Trying to solve the two questions attached, for the first one:(adsbygoogle = window.adsbygoogle || []).push({});

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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# Homework Help: Partial Differential du/dx Equations

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