Differential Equations help???

Using the substitution y=1/u, show that the general solution to (1/y^2)dy/dx + 1/y = e^(2x) is y=1/(Ae^x - e^(2x)???

What I’ve got so far:
dy/du=-1/u^2
u^2(dy/dx) + u = e^(2x)
-du/dx + u = e^(2x)
-du + u dx = e^(2x) dx
But I’m stuck here because you cannot integrate (u dx).

Many thanks

Using the substitution y=1/u, show that the general solution to (1/y^2)dy/dx + 1/y = e^(2x) is y=1/(Ae^x - e^(2x)???

What I’ve got so far:
dy/du=-1/u^2
u^2(dy/dx) + u = e^(2x)
-du/dx + u = e^(2x)
-du + u dx = e^(2x) dx
But I’m stuck here because you cannot integrate (u dx).

Many thanks

No I’ve never seen the use of integrating factors - probably why I’m unable to progress then. Please help?

So from my understanding it seems that the integrating factor is e^(-x) (hope this is correct?). So starting again:

du/dx - u = -e^(2x)
So e^(-x)(du/dx) - ue^(-x) = - e^x
Can anyone give me any pointers from here?

Recall the product rule,

$(uv)' = u'v + uv'$

Also that $\displaystyle \frac{\mathrm d}{\mathrm dx} u = \frac{\mathrm du}{\mathrm dx}$. (you should also be familiar with the derivative of e^x) Does this help?