5.2: Reduction of Order

In this section we give a method for finding the general solution of the homogeneous equation

\[\label{eq:5.6.1} P_0(x)y''+P_1(x)y'+P_2(x)y=0\]

if we happen to know one nontrivial solution \(y_1\) (don't ask how we know this solution - it will make sense later, for now let's just call it magic).

The method is called reduction of order because it reduces the task of solving Equation \ref{eq:5.6.1} to solving a first order equation.

A common method used in finding additional solutions is to multiply a known solution by an unknown function, such as

\[\label{eq:5.6.2} y=uy_1\]

where \(u\) is to be determined so that \(y\) satisfies Equation \ref{eq:5.6.1}. Substituting Equation \ref{eq:5.6.2} and both

\[\begin{align*} y'&= u'y_1+uy_1' \\[4pt] y'' &= u''y_1+2u'y_1'+uy_1'' \end{align*}\]

into Equation \ref{eq:5.6.1} yields

\[P_0(x)(u''y_1+2u'y_1'+uy_1'')+P_1(x)(u'y_1+uy_1')+P_2(x)uy_1=0. \nonumber\]

Collecting the coefficients of \(u\), \(u'\), and \(u''\) yields

\[\label{eq:5.6.3} (P_0y_1)u''+(2P_0y_1'+P_1y_1)u'+(P_0y_1''+P_1y_1'+P_2y_1) u=0.\]

However, the coefficient of \(u\) is zero, since \(y_1\) satisfies Equation \ref{eq:5.6.1}. Therefore Equation \ref{eq:5.6.3} reduces to the form

\[\label{eq:5.6.4} Q_0(x)u''+Q_1(x)u'=0.\nonumber\]

Since there no longer is a "\(u\)" term we can reduce the order by making the substitution \(z=u'\), which in turn gives \(z'=u''\) and leads to

\[\label{eq:5.6.5} Q_0(x)z'+Q_1(x)z=0.\]

Since Equation \ref{eq:5.6.5} is a linear first order equation in \(z\), we can solve it for \(z\) easily and then integrate the solution to obtain \(u\), and from there obtain \(y\) from Equation \ref{eq:5.6.2}.

Example 5.2.1

Find the general solution of \[\label{eq:5.6.6} xy''-(2x+1)y'+(x+1)y=0,\] given that \(y_1=e^x\) is one solution.

Solution

If we let \(y=ue^x\), then \(y'=u'e^x+ue^x\) and \(y''=u''e^x+2u'e^x+ue^x\), so

\[\begin{align*} xy''-(2x+1)y'+(x+1)y&=x(u''e^x+2u'e^x+ue^x) -(2x+1)(u'e^x+ue^x)+(x+1)ue^x\\ &=(xu''-u')e^x.\end{align*}\]

Therefore \(y=ue^x\) is a solution of Equation \ref{eq:5.6.6} if and only if

\[(xu''-u')e^x=0,\nonumber\]

which is a first order equation in \(u'\). If rewrite it as

\[\label{eq:5.6.7} u''-{u'\over x}=0\]

and make the substitution \(z=u'\), then Equation \ref{eq:5.6.7} becomes

\[\label{eq:5.6.8} z'-{z\over x}=0.\nonumber\]

We leave it to you to show (by separation of variables) that \[z=C_1x\nonumber\]

is a solution.

Now using the fact that \(u'=z\) and integrating we get

\[u={C_1\over2}x^2+C_2\nonumber\]

and substituting this into \(y=ue^x\) we get

\[\label{eq:5.6.9} y=ue^x={C_1\over2}x^2e^x+C_2e^x.\]

Note that \(y_1=e^x\) and \(y_2=x^2e^x\) are clearly linearly independent and form a fundamental set of solutions to the homogeneous differential Equation \ref{eq:5.6.6}.

Although Equation \ref{eq:5.6.9} is a correct form for the general solution of Equation \ref{eq:5.6.6}, it is silly to leave the arbitrary coefficient of \(x^2e^x\) as \(C_1/2\) where \(C_1\) is an arbitrary constant. Moreover, it is sensible to make the subscripts of the coefficients of \(y_1=e^x\) and \(y_2=x^2e^x\) consistent with the subscripts of the functions themselves. Therefore we rewrite Equation \ref{eq:5.6.9} as

\[y=c_1e^x+c_2x^2e^x \nonumber\]

by simply renaming the arbitrary constants. This last form is what we'll call the general solution of \ref{eq:5.6.6}.

Although we could go through this procedure each time this situation arises, it would be much better if we actually had a formula to generate a second solution, and that's precisely what the next theorem gives us.

Theorem \(\PageIndex{1}\)

Consider the differential equation \[P_0(x)y''+P_1(x)y'+P_2(x)y=0.\nonumber\]

Dividing by \(P_0(x)\), assuming it is not zero, we get

\[\label{eq:5.6.10} y''+p(x)y'+q(x)y=0.\]

Now suppose we know that \(y_1\) is one solution. Then a second linearly independent solution is given by:

\[\label{eq:5.6.11}y_2=y_1\int{e^{-\int{p(x)dx}}\over y_1^2}dx\]

Proof

The proof follows the procedure we used in Example 5.2.1.

