We will expand our problem-solving journey in ODEs by diving into homogeneous equations and Bernoulli equations. These types of equations require different approaches and substitution methods when solving them.

A first-order ODE is homogeneous if it can be expressed as:

\frac{dy}{dx} = f(x, y)

1. Understanding Homogeneous and Non-Homogeneous Equations

If for every non-zero real value $k$ ,

f(kx, ky) = k^n f(x, y)

$f(x, y)$ is a homogeneous function of order $n$ .

This means that if you replace $x$ and $y$ with $kx$ and $ky$ , respectively, where $k$ is any non-zero number, the function $f(x, y) \$ remains unchanged upto a constant. Essentially, the equation still "looks the same" after scaling both $x$ and $y \$ by the same factor.

When solving problems, you can use any non-zero variable in place of $k \$ . This flexibility allows you to substitute different symbols, like $t \$ or $a \$ , as long as they represent a non-zero scaling factor. The concept remains the same: the function should stay unchanged (up to a constant) under the scaling.

Understanding what makes an equation homogeneous will make it much easier to tackle these problems later on. In this section, we'll go through some examples of homogeneous and non-homogeneous functions and demonstrate how to verify their homogeneity.

1.1. Examples

Example 1: Homogeneous Function

Consider the function:

$f(x, y) = \frac{y^2}{x}\$

To verify if it's homogeneous, we replace $x \$ with $kx \$ and $y \$ with $ky \$ , we can take any variable constant, let's take $k$ here:

$f(kx, ky) = \frac{(ky)^2}{kx} = \frac{k^2y^2}{kx} = k \cdot \frac{y^2}{x} = kf(x, y)\$

Since $f(kx, ky) = k^1 \cdot f(x, y) \$ , the function is homogeneous of degree $1$ .

Example 2: Non-Homogeneous Function

Now, consider the function:

$f(x, y) = \frac{y^2 + x}{x}\$

Replace $x$ with $kx$ and $y \$ with $ky \$ :

$f(kx, ky) = \frac{(ky)^2 + kx}{kx} = \frac{k^2y^2 + kx}{kx} = \frac{k^2y^2}{kx} + 1 = k \cdot \frac{y^2}{x} + 1\$

Here, $f(kx, ky) \$ is not a simple multiple of $f(x, y) \$ . Instead, it has an extra term $+1 \$ , making this function non-homogeneous.

Example 3: Homogeneous Function of Degree 2

Consider the function:
$f(x, y) = x^2 + y^2\$

To check for homogeneity, substitute $x \$ with $kx \$ and $y \$ with $ky \$ :
$f(kx, ky) = (kx)^2 + (ky)^2 = k^2x^2 + k^2y^2 = k^2(x^2 + y^2) = k^2f(x, y)\$

Since $f(kx, ky) = k^2 \cdot f(x, y) \$ , this function is homogeneous of degree $2$ .

Example 4: Non-Homogeneous Function

Consider the function:

$f(x, y) = x + y^2\$

Replace $x \$ with $kx \$ and $y \$ with $ky \$ :

$f(kx, ky) = kx + (ky)^2 = kx + k^2y^2\$

Here, $f(kx, ky) \$ is not a multiple of $f(x, y) \$ , as it contains terms with different powers of $k \$ . Thus, this function is not homogeneous.

Example 5: Non-Homogeneous Function

Consider the function:
$f(x, y) = \sin(x) + y\$

Replacing $x$ with $kx$ and $y$ with $ky$ :

$f(kx, ky) = \sin(kx) + ky\$

Since $\sin(kx) \$ is not a simple multiple of $\sin(x) \$ and the second term scales linearly with $k \$ , the function is not homogeneous.

Now we’ll move into solving the above types of ODEs.

2. Method of Solving Homogeneous Differential Equations

Let's go through the steps to solve a first-order differential equation of the form:

\frac{dy}{dx} = f(x, y)

where $f(kx, ky) = k^n f(x, y)\$ is a homogeneous function of order $n$ .

