H2 Maths 5.5 - Differential Equations: Separable ODEs

Step 1 - Recognise a separable ODE

We want to solve the differential equation d y by d x equals x times y minus one.

Step 1 - Recognise a separable ODE

A separable ODE is one where you can write d y by d x as a product of a function of x only and a function of y only.

Step 1 - Recognise a separable ODE

Here the right-hand side is x times open bracket y minus one close bracket, so it separates cleanly.

Step 1 - Recognise a separable ODE

The strategy is to move all y terms to the left and all x terms to the right, then integrate both sides.

Step 2 - Separate the variables

Divide both sides by y minus one, treating d y by d x as a fraction for now.

Step 2 - Separate the variables

This gives one over y minus one, d y equals x d x.

Step 2 - Separate the variables

Both sides are now in terms of one variable only, so we can integrate.

Step 3 - Integrate both sides

The left side integrates to the natural log of the absolute value of y minus one.

Step 3 - Integrate both sides

The right side integrates to x squared over two.

Step 3 - Integrate both sides

We write a single arbitrary constant c on the right.

Step 4 - Solve for y

Exponentiate both sides to remove the logarithm.

Step 4 - Solve for y

The right side becomes e to the power x squared over two plus c, which we write as A e to the x squared over two, where A is a new arbitrary constant.

Step 4 - Solve for y

Since y minus one can be positive or negative, A can be any non-zero real number.

Step 5 - Apply the initial condition

Now apply the condition $y = 3$ when $x = 0$.

Step 5 - Apply the initial condition

Substituting gives three equals one plus A times e to the zero, so A equals two.

Step 5 - Apply the initial condition

The particular solution is y equals one plus two e to the x squared over two.