# Separable Partial ODE

1. Nov 4, 2008

### Number 7

Find the general solution of:

x.du/dx - (1/2).y.du/dy=0

I know that for it to be separable u(x,y)=X(x)Y(y)

so:

x.Y(y)dX/dx + (1/2).X(x)dY/dy = 0

which cancels to:

x/X(x).(dX/dx) = - y/2Y(y).dY/dy

so:

X(x) = -c. Y(y) c is some constant

so:

x/X(x).dX/dx = c
y/2Y(y).dY/dy=c

from here i am stuck. Any help appreciated
Thanks

2. Nov 4, 2008

### Office_Shredder

Staff Emeritus

Also, separation of variables is your friend. Just because you learned a method of solving differential equations a while ago doesn't mean you should abandon it

$$\frac{x}{X(x)}\frac{dX}{dx} = c$$

So

$$\frac{dX}{X} = c \frac{dx}{x}$$

3. Nov 4, 2008

### Number 7

Thanks I get
c.ln(x) + C=ln(X) => X=Ax^C
c.ln(y) +C= 1/2 ln(Y) => Ay^2C

so u(x,y)=X.Y=A^2(y^2C)(x^C)

...the contants for integration are required right? Just wondering because the second part of the question would be easier if they weren't there:

determine u(x,y) which satisfies the boundary conditon u(1,y)= 1 +sin y

I would go about this by making 1+siny = A^2((1)^C(Y^2C))

=> A^2=[1 - Sin(y)]/[y^2C]

so u(x,y) satisfied by the boundary condtion is:

u(x,y)=[1 - Sin(y)]/[y^2C].x^C.y^2C

y^2C cancel so:

u(x,y)=[1 - Sin(y)].x^C

BUT how do you get rid of that C or am i being stupid and iv made a previous mistake/mistakes

EDIT :OK Long winded but from finding out X and Y I can find the original c and hence A by finding the ratio between X an Y which is x/y^2 so A is known. Then the boundary condtion is simply finding the constant of integration c with known A? Il leave the previous mess on the post just to check your opinion.
Thanks

Last edited: Nov 4, 2008
4. Nov 4, 2008

### Office_Shredder

Staff Emeritus
You have some sloppy coefficient carrying there, but got lucky and ended up with an equivalent formula anyway

$$\frac{dX}{X} = c \frac{dx}{x}$$

So

$$ln(X) = cln(x) + C_x$$
which gives us

$$X = x^c*A_x$$ where $$A_x = e^{C_x}$$. This is important because the constant of integration obtained in the formula for Y need not be the same (it happens not to matter in this case, but you could shoot yourself in the foot). Similarly

$$\frac{dY}{Y} = 2c\frac{dy}{y}$$ which gives us

$$ln(Y) = 2cln(y) + C_y$$

and hence
$$Y = A_yy^{2c}$$ where $$A_y = e^{C_y}$$.

So

$$u(x,y) = A_xA_yx^cy^{2c}$$ we combine the multiplicative constants to get

$$u(x,y) = Ax^cy^{2c}$$

But hark! This gives not the final solution. Because any linear combination of these is also a solution. In particular any infinite summation that happens to uniformly converge is also a solution. You know an infinite summation of 1+siny in terms of y, so try finding A's and c's such that when you sum up all the terms and set x=1 you get the power series of 1+siny