Reduction of Order

The Essentials

The solution to a second order linear differential equation takes the form: \( y(x) = y_1(x) u(x) \), where \( y_1 (x) \) is given. Note that \( y_1 (x) \) is part of the homogeneous solution, even if the equation given isn’t homogeneous. The general solution \( y (x) \) is differentiated twice and substituted into the equation, at which point the equation can be put into this form:

The term k(x) will end up being 0, so the equation will have this form:

This equation can be solved as a first order differential equation after a substitution as it doesn’t explicitly contain u. After \( u(x) \) is found, it is plugged in here \( y(x) = y_1(x) u(x) \) to get the general solution. This work for homogeneous equations (where \( h(x) = 0 \) ) and non-homogeneous equations.

Example

Solve this equation given that \( y_1(x) = x^2 \) :

First, we can verify that \( y_1(x) \) is a solution is the homogeneous equation. If we set the right hand side to 0 and plug in \( x^2 \) for y, 2x for y′ and 2 for y′′ we get:

Which is true. Our solution is going to take the form:

So we plug in our value for y1(x) and differentiate the result:

Now, we plug in these values into the original equation:

Then, we can arrange this to be factors of u, u′, and u′′:

This equation is easily solved using direct integration:

Next, we can write down our general solution:

Upon inspection, both \( x^3 \) and \( x^2 \) are homogeneous solutions (they both make the equation equal zero) and a linear combination of these two terms make a basis for the homogeneous solution. The first term \( \frac{4}{ 15} \sqrt{x} \) is the particular solution (it makes the equation equal the forcing function, in this case \( \sqrt{x} \)).

Practice