H2 Maths 5.5 - Differential Equations: First-Order Linear (Integrating Factor)

Step 1 - Write in standard form

We want to solve d y by d x plus two over x times y equals x squared, given y equals two when x equals one.

Step 1 - Write in standard form

A first-order linear ODE has the standard form d y by d x plus P of x times y equals Q of x.

Step 1 - Write in standard form

Our equation is already in this form, with P of x equal to two over x and Q of x equal to x squared.

Step 1 - Write in standard form

The integrating factor method will turn the left-hand side into an exact derivative, which we can then integrate directly.

Step 2 - Find the integrating factor

The integrating factor is e to the power of the integral of P of x with respect to x.

Step 2 - Find the integrating factor

Integrating two over x gives two ln x, so the integrating factor is e to the two ln x.

Step 2 - Find the integrating factor

Using log laws, e to the two ln x simplifies to x squared.

Step 2 - Find the integrating factor

We drop the constant of integration here - one integrating factor is all we need.

Step 3 - Multiply through by the integrating factor

Multiply every term by x squared.

Step 3 - Multiply through by the integrating factor

The key observation is that the left-hand side is now exactly the derivative of x squared times y with respect to x.

Step 3 - Multiply through by the integrating factor

This is the product rule working in reverse: d by d x of x squared y gives x squared d y by d x plus two x y, which matches our left-hand side.

Step 3 - Multiply through by the integrating factor

The right-hand side becomes x squared times x squared, which is x to the fourth.

Step 4 - Integrate both sides

Integrate both sides with respect to x.

Step 4 - Integrate both sides

The left side integrates to x squared y directly.

Step 4 - Integrate both sides

The right side integrates to x to the fifth over five, plus an arbitrary constant C.

Step 4 - Integrate both sides

Divide through by x squared to isolate y.

Step 5 - Apply the initial condition

Substitute $y = 2$ when $x = 1$.

Step 5 - Apply the initial condition

Two equals one fifth plus C, so C equals nine fifths.

Step 5 - Apply the initial condition

The particular solution is y equals x cubed over five plus nine over five x squared.