How Can I Solve This Differential Equation with an Initial Condition of y(1)=7?

[goaliejoe35]

Homework Statement

Find the particular solution of the differential equation xdy=(x+y-4)dx that satisfies the initial condition y(1)=7.

The attempt at a solution

Ok here's my first two steps...

dy = \frac{(x+y-4)}{x} dx
\frac{dy}{(x+y-4)} = \frac{dx}{x}

Now here's where I get messed up. How can I get the x out of the side with the dy? Could someone please explain how to finish this or If I'm headed in the right direction?

 

[HallsofIvy]

Is "separation of variables" the only way you know to solve differential equations? This is NOT a "separable" equation.

 

[goaliejoe35]

Yea that's the only way I know how. But I'm looking at a different way now where they give you the standard equation y' +P(x)y = Q(x), but I don't quite understand it.

 

[konthelion]

Yea that's the only way I know how. But I'm looking at a different way now where they give you the standard equation y' +P(x)y = Q(x), but I don't quite understand it.

THat's a linear first-order differential equation. Usually, textbooks show the method for solving them. Try looking in your textbook.

 

[rock.freak667]

Use a substitution of y=Vx


EDIT: nvm...now re-read your post...

 

[tiny-tim]

y' +P(x)y = Q(x)

Homework Statement

Find the particular solution of the differential equation xdy=(x+y-4)dx that satisfies the initial condition y(1)=7.

Yea that's the only way I know how. But I'm looking at a different way now where they give you the standard equation y' +P(x)y = Q(x), but I don't quite understand it.

Hi goaliejoe35!

Hint: first step: get the RHS x-only:

xdy - ydx = (x - 4)dx.

Does the LHS now remind you of anything?

If so, fiddle around with it until you get something you can integrate.

If not, go back to your book and look at y' +P(x)y = Q(x) again