Reduction of Order

Reduction of Order

Recall that the characteristic equation of $y''+4y'+4y=0$ has a repeated root of $r=-2.$

We were able to find $e^{-2x}$ as a solution, but the other linearly independent solution $xe^{-2x}$ had to be handed to us.

Today we learn a technique for finding other linearly independent solutions from a known solution.

This technique is known as reduction of order.

The Big Idea

Suppose we have a solution $y_1$ to a linear equation $Ly=0.$

We assume that a new, linearly independent solution $y_2$ can be expressed as $$ y_2(x)=v(x)y_1(x) $$

Example

Knowing that $y_1=e^{-2x}$ is a solution to the equation $$ y''+4y'+4y=0 $$ use reduction of order to find a second linearly independent solution $y_2.$

Suppose $y_2(x)=v(x)y_1(x)=v(x)e^{-2x}.$

Then, for $y_2$ to be a solution, it must be that $$ y_2''+4y_2'+4y_2=0 $$ That is, $$ \begin{array}{lll} &\displaystyle (v(x)e^{-2x})''+4(v(x)e^{-2x})'+4(v(x)e^{-2x})=0 &\mbox{}\\ \implies &\displaystyle (ve^{-2x})''+4(ve^{-2x})'+4(ve^{-2x})=0 &\mbox{omitting arguments}\\ \implies &\displaystyle (v'e^{-2x}-2ve^{-2x})'+4(v'e^{-2x}-2ve^{-2x})+4ve^{-2x}=0 &\mbox{}\\ \implies &\displaystyle v''e^{-2x}\color{blue}{-2v'e^{-2x}}-2(v'e^{-2x}-2ve^{-2x})\color{blue}{+4v'e^{-2x}}\color{red}{-8ve^{-2x}}\color{red}{+4ve^{-2x}}=0 &\mbox{}\\ \implies &\displaystyle v''e^{-2x}\color{blue}{-2v'e^{-2x}}\color{blue}{-2v'e^{-2x}}\color{red}{+4ve^{-2x}}\color{blue}{+4v'e^{-2x}}\color{red}{-8ve^{-2x}}\color{red}{+4ve^{-2x}}=0 &\mbox{}\\ \implies &\displaystyle v''e^{-2x}=0 &\mbox{}\\ \implies &\displaystyle v''=0 &\mbox{since $e^{-2x}$ is never $0$}\\ \end{array} $$ It follows that $v'$ is a constant, say $C_1,$ which makes $v=C_1x+C_2$ where $C_2$ is another constant.

Since $C_1$ and $C_2$ are arbitrary, we choose $C_1=1$ and $C_2=0,$ so that $v(x)=x.$

Therefore, $$ y_2=v(x)e^{-2x}=xe^{-2x} $$ is another linearly independent solution.

Great News!

Reduction of order works with general linear equations!

(not just those with constant coefficients)

Another Example

Consider the homogeneous, variable-coefficient linear equation $$ x^2y''-2xy'-4y=0, \,\,\,\,x \gt 0 $$ Using that $y_1=x^{-1}$ is a solution, use reduction of order to find a second linearly independent solution $y_2.$

Also, state the general solution.

Suppose $y_2=vx^{-1}$ is a solution to the equation.

