# Partial Differential Equations

1. Nov 10, 2006

### SunGod87

Trying to solve the two questions attached, for the first one:

du/dx - x.du/dy = 0

Assume u = X.Y
Y.dX/dx - x.X.dY/dy = 0
Dividing by x.X.Y and taking one term over to the other side:
dX/dx.1/(x.X) = dY/dy.1/Y

These can be equated to a constant m

dX/X = m.x.dx
Integrating;
ln(X/C) = 1/2.m.x^2
X = C.e^((1/2.m.x^2))

dY/Y = m.dy
ln(Y/C) = m.y
Y = C.e^((y.m))

u = C^2.e^((1/2.x^2 + y).m)

However the answer given is C.e^((x^2 + 2y).m)

Have they absorbed the two constants earlier into one? Since the equation is a first order one. Also they seem to have multiplied the exponent by 2 throughout, surely this isn't allowed so I assume I've made a mistake?

Second question:
x.du.dx - 2y.du/dy = 0
Assume u = X.Y
x.Y.dX/dx - 2y.X.dY/dy = 0

Dividing by 2.X.Y and taking one term to the other side:

x/2X.dX/dx = y/Y.dY/dy
These can be equated to a constant m

y/Y.dY/dy = m
dY/Y = m.dy/y
ln(Y/C) = m.ln(y)
ln(Y/C) = ln(y^m)
Y = C.y^m

x/2X.dX/dx = m
dX/2X = m.dx/x
1/2.ln(X/C) = m.ln(x)
ln(X/C) = 2.m.ln(x)
ln(X/C) = ln(x^2m)
X = C.x^2m

XY = C^2.(x^2.y)^m

However the book has it as:

C(x^2.y)^m

Again have they just combined the two constants into one since the equation is only first order?

Thanks in advance for any help :)

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Last edited: Nov 10, 2006
2. Nov 11, 2006

### HallsofIvy

Staff Emeritus
The only "mistake" you made was in assuming that the constant C in was the same in each case. you are really saying that
$$X= C_1e^{\frac{1}{2}mx^2}$$
$$Y= C_2e^{my}$$
and so
$$u= (C_1C_2)e^{\frac{1}{2}mx^2+ my}$$
Since C1 and C2 are unknown constants, their product is also an unknown constant: let C= C1C2.
As far as the factor of 2 is concerned, what is m? Suppose you had said "These can be equated to a constant 2m" at the beginning? (Someone just doesn't like fractions!)

Exactly. You have no reason to assume that C is the same in both parts. You really have
$$XY= C_1C_2(x^2y)^m$$
Since C1 and C2 are just unknown constants, so is there product. Let C= C1C2.

Last edited: Nov 11, 2006
3. Nov 11, 2006

### SunGod87

Ah okay, all makes sense. Thank you very much :)