17calculus 17calculus
First Order Second Order Laplace Transforms Additional Topics Applications, Practice
Separation of Variables
Integrating Factors (Linear)
Exact Equations
Integrating Factors (Exact)
Constant Coefficients
Reduction of Order
Undetermined Coefficients
Variation of Parameters
Polynomial Coefficients
Cauchy-Euler Equations
Chebyshev Equations
Laplace Transforms
Unit Step Function
Unit Impulse Function
Square Wave
Shifting Theorems
Solve Initial Value Problems
Classify Differential Equations
Fourier Series
Slope Fields
Existence and Uniqueness
Boundary Value Problems
Euler's Method
Inhomogeneous ODE's
Partial Differential Equations
Linear Systems
Exponential Growth/Decay
Population Dynamics
Projectile Motion
Chemical Concentration
Fluids (Mixing)
Practice Problems
Practice Exam List
Exam A1
Exam A3
Exam B2

You CAN Ace Differential Equations

17calculus > differential equations > reduction of order

Topics You Need To Understand For This Page

Differential Equations Alpha List


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Reduction of Order

classification: second-order, linear, homogeneous

\(y'' + p(t)y' + q(t)y = 0\)

This technique takes a second-order, linear differential equation with one known solution and reduces the equation to a first-order, linear equation that may possibly be solved using a first-order technique. In general, the idea can be applied to even higher order equations but, due to complexity of integration, this technique is really only used on second-order equations. However, it can still be useful in many cases.

The idea is that we have an equation of the form \(y'' + p(t)y' + q(t)y = 0\) and we know one solution, \(y_1(t)\). To get a second solution, we try \(y_2=v(t)y_1(t)\), i.e. we assume that the second solution is equal to the first solution times some function of t. This may seem unusual but it works in many cases. Now, let's work through the calculus and see where it gets us.

For simplicity, letting \(y=y_2\), we have \(y=v(t)y_1(t)\). Taking the derivative with respect to t, we get \(y'=v'y_1+vy_1'\) and \(y''=v''y_1+2v'y_1'+vy_1''\).
To summarize, we have




Next, we take these expressions and substitute them back into the original differential equation \(y'' + p(t)y' + q(t)y = 0\). After collecting terms, we get
\( y_1v'' + (2y_1'+py_1)v' + (y_1''+py_1'+qy_1)v = 0 \)
Notice that the expression in parentheses of the last term is just the original differential equation, which is zero. So we end up with
\( y_1v'' + (2y_1'+py_1)v' = 0 \)

Now, let \(w=v'\) in the last equation to get \( y_1w' + (2y_1'+py_1)w = 0 \). This is now a first-order, linear equation, which we can use to solve for w. Finally, we integrate w to get v and we are done.

Let's take a minute to watch a video that goes through these equations again.

PatrickJMT - Reduction of Order - Why It Works

Specific Case - Constant Coefficients with Real, Repeated Roots

On the second-order, linear page, we introduced the idea that, in the case of real, repeated roots, we would have one solution \(y_1=e^{rt}\). To show that the second solution is \(y_2=ty_1\), we will use reduction of order.

Let \(y_1=c_1e^{rt}\) be one solution to \(ay''+by'+cy=0\). Since r is a repeated root of the characteristic equation, we can write the differential equation as \(y''-2ry'+r^2y=0\). Use reduction of order to derive the second solution \(y_2=c_2te^{rt}\). Try this on your own before looking at the details.

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Practice Problems

Instructions - - Unless otherwise instructed, find the second solution and the general solution to each differential equation using reduction of order. If initial conditions are given, solve the initial value problem also. Give your answers in exact terms and completely factored.

Level A - Basic

Practice A01

\(x^2y''+5xy'-5y=0\), \(x > 0\); \(y_1=x\)


Practice A02

\(x^2y''-3xy'+4y=0\); \(y_1=x^2\)


Practice A03

\(t^2y''-t(t+2)y'+(t+2)y=0\); \(y_1=t\)


Practice A04

\(x^2y''+3xy'+y=0\); \(y_1=1/x\)


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