Then $$ \begin{array}{lll} &\displaystyle x^2y_2''-2xy_2'-4y_2=0&\mbox{}\\ \implies &\displaystyle y_2''-2x^{-1}y_2'-4x^{-2}y_2=0&\mbox{}\\ \implies &\displaystyle (vx^{-1})''-2x^{-1}(vx^{-1})'-4x^{-2}(vx^{-1})=0&\mbox{}\\ \implies &\displaystyle (v'x^{-1}-vx^{-2})'-2x^{-1}(v'x^{-1}-vx^{-2})-4vx^{-3}=0&\mbox{}\\ \implies &\displaystyle v''x^{-1}-v'x^{-2}-(v'x^{-2}-2vx^{-3})-2v'x^{-2}+2vx^{-3}-4vx^{-3}=0&\mbox{}\\ \implies &\displaystyle v''x^{-1}\color{blue}{-v'x^{-2}}\color{blue}{-v'x^{-2}}\color{red}{+2vx^{-3}}\color{blue}{-2v'x^{-2}}\color{red}{-2vx^{-3}}=0&\mbox{}\\ \implies &\displaystyle v''x^{-1}\color{blue}{-4v'x^{-2}}=0&\mbox{}\\ \implies &\displaystyle v''-4x^{-1}v'=0&\mbox{}\\ \end{array} $$ We now let $w=v'$ so that last of the above equations becomes $$ w'-4x^{-1}w=0 $$ Like the last example, this is a separable equation! That is, $$ \begin{array}{lll} &\displaystyle w'=4x^{-1}w=0&\mbox{}\\ \implies &\displaystyle \frac{dw}{w}=4x^{-1}\,dx&\mbox{}\\ \implies &\displaystyle \int \frac{dw}{w}=4\int x^{-1}\,dx&\mbox{}\\ \implies &\displaystyle \ln w=4\ln x&\mbox{}\\ \implies &\displaystyle \ln w=\ln x^4&\mbox{}\\ \implies &\displaystyle w=x^4&\mbox{}\\ \end{array} $$ Since, $v'=w,$ we may now find $v.$ $$ \begin{array}{lll} \displaystyle v&\displaystyle=\int w \,dx &\mbox{}\\ \displaystyle &\displaystyle=\int x^4 \,dx &\mbox{}\\ \displaystyle &\displaystyle=\frac{1}{5}x^5&\mbox{}\\ \end{array} $$ Thus, $$\displaystyle y_2=vx^{-1}=\frac{1}{5}x^5\cdot x^{-1}=\frac{1}{5}x^{4}$$ However, by linearity, any multiple of a solution is also a solution.

So, instead we take the simpler $y_2=x^{4}$ as our second linearly independent solution.

The general solution to our equation is then $$ y=C_1x^{-1}+C_2x^4 $$

Reduction of Order for General Second-Order Linear Equations

It's a fact that for any homogeneous second-order linear equation $$ y''+p(x)y'+q(x)y=0 $$ with a known solution $y_1,$ the substitution $y_2=vy_1$ will always result in an equation in $v''$ and $v'.$

That is, the $v$ terms will always cancel out just like in the above examples.

Making the substitution $w=v'$ will then result in a separable (and linear!) first-order equation.

This is the reason this technique is called reduction of order.

Reduction of Order Formula

We shall now perform the above process on a general homogeneous second-order linear equation $$ y''+p(x)y'+q(x)y=0 $$ Supposing $y_1$ is a known solution, we make the substitution $y_1=vy_2.$

Then...

Reduction of Order Formula

$$ \begin{array}{lll} &\displaystyle y_2''+p(x)y_2'+q(x)y_2=0 &\mbox{}\\ \implies &\displaystyle y_2''+py_2'+qy_2=0&\mbox{}\\ \implies &\displaystyle (vy_1)''+p(vy_1)'+q(vy_1)=0&\mbox{}\\ \implies &\displaystyle (v'y_1+vy_1')'+p(v'y_1+vy_1')+qvy_1=0&\mbox{}\\ \implies &\displaystyle v''y_1\color{blue}{+v'y_1'}\color{blue}{+v'y_1'}\color{red}{+vy_1''}\color{blue}{+pv'y_1}\color{red}{+pvy_1'}\color{red}{+qvy_1}=0&\mbox{}\\ \implies &\displaystyle v''y_1\color{blue}{+2v'y_1'}\color{blue}{+pv'y_1}\color{red}{+vy_1''}\color{red}{+pvy_1'}\color{red}{+qvy_1}=0&\mbox{}\\ \implies &\displaystyle v''y_1\color{blue}{+(2y_1'+py_1)v'}\color{red}{+(y_1''+py_1'+qy_1)v}=0&\mbox{}\\ \implies &\displaystyle v''y_1\color{blue}{+(2y_1'+py_1)v'}\color{red}{+0 \cdot v}=0&\mbox{since $y_1''+py_1'+qy_1=0$}\\ \implies &\displaystyle v''y_1+(2y_1'+py_1)v'=0&\mbox{}\\ \implies &\displaystyle v''+\left(2\frac{y_1'}{y_1}+p\right)v'=0&\mbox{}\\ \implies &\displaystyle w'+\left(2\frac{y_1'}{y_1}+p\right)w=0&\mbox{letting $w=v'$}\\ \end{array} $$ Solving the above first-order linear equation, our integrating factor is...

