Reduction of Order

Main idea

If you already know one solution $y_1$ , look for a second one shaped like $v(t)\,y_1(t)$ . Because $y_1$ already solves the equation, the terms with $v$ itself cancel, and what's left is a first-order equation for $v'$ .

y_2 = {\htmlData{sym=0}{v(t)}}\,y_1(t)

What the symbols mean

$v(t)$
Unknown multiplier
Substituting makes the v terms cancel, leaving a first-order equation for w = v′.

Explanation

Reduction of order is simply a way to find a second solution when you already know one. The name comes from what it does: it turns your second-order equation into a first-order one (it reduces the order by one), which is much easier to solve. All we're really doing is taking the solution you have, multiplying it by some changing amount, and working out what that amount has to be. In $y_2 = v(t)\,y_1(t)$ , the pieces are $y_1$ (the solution you already know) and $v(t)$ (an unknown stretch factor that's allowed to change with time). The algebra gets a little long, but it always simplifies the same way, into an easier equation for $v$ . Bear with it.

In the last lesson we pulled $te^{rt}$ out of a hat. Here's where it comes from. A constant multiple $cy_1$ is useless, since it's the same solution again. So let the constant vary and try $y_2 = v(t)\,y_1(t)$ .

Steps:

Substitute $y = vy_1$ (with $y' = v'y_1 + vy_1'$ , and so on) into the equation.
The terms with $v$ itself cancel, leaving only $v'$ and $v''$ .
Set $w = v'$ : a first-order equation. Solve, integrate for $v$ , and drop constants.

Think about why step 2 always happens. Collect everything multiplying $v$ (with no derivatives on it) and you get exactly $y_1'' + py_1' + qy_1$ , which is $0$ because $y_1$ is a solution. That cancellation is the whole trick, and it's why the order drops from two to one.

For a repeated root, try it with $y_1 = e^{rt}$ : everything collapses to $v'' = 0$ , so $v = t$ and $y_2 = te^{rt}$ . The same tool works for equations with variable coefficients, where there's no characteristic equation to lean on, as long as you know one solution.

A common mistake. After all that work finding $v$ , it's tempting to hand in $v$ as the second solution. It isn't: $v$ is only the multiplier. The second solution is $y_2 = v\,y_1$ , so don't forget to multiply back.

See it

For y″ + 2y′ + y = 0 the characteristic equation only gives y₁ = e^(−t). Reduction of order tries y₂ = v(t)e^(−t), and substituting forces v″ = 0, so v = t. Notice how y₂ = te^(−t) is just the first solution reshaped by that growing factor: it rises to a peak of about 0.37 at t = 1, then gets dragged back down by the exponential.

Worked examples

Example 1

Find a second solution of $2t^2y'' + ty' - 3y = 0$ given $y_1 = t^{-1}$ .

1
Substitute $y = vt^{-1}$ . The terms with $v$ alone cancel, since $t^{-1}$ is already a solution, and simplifying leaves $2tv'' - 3v' = 0$ .

Practice

Type your answer and check it. Solve them all to complete the lesson.

Problem 1+5 XP

Given $y_1 = t^2$ solves $t^2y'' - 3ty' + 4y = 0$ , find a second solution.

Problem 2+5 XP

Find a second solution of $2t^2y'' + ty' - 3y = 0$ given $y_1 = t^{-1}$ .

Problem 3+5 XP

Given $y_1 = e^{2t}$ solves $y'' - 4y' + 4y = 0$ , find a second solution.

Problem 4+5 XP

What form does reduction of order assume for the second solution?

Problem 5+5 XP

Given $y_1 = 1$ solves $y'' = 0$ , find a second solution.

Problem 6+5 XP

Bending problems in polar coordinates lead to equations like $x^2y'' - xy' + y = 0$ for $x > 0$ , which has the obvious solution $y_1 = x$ . Use reduction of order to find the general solution. Use $c_1$ and $c_2$ .

Challenge problem+25 XP

The hardest one here. Get it, and you've got this lesson.

Legendre's equation $(1 - x^2)y'' - 2xy' + 2y = 0$ , which describes electric potentials around a sphere, has the solution $y_1 = x$ . Find the general solution on $-1 < x < 1$ . Use $c_1$ and $c_2$ .

Where it's used

Physicists use reduction of order to find second solutions of the Bessel, Legendre and Hermite equations once one is known, which matters for waves in cylinders, spheres and quantum wells.

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Reduction of Order

Main idea

y_2 = {\htmlData{sym=0}{v(t)}}\,y_1(t)

What the symbols mean

$v(t)$
Unknown multiplier
Substituting makes the v terms cancel, leaving a first-order equation for w = v′.

Explanation

Steps:

Substitute $y = vy_1$ (with $y' = v'y_1 + vy_1'$ , and so on) into the equation.
The terms with $v$ itself cancel, leaving only $v'$ and $v''$ .
Set $w = v'$ : a first-order equation. Solve, integrate for $v$ , and drop constants.

See it

Worked examples

Example 1

Find a second solution of $2t^2y'' + ty' - 3y = 0$ given $y_1 = t^{-1}$ .

1
Substitute $y = vt^{-1}$ . The terms with $v$ alone cancel, since $t^{-1}$ is already a solution, and simplifying leaves $2tv'' - 3v' = 0$ .

Practice

Type your answer and check it. Solve them all to complete the lesson.

Problem 1+5 XP

Given $y_1 = t^2$ solves $t^2y'' - 3ty' + 4y = 0$ , find a second solution.

Problem 2+5 XP

Find a second solution of $2t^2y'' + ty' - 3y = 0$ given $y_1 = t^{-1}$ .

Problem 3+5 XP

Given $y_1 = e^{2t}$ solves $y'' - 4y' + 4y = 0$ , find a second solution.

Problem 4+5 XP

What form does reduction of order assume for the second solution?

Problem 5+5 XP

Given $y_1 = 1$ solves $y'' = 0$ , find a second solution.

Problem 6+5 XP

Challenge problem+25 XP

The hardest one here. Get it, and you've got this lesson.

Where it's used

Physicists use reduction of order to find second solutions of the Bessel, Legendre and Hermite equations once one is known, which matters for waves in cylinders, spheres and quantum wells.

Tuning a Robot JointDiff Eq · Robotics