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)
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>`
`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! :)