Reduction of Order Formula

$$ \begin{array}{lll} \displaystyle \mu&\displaystyle=e^{\int 2\frac{y_1'}{y_1}+p \, dx} &\mbox{}\\ %\displaystyle &\displaystyle=e^{2\ln y_1+\int p \, dx} &\mbox{}\\ %\displaystyle &\displaystyle=e^{\ln y_1^2+\int p \, dx} &\mbox{}\\ %\displaystyle &\displaystyle=e^{\ln y_1^2}e^{\int p \, dx} &\mbox{}\\ %\displaystyle &\displaystyle=y_1^2 e^{\int p \, dx} &\mbox{}\\ \end{array} $$ which we presently leave in its less-simplified form.

We then have $$ e^{\int 2\frac{y_1'}{y_1}+p \, dx} w'+e^{\int 2\frac{y_1'}{y_1}+p \, dx}\left(2\frac{y_1'}{y_1}+p\right)w=0 $$ which gives $$ \left(e^{\int 2\frac{y_1'}{y_1}+p \, dx} w\right)'=0 $$

Reduction of Order Formula

Now, since $$ \begin{array}{lll} \displaystyle \mu&\displaystyle=e^{\int 2\frac{y_1'}{y_1}+p \, dx} &\mbox{}\\ \displaystyle &\displaystyle=e^{2\ln y_1+\int p \, dx} &\mbox{}\\ \displaystyle &\displaystyle=e^{\ln y_1^2+\int p \, dx} &\mbox{}\\ \displaystyle &\displaystyle=e^{\ln y_1^2}e^{\int p \, dx} &\mbox{}\\ \displaystyle &\displaystyle=y_1^2 e^{\int p \, dx} &\mbox{}\\ \end{array} $$ the last equation above becomes $$ \left(w y_1^2 e^{\int p \, dx} \right)'=0 $$ Then...

Reduction of Order Formula

$$ \begin{array}{lll} &\displaystyle w y_1^2 e^{\int p \, dx}=C &\mbox{where $C$ is an arbitrary constant}\\ \implies &\displaystyle w y_1^2 =Ce^{-\int p \, dx}&\mbox{}\\ \implies &\displaystyle w =C\frac{e^{-\int p \, dx}}{y_1^2}&\mbox{}\\ \implies &\displaystyle v' =C\frac{e^{-\int p \, dx}}{y_1^2}&\mbox{since $w=v'$}\\ \implies &\displaystyle v =C\int \frac{e^{-\int p \, dx}}{y_1^2}\,dx &\mbox{}\\ \implies &\displaystyle vy_1 =Cy_1\int \frac{e^{-\int p \, dx}}{y_1^2}\,dx &\mbox{}\\ \implies &\displaystyle y_2 =Cy_1\int \frac{e^{-\int p \, dx}}{y_1^2}\,dx &\mbox{}\\ \end{array} $$

Reduction of Order Formula

Since $C$ can be any constant, our second linearly independent solution $y_2$ to $$ y''+p(x)y'+q(x)y=0 $$ is given by the reduction of order formula: $$ y_2 =y_1\int \frac{e^{-\int p(x) \, dx}}{y_1^2}\,dx $$ where $y_1$ is our known solution.

Example

Knowing that $y_1=e^x$ is a solution to the equation $$ xy''+(1-2x)y'+(x-1)y=0, \,\,\,\, x \gt 0 $$ use reduction of order formula to find a second linearly independent solution $y_2.$

Also, state the general solution.

To use the reduction of order formula, we rewrite the equation in the form $$ y''+(x^{-1}-2)y'+(1-x^{-1})y=0 $$ Then $p(x)=x^{-1}-2$ so that $$ \begin{array}{lll} \displaystyle y_2&\displaystyle=y_1\int \frac{e^{-\int p(x) \, dx}}{y_1^2}\,dx &\mbox{}\\ \displaystyle &\displaystyle=e^{x}\int \frac{e^{-\int x^{-1}-2 \, dx}}{(e^x)^2}\,dx &\mbox{}\\ \displaystyle &\displaystyle=e^{x}\int \frac{e^{\int 2-x^{-1} \, dx}}{e^{2x}}\,dx &\mbox{}\\ \displaystyle &\displaystyle=e^{x}\int \frac{e^{2x-\ln x}}{e^{2x}}\,dx &\mbox{}\\ \displaystyle &\displaystyle=e^{x}\int e^{2x-\ln x-2x}\,dx &\mbox{}\\ \displaystyle &\displaystyle=e^{x}\int e^{-\ln x}\,dx &\mbox{}\\ \displaystyle &\displaystyle=e^{x}\int e^{\ln x^{-1}}\,dx &\mbox{}\\ \displaystyle &\displaystyle=e^{x}\int x^{-1}\,dx &\mbox{}\\ \displaystyle &\displaystyle=e^{x}\ln x &\mbox{}\\ \end{array} $$ Thus, the general solution to the equation is $$ y=C_1 e^x + C_2 e^x \ln x $$

