# Higher order linear differential equations

## Solution by reduction of order for linear DE with non-constant coefficients

Example: we want to solve

$$x^2y''+xy'-y = 0$$

Such DE might be difficult to solve, since it has non-constant coefficients. However, sometimes it is possible to find at least one solution (in above case $y_1(x) = x$) and then use method described further to derive the second solution $\boldsymbol{y_2(x)}$ from the first one.

The method can be used for higher order DE and will be shown on 2nd order DE. The aim of this chapter is to bring a generalized convenient formula, which can be used to get $y_2(x)$ readily from $y_1(x)$.

The differential equation is linear but with non-constant coefficients. First, we bring DE $a_2(x)y'' + a_1(x)y' + a_0(x) y = 0$ into standard form

$$$$y''+P(x)y' + Q(x)y = 0 \label{ref:stand}$$$$

Let us suppose that we have one solution already ($y_1(x)$). Either it is given or easy to find. We want to find the other solution $y_2(x)$ of given DE. Since $y_1(x)$ and $y_2(x)$ are linearly independent, we can express $y_2$ as

$$$$y_2(x) = u(x) \cdot y_1(x). \label{ref:expr1}$$$$

Let us differentiate $y_2(x)$ two times

\begin{align} y_2'(x) &= u'y_1 + uy_1' \nonumber\\ y_2''(x) &= u''y_1 + u'y_1' + u'y_1 +uy_1'' = u''y_1 + 2u'y_1' + uy_1'' \nonumber \end{align}

and place the products into standard form $(\ref{ref:stand})$ of DE:

\begin{align} [u''y_1 + 2u'y_1' + \color{silver}{uy_1''}] + [P(x)(u'y_1 + \color{silver}{uy_1'})] + [Q(x)\color{silver}{uy_1}] &= 0 \\ u(\color{blue}{y_1'' + Py_1' +Q(x)y_1}) + u''y_1 + 2 u'y_1' + P(x)u'y_1 &= 0 \label{ref:blue} \end{align}

The content of parentheses in $(\ref{ref:blue})$ is known to be zero, because $y_1$ is a solution of homogeneous DE $(\ref{ref:stand})$.

\begin{align} y_1u'' + 2y_1'u' + P(x)y_1u' = 0 \nonumber\\ y_1u'' + u'(2y_1' + P(x)y_1) = 0 \end{align}

Reminder: $y_1$ and $y_1'$ are known, we are solving $u$. Let us use substitution $w=u'$:

\begin{align} y_1w' + w(2y_1' + P(x)y_1) = 0 \label{ref:subst} \end{align}

The last equation $(\ref{ref:subst})$ can be solved by separating variables:

\begin{align} y_1 \frac{dw}{dx} + w(2y_1' + P(x)y_1) &= 0 \nonumber\\ dw\frac 1 w + dx(2\frac{y_1'}{y_1} + P(x)) &= 0 \nonumber\\ \log w + 2 \intop\frac{y_1'}{y_1}dx + \intop P(x)dx & =0\label{ref:cont} \end{align}

The integral $\intop (y'/y) dx$ might be something unexpected to solve, but is not difficult. Since $y' = (dy/dx)$

$$\intop \frac{y'}{y} dx = \intop \frac{dy}{dx} \frac{1}{y} dx = \intop \frac{1}{y} dy = \log y.$$

Let us continue from $(\ref{ref:cont})$:

\begin{align} \log(w y_1^2) &= -\intop P(x)dx + c \nonumber \\ wy_1^2 &= e^{-\intop P(x)dx + c} = \nonumber\\ &= ce^{-\intop P(x)dx} \nonumber\\ w &= c \frac{e^{-\intop P(x) dx}}{y_1^2} \nonumber\\ u &= c_1 \intop \frac{e^{-\intop P(x) dx}}{y_1^2} + c_2\nonumber\\ \end{align}