Steps to Solve:

1. Determine whether the equation is homogeneous:

- Check if the function $f(x, y)\$ satisfies the condition:

$f(kx, ky) = k^n f(x, y) \$

for some integer $n$ . If this is true, the equation is homogeneous.

2. Use Substitution $y = vx \$ :

- Set $y = vx \$ , where $v$ is a function of $x$ .

- This implies that $\frac{dy}{dx} = v + x\frac{dv}{dx} \$ .

- Substitute these into the original differential equation.

3. Convert the Equation into a Separable Form:

- After substituting $y = vx \$ , the equation should become separable, you can rearrange the terms to isolate $v$ and $x$ on different sides:

$v + x\frac{dv}{dx} = f(x, vx) \$

4. Integrate Both Sides:

- Integrate the equation with respect to $x$ (and $v$ if necessary).

- This should give you an implicit or explicit solution involving $v$ and $x$ .

5. Use Initial Conditions (if Given):

- If the problem provides initial conditions (like $y(x_0) = y_0 \$ ), use them to determine the constant of integration.

By following these steps, you can solve homogeneous first-order differential equations systematically.

2.1. Examples: Solve the following equations.

Example 1:

Solve

\frac{dy}{dx} = \frac{y}{x}

Solution:

1. Substitute $y = vx \$ :

$\frac{dy}{dx} = v + x\frac{dv}{dx} \$

So,

$\ v + x\frac{dv}{dx} = \frac{vx}{x} = v \$

Thus,

$x\frac{dv}{dx} = 0 \$

2. Integrate:

$\frac{dv}{dx} = 0 \quad \Rightarrow \quad v = C \$

$y = Cx \$

Example 2:

Solve

\frac{dy}{dx} = \frac{x^2 + y^2}{2xy}

Solution:

1. Substitute $y = vx \$ :

$\frac{dy}{dx} = v + x\frac{dv}{dx} \$

So,
$v + x\frac{dv}{dx} = \frac{x^2 + v^2x^2}{2vx^2} = \frac{1 + v^2}{2v} \$

2. Separate variables:

$x\frac{dv}{dx} = \frac{1 + v^2}{2v} - v = \frac{1 - v^2}{2v} \$

$\frac{2v}{1 - v^2} \, dv = \frac{dx}{x} \$

3. Integrate:

$\int \frac{2v}{1 - v^2}dv = \int \frac{dx}{x} \$

$u$ -sustitution, $\ u = 1 - v^2 \Rightarrow du = -2v \, dv \$ :

$-\int \frac{du}{u} = \ln|x| + C \$

$-\ln{|u|} = \ln{|x|} + C$

$-\ln|1 - v^2| = \ln|x| + C \$

$\ln|1 - v^2| = -\ln|x| + C \$

$e^{\ln{|1 - v^2|}} = e^{-\ln{|x|} + C}$

$1 - v^2 = K \frac{1}{x}$

$1 - v^2 = \frac{K}{x} \$

$1 - \frac{y^2}{x^2} = \frac{K}{x}$

$\frac{y^2}{x^2} = 1 - \frac{K}{x}$

$y^2 = x^2 - K x$

Final Solution:

$y = \pm \sqrt{x^2 - Kx}\$

Example 3:

Solve

\frac{dy}{dx} = \frac{3x - y}{x + y}

Solution:

1. Substitute $y = vx \$ :

$\frac{dy}{dx} = v + x\frac{dv}{dx}\$

So,

$v + x\frac{dv}{dx} = \frac{3x - vx}{x + vx} = \frac{3 - v}{1 + v} \$

2. Separate variables:

$x\frac{dv}{dx} = \frac{3 - v - v(1 + v)}{1 + v} = \frac{3 - 2v - v^2}{1 + v}$

$\frac{1 + v}{(3 + v)(1 - v)} \, dv = \frac{dx}{x} \$

3. Integrate:

Use partial fractions to integrate:

