Question
How do I solve a differential equation with initial conditions using separation of variables?
Answers
GPT-4o mini
To solve a differential equation using the method of separation of variables, follow these steps:
### Step 1: Identify the Differential Equation
Ensure that you have a first-order differential equation that can be expressed in the form:
\[
\frac{dy}{dx} = g(x) h(y)
\]
where \( g(x) \) is a function of \( x \) only and \( h(y) \) is a function of \( y \) only.
### Step 2: Rearrange the Equation
Rearrange the equation to separate the variables \( y \) and \( x \):
\[
\frac{1}{h(y)} dy = g(x) dx
\]
### Step 3: Integrate Both Sides
Integrate both sides:
\[
\int \frac{1}{h(y)} dy = \int g(x) dx
\]
### Step 4: Solve the Integrals
After integrating, you'll get two expressions that may include an arbitrary constant \( C \):
\[
H(y) = G(x) + C
\]
where \( H(y) \) is the integral of \( \frac{1}{h(y)} \) with respect to \( y \) and \( G(x) \) is the integral of \( g(x) \) with respect to \( x \).
### Step 5: Solve for y
If possible, solve for \( y \) in terms of \( x \). This might involve algebraic manipulation or isolating \( y \).
### Step 6: Apply Initial Conditions
If an initial condition is provided (e.g., \( y(x_0) = y_0 \)), substitute \( x = x_0 \) and \( y = y_0 \) into your equation to find the specific constant \( C \).
### Step 7: Write the Particular Solution
Substitute \( C \) back into your solution to get the particular solution that satisfies the initial conditions.
### Example
Let's solve the differential equation:
\[
\frac{dy}{dx} = y(1 - y)
\]
with the initial condition \( y(0) = 0.5 \).
1. **Separate the Variables**:
\[
\frac{1}{y(1-y)} dy = dx
\]
2. **Integrate Both Sides**:
To integrate the left side, use partial fraction decomposition:
\[
\frac{1}{y(1-y)} = \frac{1}{y} + \frac{1}{1-y}
\]
Now integrate:
\[
\int \frac{1}{y} dy + \int \frac{1}{1 - y} dy = \int dx
\]
This results in:
\[
\ln |y| - \ln |1 - y| = x + C
\]
Which simplifies to:
\[
\ln \left| \frac{y}{1 - y} \right| = x + C
\]
Exponentiating both sides gives:
\[
\frac{y}{1 - y} = e^{x + C}
\]
Let \( e^C = K \) (a new constant):
\[
\frac{y}{1 - y} = K e^x
\]
Hence:
\[
y = \frac{K e^x}{1 + K e^x}
\]
3. **Apply the Initial Condition**:
Now, apply \( y(0) = 0.5 \):
\[
0.5 = \frac{K}{1 + K} \quad \Rightarrow \quad 1 + K = 2K \quad \Rightarrow \quad K = 1
\]
4. **Write the Particular Solution**:
So the particular solution is:
\[
y = \frac{e^x}{1 + e^x}
\]
This completes the solution using separation of variables with the given initial condition!
### Step 1: Identify the Differential Equation
Ensure that you have a first-order differential equation that can be expressed in the form:
\[
\frac{dy}{dx} = g(x) h(y)
\]
where \( g(x) \) is a function of \( x \) only and \( h(y) \) is a function of \( y \) only.
### Step 2: Rearrange the Equation
Rearrange the equation to separate the variables \( y \) and \( x \):
\[
\frac{1}{h(y)} dy = g(x) dx
\]
### Step 3: Integrate Both Sides
Integrate both sides:
\[
\int \frac{1}{h(y)} dy = \int g(x) dx
\]
### Step 4: Solve the Integrals
After integrating, you'll get two expressions that may include an arbitrary constant \( C \):
\[
H(y) = G(x) + C
\]
where \( H(y) \) is the integral of \( \frac{1}{h(y)} \) with respect to \( y \) and \( G(x) \) is the integral of \( g(x) \) with respect to \( x \).
### Step 5: Solve for y
If possible, solve for \( y \) in terms of \( x \). This might involve algebraic manipulation or isolating \( y \).
### Step 6: Apply Initial Conditions
If an initial condition is provided (e.g., \( y(x_0) = y_0 \)), substitute \( x = x_0 \) and \( y = y_0 \) into your equation to find the specific constant \( C \).
### Step 7: Write the Particular Solution
Substitute \( C \) back into your solution to get the particular solution that satisfies the initial conditions.
### Example
Let's solve the differential equation:
\[
\frac{dy}{dx} = y(1 - y)
\]
with the initial condition \( y(0) = 0.5 \).
1. **Separate the Variables**:
\[
\frac{1}{y(1-y)} dy = dx
\]
2. **Integrate Both Sides**:
To integrate the left side, use partial fraction decomposition:
\[
\frac{1}{y(1-y)} = \frac{1}{y} + \frac{1}{1-y}
\]
Now integrate:
\[
\int \frac{1}{y} dy + \int \frac{1}{1 - y} dy = \int dx
\]
This results in:
\[
\ln |y| - \ln |1 - y| = x + C
\]
Which simplifies to:
\[
\ln \left| \frac{y}{1 - y} \right| = x + C
\]
Exponentiating both sides gives:
\[
\frac{y}{1 - y} = e^{x + C}
\]
Let \( e^C = K \) (a new constant):
\[
\frac{y}{1 - y} = K e^x
\]
Hence:
\[
y = \frac{K e^x}{1 + K e^x}
\]
3. **Apply the Initial Condition**:
Now, apply \( y(0) = 0.5 \):
\[
0.5 = \frac{K}{1 + K} \quad \Rightarrow \quad 1 + K = 2K \quad \Rightarrow \quad K = 1
\]
4. **Write the Particular Solution**:
So the particular solution is:
\[
y = \frac{e^x}{1 + e^x}
\]
This completes the solution using separation of variables with the given initial condition!