8. ODE Methods - Reduction of Order

Whilst we can feel confident about solving first order ODEs that are linear (and in some cases non-linear) as well as second order ODEs with constant coefficients, there clearly remain many classes of ODEs which we cannot currently solve. We can however use a series of generalisations of our current methods to solve more complicated problems. Here we will focus on two such methods, reduction of order and variation of parameters.

Definition

Recall that in general a linear second order differential equation is of the form:

(8.1)\[a\,y^{\prime\prime}(x)+b\,y^{\prime}(x)+c\,y(x)=f(x)\]

where \(a,\,b,\,c\) are arbitrary constants.

8.1. Euler Equations

As a warm up, lets try to solve problems of the form:

(8.2)\[ax^2\,y′′+bx\,y′+cy=0\]

We do so by using an ansatz of the form \(y = x^n\), which means that:

\[x^{n}\left( an(n-1) + bn + c\right) = 0\]

A quadratic (which we call the characteristic equation) that we can solve:

\[an^2 + (b-a)n + c = 0 \Rightarrow n = \frac{a-b \pm \sqrt{b^2-2ab + a^2 - 4ac}}{2a}\]

which means there are three distinct cases to solve for here (to begin with assume \(x > 0\) to avoid any additional complications with complex solutions)

1. Two distinct, real roots \(n = n_1,\, n_2\)

Since the ODE (8.2) here is linear, we can find the superposition of solutions:

\[y = A\,x^{n_1} + B\,x^{n_2},\, \qquad n_1,\, n_2 \in \mathbb{R}\]

2. One repeated, real root \(n\)

We find that solving the characteristic equation, the only root is:

\[n = \frac{a-b}{2a}\]

giving a solution \(y_1 = Ax^n\). But there should be two solutions to the ODE (8.2), so we can use an ansatz of the form:

\[y_2 = x^{n}\,f(x)\]

to find \(y_2\), by solving for \(f(x)\). We can think of \(f(x)\) here is a bridging function between the different solutions \(y_1,\,y_2\).

Using this ansatz, the ODE takes the form:

\[\begin{split}y &= x^n\,f \\ y' &= nx^{n-1}\,f + x^n\,f' \\ y'' &= n(n-1)x^{n-2}\,f + 2nx^{n-1}\,f' + x^{n}f''\\ \Rightarrow ax^2\,y′′+bx\,y′+cy &= ax^{n+1}\,f'' + (2an+b)\,x^{n+1}\,f' + (an(n-1) + bn + c)\,x^n\,f = 0\end{split}\]

This can be simplified straight away since we are using the value of \(n\) which satisfies the characteristic equation \(an(n-1) + bn + c = 0\), therefore the equation reduces to:

\[ax^{n+1}\,f'' + (2an+b)\,x^{n+1}\,f' = 0 \]

using the fact that \(\displaystyle n = \frac{a-b}{2a} \Rightarrow 2an + b = a\) means:

\[ax^{n+1}\left(xf'' + f'\right) = 0\]

which can be solved as a linear 1st order ODE in \(f'\):

\[\frac{\mathrm{d}}{\mathrm{d}x}\left( xf'\right) = 0 \Rightarrow xf' = A \]

which means the form of \(f(x)\) is given by:

\[f(x) = A\ln(x) + B\]

and therefore the solution here is:

\[y = x^n\,\left(A \ln(x) + B\right)\]

3. Two complex roots \(n = \lambda \pm i\mu\)

We can solve the characteristic equation here, with the proviso \((b-a)^2 - 4ac < 0\):

\[n = \frac{(a-b)\pm i\sqrt{4ac-(b-a)^2}}{2a} = \lambda \pm i \mu,\qquad \lambda,\, \mu \in \mathbb{R}\]

If we look at the solutions here, lets start with \(n = \lambda + i \mu\):

\[y_1 = Ax^{\lambda + i\mu} = Ae^{\ln(x^{\lambda + i \mu}} = Ae^{(\lambda + i\mu)\ln(x)} = Ax^{\lambda}\left( \cos(\mu\ln(x)) + i \sin(\mu\ln(x))\right)\]

where we have used the Ruler function to convert this expression into one with trigonometric and power law functions. We could do a similar exercise with \(y_2\), the only difference here being a \(-\) sign in the middle, so the overal solution \(y = y_1 + y_2\) can be written as:

\[y = y_1 + y_2 = x^{\lambda}\left(A\cos(\mu\ln(x)) + B \sin(\mu\ln(x))\right)\]

where \(A,\, B \in \mathbb{C}\) in general, but with real boundary conditions they will turn out to be real.

