x)
but treated it like
h(y) dy/dx + f(y) dy/dx = g(x)

Edit - a simple example. If you had
dy/dx = 1 + y
Its wrong to subtract y from each side
-y dy/dx = 1
and then integrate which is baically what you did. Rather you should divide by (1+y) so
dy/dx = 1(1+y)
1/(1+y) dy/dx = 1
then integrate, which is what the ms did.

so basically dont substract y from both sides in differential equations?

Sort of. In your original post you assume dy/dx multiples the xe^y term you take over, but it doesnt. If you did
dy/dx = 1 + y
-y + dy/dx = 1
Integrate wrt x, then
- Int y dx + y = x + c
Its correct but youve not made any progress as the first term can't really be solved. If you did CF and PI (or integrating factors or ...) for solving odes, then they assume a representation like
-y + dy/dx = 1
and you can use it.

Doing
1/(y+1) dy/dx = 1
so integrating wrt x gives
Int 1/(y+1) dy = Int dx
which is the usual ln(...) = x ... is what you do with seperation of variables.

You just need to be precise with your algebra and going from line 3 to line 2 does not follow as its not xe^y dy/dx.

There is a fairly storng hint to use part a), but you go wrong from line 1 to line 2 (basic algebra). If you take 3sqrt(y) over, its -3sqrt(y) + dy/dx on the left, which isnt useful, nevemind wrong. You have multiply/divide by a term to attach it to dy/dx on the left.

Edit - when you do seperation of variables you essentially want to get a "reverse chain rule" on the left. Your basic equation should be something like
f(y) dy/dx = g(x)
where you know y is a function of x so y(x).

If jump ahead and imagine the integral of that (whatever it is) to be
F(y) = G(x)
then differentiating both sides with respect to x you get
dF/dy dy/dx = dG/dx
so
f = dF/dy
g = dG/dx
Note that the dF/dy term must be multipied by dy/dx. Treating integration as reverse differentiation, then the integral of f(y) dy/dx with respect to x is simply the integral of f(y) with respect to y.

ahhhhh I cant people I did that first mistake. thank you