Implicit and explicit solution for a given initial-value problem

[Cheruby]

The equation is:
x² dy/dx= y - xy
IC (initial conditions): y(-1) = -1 (This is used to solve for C)
Must first separate the variables x and y and then integrate them and solve for y, but I got stuck...

x²dy = (y - xy)dx
x²dy = y(1-x)dx
dy= y(1-x)dx/x² <-- not sure what to do with the y now... I figure I could divide everything by y so I could bring it to the left side but the right side would have dx/y

The answer is y = e^-(1+1/x)/x

 

[Jackx]

The equation is:
x² dy/dx= y - xy
IC (initial conditions): y(-1) = -1 (This is used to solve for C)
Must first separate the variables x and y and then integrate them and solve for y, but I got stuck...

x²dy = (y - xy)dx
x²dy = y(1-x)dx
dy= y(1-x)dx/x² <-- not sure what to do with the y now... I figure I could divide everything by y so I could bring it to the left side but the right side would have dx/y

The answer is y = e^-(1+1/x)/x

You are doing it right, but when you divide by Y you will have dy/y, and that is perfectly acceptable. The right side you will have (1-x)/x^2 dx. Then integrate both sides.

 

[Cheruby]

You are doing it right, but when you divide by Y you will have dy/y, and that is perfectly acceptable. The right side you will have (1-x)/x^2 dx. Then integrate both sides.

But wouldn't the right side have dx/y? That's why I'm stuck, I'm not sure if it's acceptable!

 

[Jackx]

dx can be treated as just another factor on the right side. So since they are all factors when you divide by y the y just goes away on the right side.

Now if it was y + (whatever)dx, then yes if you divided by y you would have a dx/y.

 

[Cheruby]

Alright thank you Jackx