8.1.1. Other ranges of \(x\)

So far we have taken \(x>0\) to ensure all of our solutions here are real, however if we look again at (8.2), taking a coordinate transformation \(t = -x\) and defining a function \(z\) such that:

\[z(t) = y(x) = y(-t)\]

then this means that:

\[\begin{split}\frac{\mathrm{d}}{\mathrm{d}t} z(t) = \dot{z}(t) &= - \frac{\mathrm{d}}{\mathrm{d}x} y(x) = y'(x)\\ \frac{\mathrm{d}^2}{\mathrm{d}t^2} z(t) = \ddot{z}(t)&= + \frac{\mathrm{d}^2}{\mathrm{d}x^2} y(x) = y''(x)\end{split}\]

and therefore (8.2) reads as:

\[\begin{split}a(-t)^2\ddot{z} + b(-t)(-\dot{z}) + cz &= 0 \\ \Rightarrow at^2\ddot{z} + bt\dot{z} + cz = 0\end{split}\]

This shows that for \(t>0\) we can still have real solutions to this ODE, there would follow the same form as those found for \(>0\) - but this range corresponds to \(x<0\)!

Looking at how this could change the solutions, say for case (1) with real, distinct roots in \(n\):

\[z(t) = At^{n_1} + Bt^{n_1} \Rightarrow y(x) = A(-x)^{n_1} + B(-x)^{n_2}, \qquad x < 0\]

Therefore the argument of these solutions is always \(|x|\) irrespective of whether \(x<0\) or \(x > 0\), therfore we modify our solutions to be of the form:

1. \(y = A\,x^{n_1} + B\,x^{n_2} \longrightarrow y = A\,|x|^{n_1} + B\,|x|^{n_2}\)

2. \(y = x^n\,\left(A \ln(x) + B\right) \Longrightarrow y = |x|^n\,\left(A \ln|x| + B\right)\)

3. \(y = x^{\lambda}\left(A\cos(\mu\ln(x)) + B \sin(\mu\ln(x))\right) \longrightarrow y = |x|^{\lambda}\left(A\cos(\mu\ln|x|) + B \sin(\mu\ln|x|)\right)\)

We notice that \(x=0\) is not covered by these solutions as it corresponds to \(cy=0\).

By a similar principle, we could define a variable of the form \(t = x - s\), with predicably similar behaviour, therefore we can also shift the ODe to be of the form:

\[a(x-s)^2y'' + b(x-s)y' + cy = 0\]

giving solutions of the form: 1. \(y = A\,x^{n_1} + B\,x^{n_2} \longrightarrow y = A\,|x-s|^{n_1} + B\,|x-s|^{n_2}\)

2. \(y = x^n\,\left(A \ln(x) + B\right) \Longrightarrow y = |x-s|^n\,\left(A \ln|x-s| + B\right)\)

3. \(y = x^{\lambda}\left(A\cos(\mu\ln(x)) + B \sin(\mu\ln(x))\right) \longrightarrow y = |x-s|^{\lambda}\left(A\cos(\mu\ln|x-s|) + B \sin(\mu\ln|x-s|)\right)\)

however it is also clear here that this solution cannot contain the point \(x=s\).

Worked example

Solve the following ODE:

\[x^2\,y'' + xy' - y = 0\]

Using the ansatz \(y = x^n\), then:

\[\begin{split}y(x) &= x^n \\ y'(x) &= nx^{n-1}\\ y''(x) &= n(n-1)x^{n-2}\end{split}\]

which means that the ODE can be written as:

\[x^{n}\Big(n(n-1)+n - 1 \Big) = 0 \Rightarrow n^2 - 1 = 0\]

which means that \(n = \pm 1\) and therefore the solution will be of the form:

\[y = Ax^1 + Bx^{-1}\]