In Summary

Knowing one solution $y_1$ to a homogeneous, second-order linear equation $$ y''+p(x)y'+q(x)y=0 $$ we have two options for finding a second linearly independent solution $y_2$:

Option #1: Make the substitution $y_2=v(x)y_1$ which will result in an equation in $v''$ and $v'.$ Taking $w=v'$ we can then solve the resulting first-order equation in $w$ and then integrate to obtain $v$ which gives $y_2.$

Pro: easy to remember. Con: more computationally intensive.

Option #2: Use the reduction of order formula $$ y_2 =y_1\int \frac{e^{-\int p(x) \, dx}}{y_1^2}\,dx $$ Pro: computationally easier. Con: more difficult to remember.

Non-Homogeneous Example

Given that $y_1=x^3$ solves the linear equation $$ x^2y''+xy'-9y=16x+27, \,\,\,\, x \gt 0 $$ find the general solution.

Getting the homogeneous equation into standard form $$ y''+x^{-1}y'-9x^{-2}y=0 $$ we first use the reduction of order formula to find a second linearly independent solution $y_2.$ $$ \begin{array}{lll} \displaystyle y_2&\displaystyle= y_1\int \frac{e^{-\int p(x) \, dx}}{y_1^2}\,dx&\mbox{}\\ \displaystyle &\displaystyle=x^3\int \frac{e^{-\int x^{-1} \, dx}}{(x^3)^2}\,dx &\mbox{}\\ \displaystyle &\displaystyle=x^3\int \frac{e^{-\ln x}}{x^6}\,dx &\mbox{}\\ \displaystyle &\displaystyle=x^3\int \frac{e^{\ln x^{-1}}}{x^6}\,dx &\mbox{}\\ \displaystyle &\displaystyle=x^3\int \frac{x^{-1}}{x^6}\,dx &\mbox{}\\ \displaystyle &\displaystyle=x^3\int x^{-7}\,dx &\mbox{}\\ \displaystyle &\displaystyle=x^3\left(-\frac{1}{6}x^{-6}\right) &\mbox{}\\ \displaystyle &\displaystyle=-\frac{1}{6}x^{-3} &\mbox{}\\ \end{array} $$ Thus, we take our second linearly independent solution to be $y_2=x^{-3}.$

The general solution to the homogeneous equation is, therefore, $$ y_h=C_1x^3+C_2x^{-3} $$ We now find a particular solution $y_p$ using undetermined coefficients.

For this second order equation we know that $y_p$ will have the form $y_p=Ax^2+Bx+C$

So, $$ \begin{array}{lll} &\displaystyle x^2y_p''+xy_p'-9y_p=16x+27&\mbox{}\\ \implies &\displaystyle x^2(Ax^2+Bx+C)''+x(Ax^2+Bx+C)'-9(Ax^2+Bx+C)=16x+27 &\mbox{}\\ \implies &\displaystyle x^2(2Ax+B)'+x(2A+B)-9Ax^2-9Bx-9C=16x+27 &\mbox{}\\ \implies &\displaystyle \color{blue}{x^2\cdot 2A}\color{red}{+2Ax}\color{red}{+Bx}\color{blue}{-9Ax^2}\color{red}{-9Bx}-9C=16x+27 &\mbox{}\\ \implies &\displaystyle -7Ax^2+(2A-8B)x-9C=16x+27 &\mbox{}\\ \implies &\displaystyle \begin{cases}-7A=0\\2A-8B=16\\-9C=27\end{cases} &\mbox{}\\ \implies &\displaystyle \begin{cases}A=0\\-8B=16\\C=-3\end{cases} &\mbox{}\\ \implies &\displaystyle \begin{cases}A=0\\B=-2\\C=-3\end{cases} &\mbox{}\\ \end{array} $$ Thus, our particular solution is $$ y_p=-2x-3 $$ and our general solution is, therefore, $$ y=y_h+y_p=C_1x^3+C_2x^{-3}-2x-3 $$