# Partial Differential du/dx Equations

• SunGod87
In summary, the conversation is about solving two given questions using the assumption u = X.Y. The solution involves equating expressions to a constant m and integrating. The only mistake made was assuming that the constant C was the same in both parts, when in fact it is just an unknown constant. The book's answer combines the two unknown constants into one, making it simpler.
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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SunGod87 said:
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?
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!)

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

## 1. What is a partial differential equation (PDE)?

A PDE is a mathematical equation that involves multiple independent variables and their partial derivatives. It is used to describe the relationship between these variables and how they change in relation to each other.

## 2. What is the difference between a PDE and an ordinary differential equation (ODE)?

A PDE involves multiple independent variables, while an ODE only involves one independent variable. This means that the solutions to a PDE are functions of multiple variables, while the solutions to an ODE are functions of a single variable.

## 3. What are the applications of PDEs in science?

PDEs are used to model various physical phenomena in fields such as physics, engineering, and mathematics. Some examples include heat transfer, fluid dynamics, and quantum mechanics.

## 4. How are PDEs solved?

The method of solving a PDE depends on its type and complexity. Some common techniques include separation of variables, Fourier transforms, and numerical methods such as finite difference and finite element methods.

## 5. What are the challenges in solving PDEs?

PDEs are often difficult to solve analytically, meaning a closed-form solution cannot be obtained. In these cases, numerical methods must be used, which can be time-consuming and require significant computational resources. Additionally, PDEs can have multiple solutions or no solution at all, making it challenging to determine the correct solution for a given problem.

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