Further worked examples

1. Solve the ODE \(x^2y'' - 2y = 0, \qquad x >0\)

Using the ansatz \(y = x^n\):

\[x^n \Big(n(n-1) - 2\Big) = 0 \Rightarrow n^2 - n - 2 = (n-2)(n+1) = 0\]

which has roots of \(n = -1,\, 2\) and hence

\[y = Ax^2 + Bx^{-1}\]

2. Solve the ODE \(x^2\,y'' - 3x\,y' + 4y = 0, \qquad x >0\)

Using the ansatz \(y = x^n\):

\[\begin{split}x^{n}\Big(n(n-1)-3n+4 \Big) &= 0 \Rightarrow n^2-4n+4 = 0 \\ (n-2)^2 &= 0\end{split}\]

which has a repeated root of \(n=2\), so one of the solutions is \(y = Ax^2\). We can then use:

\[\begin{split}y &= x^2 \,f(x) \\ y' &= 2x\,f + x^2\,f' \\ y'' &= 2\,f + 2x\,f' + 2x\,f' + x^2\,f'' = 2\,f + 4x\,f' + x^2\,f''\end{split}\]

substituting these results in we find:

\[x^2\,y'' - 3x\,y' + 4y = x^4\,f''+ (4x^3-3x^3)\,f' + (2x^2-6x^2+4x^2)\,f = 0 \]

which means we have to solve:

\[x^4\,f'' + x^3\,f' = 0 \Rightarrow f'' + \frac{1}{x}f' = 0\]

Using the integrating factor method, with \(\mu = e^{\int \frac{1}{x}\,\mathrm{d}x} = x\) we find:

\[\frac{\mathrm{d}}{\mathrm{d}x}\left(x\,f' \right) = 0 \Rightarrow x\,f' = C \]

and so the solution is found as:

\[f' = \frac{C}{x} \Rightarrow f(x) = C\ln(x)+D\]

where \(C,\, D\) are constants to be found, so the final solution is:

\[y = x^2(A + B\ln(x))\]

3. Find the solution to the following differential equation on any interval not containing \(x = - 6\):

\[3 ( x + 6 )^2 y'' + 25 ( x + 6 ) y' - 16 y = 0\]

Using the ansatz \(y = |x+6|^n\) we find:

\[|x+6|^n\left(3n(n-1) + 25n - 16 \right) = 0 \Rightarrow 3n^2 + 22n - 16 = 0\]

which is solved for \((3n - 2)(n + 8) = 0\) nd so \(n = \frac{2}{3}, \,-8\), meaning that the general solution is of the form:

\[y = A|x+6|^{2/3} + B|x+6|^{-8}\]

8.2. Reduction of Order

Recall the following linear homogeneous second order ODE:

\[a\,y^{\prime\prime}(x)+b\,y^{\prime}(x)+\frac{b^2}{4a}\,y(x)=0\]

which if we solve using our usual ansatz \(y = e^{\lambda x}\), we find that we have to solve \(a\lambda^2 + b\lambda + \frac{b^2}{2a}=0\) and since the discriminant of this quadratic is zero, this equation has repeated roots, giving one solution of the form:

\[y_1 = Ae^{-b x/2a}\]

Since we must have two linearly independent solutions to this problem, we think about the form of the final solution as:

\[y = y_1 + y_2 = f(x)\,e^{-bx/2a}\]

where \(f(x)\) is a function to be found and must have a linear component and a non-linear component. Using this asumption, we can solve for \(f(x)\) to find that:

\[y = (A + Bx)\,e^{-b x/2a}\]

But can we generalise this method to a more complicated form of second order ODE? Yes, this method is known as reduction of order.

We can apply this method if we know one of the solutions \(y_1(x)\) and are searching for a second linearly independent solution \(y_2(x)\).

Definition

We employ this method to solve inhomogeneous linear differential equations of the form:

(8.3)\[a(x)\,y'' + b(x)\,y' + c(x)\,y = f(x) \longrightarrow y'' + p(x)\,y' + q(x)\,y = r(x)\]

where we have relabelled \(\displaystyle p(x) = \frac{b(x)}{a(x)},\, q(x) = \frac{c(x)}{a(x)},\, r(x) = \frac{f(x)}{a(x)}\) for simplicity.