Suppose \(y_1\) is a solution to \ref{eq:5.6.10}

Let \(y=uy_1\), then \(y'=u'y_1+uy_1'\) and \(y''=u''y_1+2u'y_1'+uy_1''\), so \ref{eq:5.6.10} becomes

\[\begin{align*} y''+p(x)y'+q(x)y&=(u''y_1+2u'y_1'+uy_1'') +p(x)(u'y_1+uy_1')+q(x)uy_1=0\end{align*}\nonumber\]

\[=u''y_1+2u'y_1' +u'p(x)y_1+u(y_1''+p(x)y_1'+q(x)y_1).\nonumber\]

However, we know \(y_1\) is a solution to \ref{eq:5.6.10}, so the coefficient of \(u\), \(y_1''+p(x)y_1'+q(x)y_1=0\), and we get

\[u''y_1+2u'y_1' +u'p(x)y_1=0.\nonumber\]

Since there is no "\(u\)" term it is easy to reduce the order by substituing \(z=u'\) giving us

\[z'y_1+2zy_1' +zp(x)y_1=0. \nonumber \]

Rewriting this as

\[y_1z'+(2y_1' +p(x)y_1)z=0\nonumber\]

and diving by \(y_1\) we get the standard linear form

\[\label{eq:5.6.12} z'+{2y_1' +p(x)y_1\over y_1}z=0\]

So, the integrating factor becomes \[e^{\int{2y_1' +p(x)y_1\over y_1}dx}\nonumber\]

\[=e^{\int{2y_1'\over y_1}dx+\int{p(x)y_1\over y_1}dx}\nonumber\]

\[=e^{\int{2y_1'\over y_1}dx}e^{\int{p(x)}dx}\nonumber\]

\[=y_1^2e^{\int{p(x)}dx}\nonumber\]

and multiplying both sides of \ref{eq:5.6.12} by the integrating factor we get \[d[y_1^2e^{\int{p(x)}dx}z]=0\nonumber\]

Integrating this we get \[y_1^2e^{\int{p(x)}dx}z=c_1\nonumber\]

so \[z=c_1{1\over y_1^2e^{\int{p(x)dx}}}\nonumber\]

Now substituting \(u'\) back in for \(z\) we have \[u'=c_1{1\over y_1^2e^{\int{p(x)dx}}}\nonumber\]

Integrating this we get \[u=c_1\int{dx\over y_1^2e^{\int{p(x)dx}}}+c_2\nonumber\]

Therefore, \[y=uy_1=(c_1\int{dx\over y_1^2e^{\int{p(x)dx}}}+c_2)y_1=c_1y_1\int{dx\over y_1^2e^{\int{p(x)dx}}}+c_2y_1.\nonumber\]

So, we get that the second solution must be \[y_2=y_1\int{e^{-\int{p(x)dx}}\over y_1^2}dx\nonumber\]

which is clearly independent of \(y_1\) as it is not a constant multiple of \(y_1\).

If we wanted to check independence using the Wronskian we would get

\[W(x)=\left| \begin{array}{rr} y_1 &&&& y_1\int{e^{-\int{p(x)dx}}\over y_1^2}dx \\ y_1' &&&& y_1'\int{e^{-\int{p(x)dx}}\over y_1^2}dx+y_1{e^{-\int{p(x)dx}}\over y_1^2} \end{array} \right|=e^{-\int{p(x)dx}}\not\equiv0\nonumber\] for all x.

Example 5.2.2

1. Find a second linearly independent solution of \[\label{eq:5.6.13} x^2y''+xy'-y=0\] given that \(y_1=x\) is one solution.
2. Find the general solution of \ref{eq:5.6.13}
3. Solve the initial value problem \[\label{eq:5.6.14} x^2y''+xy'-y=0, \quad y(1)=2,\; y'(1)=-3.\]

Solution a

We first need to divide by the lead coefficient to get the equation in the form \ref{eq:5.6.10}:

\[y''+{1\over x}y'-{1\over x^2}y=0. \nonumber \]

So, \(p(x)={1\over x}\) and using \ref{eq:5.6.11} we get

\[y_2=x\int{e^{-\int{1\over x}dx}\over x^2}dx\nonumber\]

\[=x\int{1\over x^3}dx\nonumber\]

\[={-1\over 2x}\nonumber\]

Since any constant will be absorbed into an arbitrary constant in the general solution we will take

\[y_2={1\over x}\nonumber\]

Since we used \ref{eq:5.6.11} we know \(y_2\) is independent of \(y_1\).

Solution b

Using the given \(y_1\) and what we just found for \(y_2\) we can form the general solution of \ref{eq:5.6.13}

\[y=c_1x+c_2{1\over x}\nonumber\]

Solution c

Using the general solution from part b and the initial conditions from \ref{eq:5.6.14} we get

\[c_1+c_2=2\nonumber\]

and

\[c_1-c_2=-3\nonumber\]

Solving this system we get \(c_1=-1/2\) and \(c_2=5/2\) which gives us our solution to \ref{eq:5.6.14}

\[y={-x\over 2}+{5\over 2x}\nonumber\] which is unique on \((0,\infty)\).

Why is it unique? Why is the interval of validity \((0,\infty)\)?

Example 5.2.3

Find the general solution of

\[\label{eq:5.6.19} x^2y''-3xy'+3y=0,\]

given that \(y_1=x\) is a solution.

Solution

\[y_2=x\int{e^{\int{3\over x}dx}\over x^2}dx\nonumber\]

\[=x\int{x}dx\nonumber\]

\[={x^3\over 2}\nonumber\]

So, we will take \(y_2=x^3\) and our general solution is

\[y=c_1x+c_2x^3\nonumber\]