We are looking for linear independent function and we will place arbitrary constant with $y_1$ and $y_2$. So $c_1,\ c_2$ are not much of our concern:

\begin{align} \color{red}{u(x)} &\color{red}{= \intop \frac{e^{-\intop P(x) dx}}{y_1^2(x)}dx} \label{ref:form2} \hskip1em \text{and}\\ \color{red}{y_2(x)} &= \color{red}{u(x)\cdot y_1(x)} \label{ref:form1}. \end{align}

Equation $(\ref{ref:form1})$ above is our expression $(\ref{ref:expr1})$ of $y_2(x)$ in terms of $u(x)$ and $y_1(x)$ from beginning of the lesson. Formula ($\ref{ref:form2}$) is then an important equation expressing $u(x)$ in order to find the other solution $y_2(x)$.

### Example

Solve $x^2 y'' + xy' - y = 0$ if $y_1 = x$ is known to be first solution.

The reduction of order method is able to find $y_2(x)$ from first solution $y_1(x)$. First we need to bring DE into expected standard form:

\begin{align} y'' + \frac{x}{x^2}y' - \frac{1}{x^2}y &= 0 \hskip2em x \neq 0 \nonumber \\ y'' + \color{red}{\frac 1 x} y' - \frac{1}{x^2}y &= 0 \hskip2em \color{red}{\frac 1 x} = P(x)\nonumber \end{align}

Let us use $P(x)$ and $y_1(x)$ within $(\ref{ref:form2})$ in order to determine $y_2(x)$ from $(\ref{ref:form1})$ : \begin{align} u(x) &= \intop \frac{e^{-\intop 1/x\ dx}}{y_1^2(x)}dx = \intop \frac{e^{-\log x}}{x^2}dx = \nonumber \\ &=\intop \frac{-x}{x^2}dx = -\intop x^{-3} dx = -\frac 1 2 x^{-2} \nonumber \\ y_2(x) & = u(x) \cdot y_1(x) = - \frac 1 2x^{-2} x = - \frac 1 2 x^{-1} = x^{-1} \nonumber \end{align}

Since we have both $y_1(x)$ and $y_2(x)$ we can collect complete complementary solution

$$\underline{\underline{y_c(x) = c_1 x + c_2 x^{-1}}}$$

### Example

Solve DE $(2x^2 + 1)y'' - 4xy' + 4y = 0$ if $y_1(x) = x$ is known to be first solution.

This DE is linear with non-constant coefficients and is hard to solve. However because we know the first solution $y_1(x)$, we can use above method to find $y_2(x)$ and complete the solution.

The method was derived from DE in standard form, let us bring DE into standard form then:

\begin{align} & y'' - \frac{4x}{(2x^2 +1)}y' + \frac{4}{(2x^2+1)} y = 0 \nonumber\\ & u = \intop \frac{e^{-\intop - {4x}/{(2x^2+1)}\ dx}}{y_1^2}dx = \intop \frac{e^{\log (2x^2+1)}}{x^2}dx = \intop \frac{2x^2+1}{x^2}dx \nonumber\\ & u = \intop 2 dx + \intop \frac{1}{x^2} dx = 2x - x^{-1} \nonumber \\ & y_2(x) = x \cdot (2x - x^{-1}) = 2x^2 -1 \nonumber \\ & \underline{\underline{y_c(x)= c_1 x + c_2 (2x^2 -1)}} \nonumber \end{align}

### Example

Use the reduction of order method to find general solution of DE $x^2y'' - xy'+y = 0$, $y_1 = x$.

$$y'' - \frac{x}{x^2} y' + \frac{1}{x^2}y=0$$

We have to find $y_2(x) = u(x)\cdot y_1(x)$. From above DE, which was brought into standard form we will take $P(x) = -1/x$.

