Question

Solve the following first-order partial differential equations by separating the variables: (a) \(\frac{\partial u}{\partial x}-x \frac{\partial u}{\partial y}=0\); (b) \(x \frac{\partial u}{\partial x}-2 y \frac{\partial u}{\partial y}=0\).

Short answer

Solutions are: (a) \(u(x,y) = Ce^{kx}y^k \); (b) \(u(x,y) = Ce^{kx}y^{k/2} \).

Step-by-step solution

Introduction to Separating Variables

To solve a first-order partial differential equation by separating variables, assume the solution is in a separable form, such as: \( u(x,y) = X(x)Y(y) \)

Apply Separation of Variables to Equation (a)

For equation (a) \( \frac{\partial u}{\partial x}-x \frac{\partial u}{\partial y}=0 \), assume \( u(x,y) = X(x)Y(y) \). Then, compute the partial derivatives: \( \frac{\partial u}{\partial x} = X'(x)Y(y) \) and \( \frac{\partial u}{\partial y} = X(x)Y'(y) \). Substitute these into the equation: \( X'(x)Y(y) - xX(x)Y'(y) = 0 \)

Separate Variables for Equation (a)

Factor out common terms from the resulting equation: \( Y(y)X'(x) = xX(x)Y'(y) \). Rearrange to isolate functions of \( x \) and \( y \) on opposite sides: \( \frac{X'(x)}{X(x)} = x \frac{Y'(y)}{Y(y)} \). Since the left side depends only on \( x \) and the right only on \( y \), each side must equal a constant, say \( k \): \( \frac{X'(x)}{X(x)} = k \) and \( x \frac{Y'(y)}{Y(y)} = k \)

Solve for Functions of x and y in Equation (a)

Solve \( \frac{X'(x)}{X(x)} = k \) by integrating: \( X(x) = Ae^{kx} \). Solve \( x \frac{Y'(y)}{Y(y)} = k \) by integrating: \( Y(y) = By^{k} \). Therefore, the solution to the PDE is: \( u(x,y) = Ae^{kx}By^{k} = C e^{kx} y^{k} \), where \( C = AB \)

Apply Separation of Variables to Equation (b)

For equation (b) \( x \frac{\partial u}{\partial x}-2 y \frac{\partial u}{\partial y}=0 \), assume \( u(x,y) = X(x)Y(y) \). Then, compute the partial derivatives: \( x \frac{\partial u}{\partial x} = xX'(x)Y(y) \) and \( \frac{\partial u}{\partial y} = X(x)Y'(y) \). Substitute these into the equation: \( xX'(x)Y(y) - 2yX(x)Y'(y) = 0 \)

Separate Variables for Equation (b)

Factor out common terms from the resulting equation: \( xX'(x)Y(y) = 2yX(x)Y'(y) \). Rearrange to isolate functions of \( x \) and \( y \) on opposite sides: \( \frac{X'(x)}{X(x)} = \frac{2y}{x} \frac{Y'(y)}{Y(y)} \). Since the left side depends only on \( x \) and the right only on \( y \), each side must equal a constant, say \( k \): \( \frac{X'(x)}{X(x)} = k \) and \( \frac{2y}{x} \frac{Y'(y)} = k \)

Solve for Functions of x and y in Equation (b)

Solve \( \frac{X'(x)}{X(x)} = k \) by integrating: \( X(x) = Ae^{kx} \). Solve \( \frac{2y}{x} \frac{Y'(y)} = k \) by first separating: \( 2 \frac{Y'(y)}{Y(y)} = \frac{kx}{y} \). Integrate both sides: \( Y(y) = By^{k/2} \). Therefore, the solution to the PDE is: \( u(x,y) = Ae^{kx}By^{k/2} = C e^{kx} y^{k/2} \), where \( C = AB \)

