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Reduction of order is a technique in mathematics for solving second-order linear ordinary differential equations. It is employed when one solution $y_1(x)$ is known and a second linearly independent solution $y_2(x)$ is desired. The method also applies to n-th order equations. In this case the ansatz will yield a (n-1)-th order equation for $v$.

Second-order linear ordinary differential equations

An Example

Consider the general homogeneous second-order linear constant coefficient ODE

$a{y}^{″}(x)+b{y}^{\prime }(x)+cy(x)=0,$

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, $b^2-4ac$, vanishes). Thus we have

${\lambda }_1={\lambda }_2=-\frac{b}{2a}.$

Thus our one solution to the ODE is

$y_1(x)=e^{-\frac{b}{2a}x}.$

To find a second solution we take as a guess

$y_2(x)=v(x)y_1(x)$

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

$a(v^{″}{y}_1+2v^{\prime }{y_1}^{\prime }+v{y_1}^{″})+b(v^{\prime }{y}_1+v{y_1}^{\prime })+cv{y}_1=0.$

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

$\left(a{y}_1\right)v^{″}+(2a{y_1}^{\prime }+b{y}_1)v^{\prime }+(a{y_1}^{″}+b{y_1}^{\prime }+c{y}_1)v=0.$

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)

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

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

$y_2(x)=x{y}_1(x)=x{e}^{-\frac{b}{2a}x}.$

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

$W(y_1,y_2)(x)=\left|\begin{array}{cc}{y}_1& x{y}_1\\ {y_1}^{\prime }& y_1+x{y_1}^{\prime }\end{array}\right|=y_1(y_1+x{y_1}^{\prime })-x{y}_1{y_1}^{\prime }=y_1^2+x{y}_1{y_1}^{\prime }-x{y}_1{y_1}^{\prime }=y_1^2=e^{-\frac{b}{a}x}\ne 0.$

Thus $y_2(x)$ is the second linearly independent solution we were looking for.

General method

Given the general non-homogeneous linear differential equation

$y^{″}+p(t)y^{\prime }+q(t)y=r(t)$

and a single solution $y_1(t)$ of the homogeneous equation [$r(t)=0$], let us try a solution of the full non-homogeneous equation in the form:

$y_2=v(t)y_1(t)$

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

${y_2}^{\prime }=v^{\prime }(t)y_1(t)+v(t){y_1}^{\prime }(t)$

and

${y_2}^{″}=v^{″}(t)y_1(t)+2v^{\prime }(t){y_1}^{\prime }(t)+v(t){y_1}^{″}(t).$

If these are substituted for $y$, $y^{\prime }$, and $y^{″}$ in the differential equation, then

$y_1(t)v^{″}+(2{y_1}^{\prime }(t)+p(t)y_1(t))v^{\prime }+({y_1}^{″}(t)+p(t){y_1}^{\prime }(t)+q(t)y_1(t))v=r(t).$

Since $y_1(t)$ is a solution of the original homogeneous differential equation, ${y_1}^{″}(t)+p(t){y_1}^{\prime }(t)+q(t)y_1(t)=0$, so we can reduce to

$y_1(t)v^{″}+(2{y_1}^{\prime }(t)+p(t)y_1(t))v^{\prime }=r(t)$

which is a first-order differential equation for $v^{\prime }(t)$ (reduction of order). Divide by $y_1(t)$, obtaining

$v^{″}+(\frac{2{y_1}^{\prime }(t)}{y_1(t)}+p(t))v^{\prime }=\frac{r(t)}{y_1(t)}$.

Integrating factor: $\mu (t)=e^{\int (\frac{2{y_1}^{\prime }(t)}{y_1(t)}+p(t))dt}=y_1^2(t)e^{\int p(t)dt}$.

Multiplying the differential equation with the integrating factor $\mu (t)$, the equation for $v(t)$ can be reduced to

$\frac{d}{dt}(v^{\prime }(t)y_1^2(t)e^{\int p(t)dt})=y_1(t)r(t)e^{\int p(t)dt}$.

After integrating the last equation, $v^{\prime }(t)$ is found, containing one constant of integration. Then, integrate $v^{\prime }(t)$ to find the full solution of the original non-homogeneous second-order equation, exhibiting two constants of integration as it should:

$y_2(t)=v(t)y_1(t)$.

See also

• Variation of parameters

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.
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• Eric W. Weisstein, Second-Order Ordinary Differential Equation Second Solution, From MathWorld—A Wolfram Web Resource.