1. ## Differential Equations

Use the substiution y=xu to find the general solution of the differential equation:

(x+y)dy/dx=y

dy/dx = u+x(du/dx)

(x+xu)(u+x(du/dx)=xu

where do i go from here?

thanks

2. To guide you, why are you making that substitution? (review your class notes on homogeneous equations)

3. I only have one example.
and its my first week of Differential Equations.
I just want to make sure i understand it before my next lesson which is tomorrow.

Do i expand the brackets?

4. Originally Posted by wonderboy1953
To guide you, why are you making that substitution? (review your class notes on homogeneous equations)
I would bet he/she is making the substitution because

Originally Posted by BabyMilo
Use the substiution y=xu to find the general solution of the differential equation:
BabyMilo,

$(x+xu)dy-xudx=0$

$dy=udx+xdu$

$(x+xu)(udx+xdu)-xudx=0\Rightarrow x(1+u)(udx+xdu)-xudx=0$

$\Rightarrow (1+u)(udx+xdu)-udx=0$

Distribute and group us with du and xs with dx.

Then integrate and back sub.

5. not sure what Distribute and group us with du and xs with dx. means.

should i come to u^2+(x+xu)(du/dx)=0 ?

6. I am sorry I made a clerical error.

$\displaystyle (x+y)\frac{dy}{dx}=y\Rightarrow(x+y)dy=ydx$

$y=ux \ \ \ dy=xdu+udx$

$xdy+ydy=ydx\Rightarrow x(xdu+udx)+ux(xdu+udx)=uxdx$

$x^2du+xudx+ux^2du+u^2xdx=uxdx$

$x^2du+ux^2du+u^2xdx=0\Rightarrow (x^2+ux^2)du=-u^2xdx$

$\displaystyle \frac{1+u}{u^2}du=-\frac{dx}{x}$

Sorry about the mistake earlier. Do you understand now?