H2 Maths 5.5 - Differential Equations: Second-Order Linear with Constant Coefficients

Step 1 - Write the auxiliary equation

We want to solve y double prime plus three y prime plus two y equals six e to the x, given y equals one and y prime equals zero when x equals zero.

Step 1 - Write the auxiliary equation

For a second-order linear ODE with constant coefficients, the first task is always to find the complementary function by solving the auxiliary equation.

Step 1 - Write the auxiliary equation

Replace y double prime with m squared, y prime with m, and y with one.

Step 1 - Write the auxiliary equation

This gives m squared plus three m plus two equals zero.

Step 2 - Write the complementary function

When the auxiliary equation has two distinct real roots m one and m two, the complementary function is A e to the m one x plus B e to the m two x.

Step 2 - Write the complementary function

Here the roots are minus one and minus two, so the complementary function is A e to the minus x plus B e to the minus two x.

Step 2 - Write the complementary function

A and B are arbitrary constants that will be fixed later by the initial conditions.

Step 3 - Find the particular integral

The particular integral is a specific solution to the full equation with the right-hand side included.

Step 3 - Find the particular integral

Since the right-hand side is six e to the x, we try a particular integral of the form y p equals k e to the x.

Step 3 - Find the particular integral

Differentiating: y p prime equals k e to the x and y p double prime equals k e to the x.

Step 3 - Find the particular integral

Substituting into the ODE: k e to the x plus three k e to the x plus two k e to the x equals six k e to the x equals six e to the x, so k equals one.

Step 4 - Write the general solution

The general solution is the sum of the complementary function and the particular integral.

Step 4 - Write the general solution

This is always the structure for a non-homogeneous linear ODE: y equals y c plus y p.

Step 4 - Write the general solution

So the general solution is y equals A e to the minus x plus B e to the minus two x plus e to the x.

Step 5 - Apply the initial conditions

Differentiate to get $y'$, then apply both initial conditions simultaneously.

Step 5 - Apply the initial conditions

From y equals one at x equals zero: A plus B plus one equals one, so A plus B equals zero.

Step 5 - Apply the initial conditions

From y prime equals zero at x equals zero: minus A minus two B plus one equals zero, so minus A minus two B equals minus one.

Step 5 - Apply the initial conditions

Adding the two equations gives minus B equals minus one, so B equals one and A equals minus one.