\begin{align} & u(x) = \intop \frac{e^{-\intop P(x)dx}}{y_1^2} dx = \intop \frac{e^{-\intop -1/x\ dx}}{x^2} dx = \intop \frac{e^{\log x}}{x^2} dx = \intop \frac{1}{x} dx = \log |x| \nonumber \\ & y_2(x) = u(x)\cdot y_1(x) = \log {|x|}x \nonumber \\ & \underline{\underline{y_c(x) = c_1x + c_2 \log{|x|}\cdot x}} \nonumber \end{align}

### Example

Use the reduction of order method to find general solution of DE $y'' +9y = 0$, $y_1 = \sin 3x$.

$$y'' - \frac{x}{x^2} y' + \frac{1}{x^2}y=0$$

We have to find $y_2(x) = u(x)\cdot y_1(x)$. From above DE, which was brought into standard form we will take $P(x) = 0$, becaue member with $y'$ is missing.

\begin{align} & u(x) = \intop \frac{e^{-\intop P(x)dx}}{y_1^2} dx = \intop \frac{1}{\sin^2 3x} dx = \frac{1}{3 \tan 3x} \nonumber \\ & y_2(x) = u(x)\cdot y_1(x) =\frac{1}{3 \tan 3x} \cdot \sin 3x = -\frac 1 3 \frac{\cos 3x}{\sin 3x} \sin 3x = -\frac{1}{3} \cos 3x \nonumber \\ & \underline{\underline{y_c(x) = c_1 \sin 3x + c_2 \cos 3x}} \nonumber \end{align}

### Example

Use the reduction of order method to find general solution of DE $x^2y'' + xy' - 4y = x^3$, $y_1 = x^2$.

Note: you might be not able to understand this example, because although the example is solved by the method explained above, it also expects some knowledge from chapters behind.

You may wonder where the solution $x^2$ comes from. But if we study the associated homogeneous DE, it has to be not too difficult to try a few simple solutions like $y=x,\ y=c_1,\ y=x^2$ and test, whether they are the solution or not. Then we can use the method to derive the next solution $y_2$.

Note: it can be also observed that the DE can be solved as a Cauchy-Euler DE.

To use the reduction of order method we have to bring the DE into standard form to find $P(x)$.

$$y'' + \color{red}{\frac{1}{x}}y' - \frac{4}{x^2}y=x \implies P(x) = \frac 1 x$$

Now we can evaluate $u(x)$, once we have $u(x)$ we use it to multiply $y_1$ in order to get the other solution $y_2= u(x)\cdot y_1(x)$.

$$u(x) = \int \frac{e^{-\intop P(x) dx}}{y_1^2(x)}dx = \int {\frac{e^{-\int{1/x\ dx}}}{x^4}dx} = \int {\frac{1/x}{x^4} dx} = \int{x^{-5}\ dx} = \color{red}{-\frac 1 4 }x^{-4}\\$$

The constant $-\frac 1 4$ can be lost, because an arbitrary constant $c_2$ will be included anyway.

\begin{align} &y_2 =u(x)\cdot y_1(x)= \color{silver}{-\frac{1}{4}}x^{-4} \cdot x^2 = \color{silver}{-\frac{1}{4}}x^{-2} \nonumber \\ &y = \color{blue}{c_1x^2} + c_2x^{-2} + y_p \nonumber \end{align}

The task remains to find $y_p$ since DE given to solve is nonhomogeneous. That can be done by method of undetermined coefficients. We might assume that the particular solution is going to be in the form of $y_p = \color{blue}{Ax^2} + Bx + C$, then evaluate $y_p',\ y_p''$, substitute to DE in order to find coefficients $A,\ B,\ C$. But you will observe soon that it does not lead to a solution. That is because we deal with the second type (the member $g(x)$ contains a member from $y_c$). The next attempt to assume $y_p = Ax^3 \color{silver}{+ Bx^2} + Cx + D$ brings the constants $A = 1/5,\ C = 0,\ D = 0$ and then the general solution is:

$$\underline{\underline{y = {c_1x^2} + c_2x^{-2} + \frac 1 5 x^3}}.$$

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