Key concepts

separation of variables

The method of separation of variables is a powerful technique for solving first-order partial differential equations (PDEs). This method assumes that the solution can be written as a product of functions, each depending on a single variable. For example, if we have a function of two variables, say \(u(x,y)\), we assume it can be expressed as \(u(x, y) = X(x)Y(y)\), where \(X(x)\) is a function of \(x\) alone and \(Y(y)\) is a function of \(y\) alone.
The key idea is to break down a complicated problem involving multiple variables into simpler problems involving only one variable each. Thus, our goal is to rewrite the PDE in a form where we can separate the variables.
For example, let's take a typical step: consider the partial derivatives \(\frac{\partial u}{\partial x} = X'(x)Y(y)\) and \(\frac{\partial u}{\partial y} = X(x)Y'(y)\). When substituted back into the original PDE, we can often arrange the terms to isolate functions of each variable on opposite sides of the equation. This gives us something like \( \frac{X'(x)}{X(x)} = f(x)\) and \(\frac{Y'(y)}{Y(y)} = g(y)\), where \(f(x)\) and \(g(y)\) are functions of \(x\) and \(y\), respectively.
Since the left side of each equation depends only on one variable and the right side only on the other, each side must equal the same constant, usually denoted by \(k\). This reduces the problem to solving simpler ordinary differential equations (ODEs). Finally, by solving these ODEs, you find functions for \(X(x)\) and \(Y(y)\), leading you to the general solution of the PDE.

partial derivatives

Partial derivatives are a core concept when dealing with functions of multiple variables. If you have a function \(u(x,y)\), the partial derivatives represent how the function changes as you vary one of the variables while keeping the other constant.
For instance, the partial derivative \(\frac{\partial u}{\partial x}\) measures the rate of change of \(u\) as \(x\) changes, keeping \(y\) fixed. Similarly, \(\frac{\partial u}{\partial y}\) measures the rate of change of \(u\) with respect to \(y\), keeping \(x\) fixed.
Mathematically, if \(u(x,y)\) is our function, these can be written as:
- \(\frac{\partial u}{\partial x} = \lim_{h \to 0} \frac{u(x+h,y) - u(x,y)}{h}\)
- \(\frac{\partial u}{\partial y} = \lim_{h \to 0} \frac{u(x,y+h) - u(x,y)}{h}\)
When we use the method of separation of variables, we rely heavily on computing these partial derivatives assuming the solution can be factored into products of single-variable functions. For example, if \(u(x,y) = X(x)Y(y)\), then \(\frac{\partial u}{\partial x} = X'(x)Y(y)\) and \(\frac{\partial u}{\partial y} = X(x)Y'(y)\). Substituting these back into the original partial differential equations gives us equations that can be separated into simpler, ordinary differential equations.

integration

Integration is the process of finding the antiderivative of a function. In solving PDEs using separation of variables, after separating the variables and equating them to a constant, you will typically end up with ordinary differential equations (ODEs). These ODEs require integration to solve.
For example, suppose you have the equation \(\frac{X'(x)}{X(x)} = k\). To solve for \(X(x)\), you integrate both sides with respect to \(x\):
\[\int \frac{1}{X(x)} X'(x) \, dx = \int k \, dx\].
This simplifies to:\[\ln|X(x)| = kx + C\].
Exponentiating both sides to solve for \(X(x)\), we get:
\[X(x) = Ae^{kx}\], where \(A = e^C\) is the integration constant.
Similarly, for \(Y(y)\), if you have \(\frac{Y'(y)}{Y(y)} = \frac{k}{y}\), you integrate:
\[\int \frac{1}{Y(y)} Y'(y) \, dy = \int \frac{k}{y} \, dy\].
This results in:\[\ln|Y(y)| = k \ln|y| + C\].
Exponentiating both sides yields:
\[Y(y) = By^k\], where \(B = e^C\) is another constant.
These steps show how integration is used repeatedly in solving differential equations, making it an essential part of the process.