# Solve the differential equation ... 10) xy + y' = 100x ... I'm supposed to make it equal y, and y' has to be cancelled out, but I kept getting stuck at this point: ... x = y' / (100 + y)

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### 1 Answer

Put the terms with x on one side, subtract xy on both sides.

`xy + y' - xy = 100x - xy`

`y' = 100x - xy`

Factor out x on left side.

`y' = x(100 - y)`

Divide both sides by (100 - y) to isolate the x.

`(y')/(100-y) = x` or `x = (y')/(100-y)`

Note that we can rewrite the y' as dy/dx.

`x = ((dy/dx))/(100-y)`

Multiplying dx on both sides.

`xdx = (dy)/(100-y)`

We are now done on separating the variables, we can now take the integral of both sides.

`int(xdx) = int((dy)/(100-y))`

Using the formulas:

`int(u^ndu) = (u^(n+1))/(n+1) + c` `<br>`

and

`int(du)/(u) = ln(u)`

`(x^2)/(2) + c = -ln(100-y)`

Multiply both sides by -1.

`-(x^2)/(2)-c = ln(100-y)`

Take the exponential of each sides.

`e^(-(x^2)/(2) - c) = e^(ln(100-y))`

Take note that:``

`e^(ln(u)) = u`

`e^((-x^2)/(2)-c) =100 -y`

Subtract both sides by 100 and then multiply both sides by -1.

`y = 100 - e^(-(x^2)/(2)-c)`

We can also write that as:

`y = 100 - (e^(-(x^2)/(2)))/(e^c)`

We can also let the c = e^c.

`y = 100 - (e^(-(x^2)/(2))/(c))`

That is it! :)