If we know a solution \(y_1(x)\) of the homogeneous problem \(y_1'' + p\,y_1'+q\,y_1 = 0\), then the solution to the homogeneous problem will be of the form:

\[y_2(x) = u(x)\,y_1(x)\]

where \(u(x)\) can be found from the complicated expression:

\[u(x) = \int \left(C\,y_1^{-2}\,e^{-\int p\,\mathrm{d}x}\right)\,\mathrm{d}x\]

and for the inhomogeneous problem will be of the form:

\[u(x) = \int \left(\frac{\int y_1\,r\,e^{\int p\,\mathrm{d}x}\,\mathrm{d}x + C}{y_1^2\,e^{\int p\,\mathrm{d}x}}\right)\,\mathrm{d}x\]

Using \(y_2(x) = u(x)\,y_1(x)\), we can find the derivatives:

\[\begin{split}y_2' &= u'\,y_1 + u\,y_1' \\ y_2'' &= u''\,y_1 + 2u'\,y_1+ f\,y_1'' \end{split}\]

and sustituting these into (8.3):

\[y_1\,u'' + (2y_1' + p\,y_1)\,u' + (y_1''+p\,y_1'+q\,y_1)\,u = r\]

but since \(y_1\) satisfies \(y_1'' + p\,y_1' + q\,y_1 = 0\), then this reduces do:

\[y_1\,u'' + (2y_1' + p\,y_1)\,u' = r\]

which is a linear first order differential equation in \(f'\):

\[u'' + \left(\frac{2y_1'}{y_1} + p\right)f' = \frac{r}{y_1}\]

We call this method reduction of order because we have reduced a second order problem down to a first order one, if we started with an \(n^{th}\) order ODE, this equation would be an \((n-1)^{th}\) order ODE. Using the relevant integrating factor:

\[\mu = \exp\left[\int\left(\frac{2y_1'}{y_1} + p\right)\,\mathrm{d}x\right] = \exp\left(2\ln|y_1| + \int p\,\mathrm{d}x\right) = y_1^2\,e^{\int p\,\mathrm{d}x}\]

which leaves us to solve:

\[\begin{split}\frac{\mathrm{d}}{\mathrm{d}x}\left( u'\,y_1^2\,e^{\int p\,\mathrm{d}x}\right) &= y_1\,r\,e^{\int p\,\mathrm{d}x}\\ \Rightarrow u'(x) &= \frac{\int y_1\,r\,e^{\int p\,\mathrm{d}x}\,\mathrm{d}x + C}{y_1^2\,e^{\int p\,\mathrm{d}x}}\\ u(x) &= \int \frac{\int y_1\,r\,e^{\int p\,\mathrm{d}x}\,\mathrm{d}x + C}{y_1^2\,e^{\int p\,\mathrm{d}x}}\,\mathrm{d}x\end{split}\]

and for the homogeneous problem, where \(r=0\), we find:

\[u(x) = \int \left(C\,y_1^{-2}\,e^{-\int p\,\mathrm{d}x}\right)\,\mathrm{d}x\]

8.3. Fundamental sets of solutions

One issue to contend with is what after all the work of bridging function we find that \(y_2 = \alpha\,y_1,\, \alpha \in \mathbb{C}\) - i.e. the two solutions are just multipes of each other.

We really need a way to consider the distinct-ness of any solutions we find or if we switch representation of solutions that they form a fundamental set, e.g. \(\sin(x),\, \cos(x)\) could be considered to be a fundamental set, as could \(e^{ix},\, e^{-ix}\), but if we mix the two \(\cos(x),\, e^{ix}\) then this is not a fundamental set and we run the risk of having duplicte solutions.

Lets start with a 2nd order system which has two fundamental solutions \(y_1,\, y_2\) and has to satisy initial conditions \(y(x_0) = y_0,\, y'(x_0) = y_0\):

\[\begin{split}y(x) = Ay_1(x) + By_2(x) \Rightarrow \,&\, y(x_0) = y_0 = Ay_1(x_0) + By_2(x_0)\\ y'(x) = A{y_1}'(x) + B{y_2}'(x) \Rightarrow \,&\, y'(x_0) = {y_0}' = A{y_1}'(x_0) + B{y_2}'(x_0)\end{split}\]

