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Differential equations – The technique of separating the variables

📘 Notes

Differential Equations — The Technique of Separating the Variables

In Year 13 Cambridge 9709, some differential equations can be solved by separating the variables. This means rearranging the equation so that all terms involving \( y \) are on one side and all terms involving \( x \) are on the other side.

After separating, we integrate both sides and then use any given condition to find the constant of integration. This is a standard exam method and it is important to show each step clearly.

Key definitions and formulae

1. Separable differential equation

A differential equation is separable if it can be written in the form

\[ \frac{dy}{dx}=f(x)g(y). \]

We then rearrange it so that the \( y \)-terms and \( dy \) are together, and the \( x \)-terms and \( dx \) are together.

2. Standard separation step

If

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

then separate the variables as

\[ \frac{1}{g(y)}\,dy=f(x)\,dx. \]

Then integrate both sides.

3. Integration step

\[ \int \frac{1}{g(y)}\,dy=\int f(x)\,dx. \]

Do not forget the constant of integration.

4. Using an initial condition

If a point such as \( (x_0,y_0) \) is given, substitute these values into the integrated equation to find the constant.

\[ y=y_0 \quad \text{when} \quad x=x_0. \]

5. General method

  1. Rearrange the differential equation so the variables are separated.
  2. Integrate both sides.
  3. Add the constant of integration.
  4. Use any given condition to find the constant.
  5. If needed, rearrange to make \( y \) the subject.

Worked examples

Example 1: Solve a simple separable differential equation

Solve

\[ \frac{dy}{dx}=3x. \]

Solution

This is already separated, so integrate both sides:

\[ dy=3x\,dx \] \[ \int dy=\int 3x\,dx \] \[ y=\frac{3}{2}x^2+c. \]
The general solution is \[ y=\frac{3}{2}x^2+c. \]

Example 2: Separate the variables and solve

Solve

\[ \frac{dy}{dx}=xy. \]

Solution

Rearrange so that all \( y \)-terms are on one side:

\[ \frac{1}{y}\frac{dy}{dx}=x \] \[ \frac{1}{y}\,dy=x\,dx. \]

Integrate both sides:

\[ \int \frac{1}{y}\,dy=\int x\,dx \] \[ \ln|y|=\frac{1}{2}x^2+c. \]
The general solution is \[ \ln|y|=\frac{1}{2}x^2+c. \]

Example 3: Solve and use an initial condition

Solve

\[ \frac{dy}{dx}=2xy \]

given that \( y=3 \) when \( x=0 \).

Solution

Separate the variables:

\[ \frac{1}{y}\,dy=2x\,dx. \]

Integrate:

\[ \int \frac{1}{y}\,dy=\int 2x\,dx \] \[ \ln|y|=x^2+c. \]

Use the condition \( y=3 \) when \( x=0 \):

\[ \ln 3=0+c \] \[ c=\ln 3. \]

So

\[ \ln|y|=x^2+\ln 3. \]

Now make \( y \) the subject:

\[ |y|=e^{x^2+\ln 3}=3e^{x^2}. \]

Since \( y=3 \) when \( x=0 \), \( y \) is positive here, so

\[ y=3e^{x^2}. \]
The particular solution is \[ y=3e^{x^2}. \]

Example 4: Solve with a negative power of \( y \)

Solve

\[ \frac{dy}{dx}=\frac{x}{y^2}. \]

Solution

Separate the variables:

\[ y^2\,dy=x\,dx. \]

Integrate both sides:

\[ \int y^2\,dy=\int x\,dx \] \[ \frac{1}{3}y^3=\frac{1}{2}x^2+c. \]

Multiply through by \( 6 \) if a simpler form is wanted:

\[ 2y^3=3x^2+c. \]
A correct general solution is \[ \frac{1}{3}y^3=\frac{1}{2}x^2+c. \]

Example 5: Solve using exponential integration

Solve

\[ \frac{dy}{dx}=e^x y \]

given that \( y=2 \) when \( x=0 \).

Solution

Separate the variables:

\[ \frac{1}{y}\,dy=e^x\,dx. \]

Integrate:

\[ \int \frac{1}{y}\,dy=\int e^x\,dx \] \[ \ln|y|=e^x+c. \]

Use \( y=2 \) when \( x=0 \):

\[ \ln 2=e^0+c=1+c \] \[ c=\ln 2-1. \]

So

\[ \ln|y|=e^x+\ln 2-1. \]

Exponentiate:

\[ y=e^{e^x+\ln 2-1} \] \[ y=2e^{e^x-1}. \]
The particular solution is \[ y=2e^{e^x-1}. \]

Common mistakes and exam tips

Common mistakes

  • Not separating the variables correctly before integrating.
  • Forgetting the constant of integration.
  • Writing \( \ln y \) instead of \( \ln|y| \) when integrating \( \dfrac{1}{y} \).
  • Using the initial condition before integrating.
  • Making algebra mistakes when rearranging to find \( y \).

Exam tips

  • Show the separation step clearly, for example \[ \frac{1}{y}\,dy=x\,dx. \]
  • Integrate both sides carefully and write the constant only once.
  • If an initial condition is given, use it to find the constant as soon as you have integrated.
  • Leave the answer in implicit form if that is simpler, unless the question asks for \( y \) explicitly.
  • Check that your final answer satisfies the given condition.

Summary

\[ \frac{dy}{dx}=f(x)g(y) \] \[ \frac{1}{g(y)}\,dy=f(x)\,dx \] \[ \int \frac{1}{g(y)}\,dy=\int f(x)\,dx \]

The method of separating the variables is one of the main techniques for solving first-order differential equations in Year 13. The key idea is to separate, integrate, add the constant, and then use any given condition carefully.