Reduction of order

A Simple Example

Consider the general second-order constant coefficient ODE

$a\; y(x)\; +\; b\; y\text{'}(x)\; +\; c\; y(x)\; =\; 0,\; ;$

where $a,\; b,\; c$ are real non-zero coefficients. Furthermore, assume that the associated characteristic equation

$a\; lambda^\{2\}\; +\; b\; lambda\; +\; c\; =\; 0\; ;$

has repeated roots (i.e. the discriminant, $b^2\; -\; 4\; a\; c$, vanishes). Thus we have

$lambda\_\{1,2\}\; =\; -frac\{b\}\{2\; a\}.$

Thus our one solution to the ODE is

$y\_1(x)\; =\; e^\{-frac\{b\}\{2\; a\}\; x\}.$

To find a second solution we take as an ansatz

$y\_2(x)\; =\; v(x)\; y\_1(x)\; ;$

where $v(x)$ is an unknown function to be determined. Since $y\_2(x)$ must satisfy the original ODE, we substitute it back in to get

$a\; left(v$ y_1 + 2 v' y_1' + v y_1 right) + b left(v' y_1 + v y_1' right) + c v y_1 = 0.

Rearranging this equation in terms of the derivatives of $v(x)$ we get

$left(a\; y\_1\; right)\; v$ + left(2 a y_1' + b y_1 right) v' + left(a y_1 + b y_1' + c y_1 right) v = 0.

Since we know that $y\_1(x)$ is a solution to the original problem, the coefficient of the last term is equal to zero. Furthermore, substituting $y\_1(x)$ into the second term's coefficient yields (for that coefficient)

$2\; a\; left(-\; frac\{b\}\{2\; a\}\; e^\{-frac\{b\}\{2\; a\}\; x\}\; right)\; +\; b\; e^\{-frac\{b\}\{2\; a\}\; x\}\; =\; left(-b\; +\; b\; right)\; e^\{-frac\{b\}\{2\; a\}\; x\}\; =\; 0.$

Therefore we are left with

$a\; y\_1\; v\; =\; 0.\; ;$

Since $a$ is assumed non-zero and $y\_1(x)$ is an exponential function and thus never equal to zero we simply have

$v\; =\; 0.\; ;$

This can be integrated twice to yield

$v(x)\; =\; c\_1\; x\; +\; c\_2\; ;$

where $c\_1,\; c\_2$ are constants of integration. We now can write our second solution as

$y\_2(x)\; =\; (c\_1\; x\; +\; c\_2\; )\; y\_1(x)\; =\; c\_1\; x\; y\_1(x)\; +\; c\_2\; y\_1(x).\; ;$

Since the second term in $y\_2(x)$ is a scalar multiple of the first solution (and thus linearly dependent) we can drop that term, yielding a final solution of

$y\_2(x)\; =\; x\; y\_1(x)\; =\; x\; e^\{-frac\{b\}\{2\; a\}\; x\}.$

Finally, we can prove that the second solution $y\_2(x)$ found via this method is linearly independent of the first solution by calculating the Wronskian

Thus $y\_2(x)$ is the second linearly independent solution we were looking for.

The General Method

Given a differential equation

$y+p(t)y\text{'}+q(t)y=0,$

and a single solution ($y\_1(t)$), let the second solution be defined

$y\_2=v(t)y\_1(t),$

where $v(t)$ is an arbitrary function. Thus

$y\_2\text{'}=v\text{'}(t)y\_1(t)+v(t)y\_1\text{'}(t),$

and

$y\_2$=v(t)y_1(t)+2v'(t)y_1'(t)+v(t)y_1(t).,

If these are substituted for $y$, $y\text{'}$, and $y$ in the differential equation, then

$y\_1(t),v$+(2y_1'(t)+p(t)y_1(t)),v'+(y_1(t)+p(t)y_1'(t)+q(t)y_1(t)),v=0.

Since $y\_1(t)$ is a solution of the original differential equation, $y\_1(t)+p(t)y\_1\text{'}(t)+q(t)y\_1(t)=0$, so we can reduce to

$y\_1(t),v+(2y\_1\text{'}(t)+p(t)y\_1(t)),v\text{'}=0$

which is a first-order differential equation for $v\text{'}(t)$. Divide by $y\_1(t)$, obtaining

$v\text{'}\text{'}+left(frac\{2y\_1\text{'}(t)\}\{y\_1(t)\}+p(t)right),v\text{'}=0$

and $v\text{'}(t)$ can be found using a general method. Once $v\text{'}(t)$ is solved, integrate it and enter into the original equation for $y\_2$:

$y\_2=v(t)y\_1(t).,$

References

• W. E. Boyce and R. C. DiPrima, Elementary Differential Equations and Boundary Value Problems (8th edition), John Wiley & Sons, Inc., 2005. ISBN 0-471-43338-1.
• Eric W. Weisstein, Second-Order Ordinary Differential Equation Second Solution, From MathWorld--A Wolfram Web Resource.