$\int \frac{1}{2(1-v)} - \frac{1}{2(3+v)} dv = \int \frac{1}{x} dx$

$-\frac{1}{2} \ln{|1-v|} - \frac{1}{2} \ln{|3+v|} = \ln{|x|} + C$

$-\frac{1}{2}\ln{|(1-v)(3+v)|} = \ln{|x|} + C$

Taking exponents on both sides,

$\frac{1}{\sqrt{(1-v)(3+v)}} = Kx$

Substituting $v = \frac{y}{x}$

$\frac{1}{\sqrt{(1-\frac{y}{x})(3+\frac{y}{x})}} = Kx$

$\frac{x}{\sqrt{(x - y)(3x+y)}} = Kx$

Cross-multiplying and squaring both sides.

$(x-y)(3x+y) = K_0$

Final Solution:

$(x-y)(3x+y) = K_0$

Example 4:

Example 4

Solve

$\frac{dy}{dx} = \frac{y}{x + y}\$

Answer

Solution:

1. Substitute $y = vx \$ :

$x \frac{dv}{dx} = \frac{v}{1 + v} - v\$

Simplifying the right-hand side:

$x \frac{dv}{dx} = \frac{v(1 + v - 1)}{1 + v} = \frac{v^2}{1 + v}\$

Thus, the equation becomes:

$\frac{dv}{dx} = \frac{v^2}{x(1 + v)}\$

Now, separating variables:

$\frac{1 + v}{-v^2} dv = \frac{dx}{x}\$

Step 3: Integrate both sides

Now, break the left side into two parts:

$-\int \left( \frac{1}{v^2} + \frac{1}{v} \right) dv = \int \frac{dx}{x}\$

Which simplifies to:

$-\int \frac{1}{v^2} dv - \int \frac{1}{v} dv = \int \frac{dx}{x}\$

The integrals on the left-hand side are:

$\frac{1}{v} - \ln|v|\$

So the equation becomes:

$\frac{1}{v} - \ln|v| = \ln|x| + C\$

Step 4: Substitute back $v = \frac{y}{x} \$

Now, substitute $v = \frac{y}{x} \$ back into the equation:

Final solution: $\frac{x}{y} - \ln\left|\frac{y}{x}\right| = \ln|x| + C\$

Example 5:

Example 5

Solve

\frac{dy}{dx} = \frac{2x^2 - y^2}{2xy}

Answer

Solution:

1. Substitute $y = vx$

$\frac{dy}{dx} = v + x\frac{dv}{dx}$

$v + x\frac{dv}{dx} = \frac{2x^2 - v^2x^2}{2vx^2} = \frac{2 - v^2}{2v}$

$x\frac{dv}{dx} = \frac{2 - v^2}{2v} - v$

2. Separate variables:

$x\frac{dv}{dx} = \frac{2 - v^2 - 2v^2}{2v} = \frac{2 - 3v^2}{2v}$
$\frac{2v}{2 - 3v^2} \, dv = \frac{dx}{x}$

3. Integrate:

Use $u$ -substitution: $u = 2 - 3v^2$

$-\frac{1}{3}\int \frac{1}{u} dv= \int \frac{dx}{x}$

$-\frac{1}{3} \ln|2 - 3v^2| = \ln|x| + C$

Final Solution:

$-\frac{1}{3} \ln \left|2 - 3 \frac{y^2}{x^2} \right| = \ln|x| + C$

3. Practice Questions

Solve the following homogeneous equations,

$\sqrt{x^2 + y^2} \, dx = x \, dy - y \, dx, \quad y(1) = 0\$
$\frac{dy}{dx} = \frac{3xy}{x^2 + y^2}\$
$\frac{dy}{dx} = \frac{x - y}{x + y}\$

Not sure about your answers? Don't worry, click the link below to check the video where the answers are explained.

Practice Questions Discussion

Related Articles

5. Substitution methods

1. Understanding Homogeneous and Non-Homogeneous Equations

1.1. Examples

2. Method of Solving Homogeneous Differential Equations

2.1. Examples: Solve the following equations.

Example 4

Example 5

3. Practice Questions