This means we have two equations to find two unknowns \(A,\,B\), we could therefore to solve for these simultaneously, however a different method is to use a 2x2 matrix:

\[\begin{split}\begin{pmatrix} y_0 \\ {y_0}' \end{pmatrix} = \begin{pmatrix} y_1 & y_2 \\ {y_1}' & {y_2}'\end{pmatrix}\,\begin{pmatrix} A \\ B \end{pmatrix}\end{split}\]

which means to find \(A,\,B\) we can use a matrix inverse:

\[\begin{split}\frac{1}{y_1\,{y_2}' - {y_1}'\,y_2}\begin{pmatrix} {y_2}' & -y_2 \\ -{y_1}' & y_1\end{pmatrix}\,\begin{pmatrix} y_0 \\ {y_0}' \end{pmatrix} = \begin{pmatrix} A \\ B \end{pmatrix}\end{split}\]

Therefore in order for \(A,\,B\) to exist, the matrix determinant must be non-zero - this is feature here that tells us if we have a funamental set of solutions:

\[\begin{split}W(y_1,\,y_2)(x) = \begin{vmatrix} y_1 & y_2 \\ {y_1}' & {y_2}'\end{vmatrix}\end{split}\]

We call this determinant the Wronskian.

Worked example

Show that the functions \(y_1 = e^{\lambda x}\) and \(y_2 = x\,e^{\lambda x}\) form a fundamental set of solutions.

\[\begin{split}W(y_1,\,y_2)(x) = \begin{vmatrix} y_1 & y_2 \\ {y_1}' & {y_2}'\end{vmatrix} &= \begin{vmatrix} e^{\lambda x} & x\,e^{\lambda x} \\ \lambda\,e^{\lambda x} & \lambda\,x\,e^{\lambda x} + e^{\lambda x}\end{vmatrix}\\ &= e^{2\lambda x} (\lambda x + 1 - \lambda x) = e^{2\lambda x} \end{split}\]

and clearly for any finite \(x\), \(W \neq 0\), hence we have a fundamental set.

Further worked examples

1. Show that the functions \(y_1 = x^{n_1}\) and \(y_2 = x^{n_2}\) form a fundamental set of solutions for \(x > 0\)

Lets switch to \(y_1 = e^{n_1\,\ln(x)},\, y_2 = e^{n_2\,\ln(x)}\)

\[\begin{split}W(y_1,\,y_2)(x) = \begin{vmatrix} y_1 & y_2 \\ {y_1}' & {y_2}'\end{vmatrix} &= \begin{vmatrix} e^{n_1\,\ln(x)} & e^{n_2\,\ln(x)} \\ \frac{n_1}{x}\,e^{n_1\,\ln(x)} & \frac{n_2}{x}\,e^{n_2\,\ln(x)} \end{vmatrix}\\ &= e^{(n_1+n_2)\,\ln(x)} \left(\frac{n_2}{x} - \frac{n_1}{x}\right) = (n_1 - n_1)\,x^{n_1+ n_2 - 1}\end{split}\]

and clearly for any finite \(x\) and \(n_1 \neq n_2\), \(W \neq 0\), hence we have a fundamental set.

2. Show that the functions \(y_1 = x^{\lambda + \mu i} = e^{(\lambda + \mu i)\ln (x)}\) and \(y_2 = x^{\lambda - \mu i}= e^{(\lambda - \mu i)\ln (x)}\) form a fundamental set of solutions, where \(x > 0\)

\[\begin{split}W(y_1,\,y_2)(x) = \begin{vmatrix} y_1 & y_2 \\ {y_1}' & {y_2}'\end{vmatrix} &= \begin{vmatrix} e^{(\lambda + \mu i)\ln (x)} & e^{(\lambda - \mu i)\ln (x)} \\ \frac{(\lambda + \mu i)}{x}e^{(\lambda + \mu i)\ln (x)} & \frac{(\lambda - \mu i)}{x}e^{(\lambda - \mu i)\ln (x)}\end{vmatrix}\\ &= e^{(\lambda^2 + \mu^2)\ln(x)}\left(\frac{\lambda - \mu i}{x} - \frac{\lambda + \mu i}{x} \right) \\ &= -\frac{2\mu\,i\,e^{(\lambda^2 + \mu^2)\ln(x)}}{x} = -2\mu\,i\,x^{\lambda^2 + \mu^2-1}\end{split}\]

and clearly for any finite \(x > 0\), \(W \neq 0\), hence we have a fundamental set.