7 Linear Ordinary Differential Equations

Introduction

• Q1: Consider the cooling model. Assume there is a cup of boiled water (100°C), environment temperature is 20°C. How does water cool down over time?
• A: According to the Newton's Law of Cooling

$$\frac{\mathrm{d}T}{\mathrm{d}t}=-k(T-20).$$

$k$ is the cooling rate constant. Initial temperature $T(0)=100$.

$$T=20+80e^{-t}$$

• Q2: Assume a ball goes from static to free fall. How to calculate the fall distance over time?
• A:

$$\begin{array}{rl}\frac{\mathrm{d}v}{\mathrm{d}t}& =g\\ \frac{\mathrm{d}x}{\mathrm{d}t}& =v\\ ⟹\frac{{\mathrm{d}}^{2}x}{\mathrm{d}{t}^2}& =g\end{array}$$

$g$ is gravity, with initial position $x(0)=0$.

$$x=\frac{1}{2}g{t}^2$$

Applications about Differential Equations include:

• neutron diffusion equation:$$D{\mathrm{\nabla }}^2\varphi -\sum _a\varphi +k_{\mathrm{\infty }}\sum _a\varphi =0$$
• decay chain model:$$\frac{\mathrm{d}{N}_i}{\mathrm{d}t}=-{\lambda }_i{N}_i+\sum _{j=1}^{i-1}{b}_{j\to i}{\lambda }_j{N}_j$$
• Tsiolkovsky rocket equation:$$\frac{\mathrm{d}v}{\mathrm{d}t}=\frac{u_e}{m}\frac{\mathrm{d}m}{\mathrm{d}t}-g$$
• Navier-Stokes equations:$$\rho (\frac{\partial \mathit{v}}{\partial t}+\mathit{v}\cdot \mathrm{\nabla }\mathit{v})=-\mathrm{\nabla }p+\mu {\mathrm{\nabla }}^2\mathit{v}+\mathit{f}.$$

A differential equation relates two or more variables via derivatives or differentials.

The simplest form:

$$\frac{\mathrm{d}y}{\mathrm{d}x}=f(x)$$

where $f(x)$ is a function of $x$. we define a solution of a differential equation to be any functional relation, not involving derivatives or integrals of unknown functions

The solution is obtained immediately by integration, in the form

$$y=\int f(x)\mathrm{d}x+C$$

Whether or not it happens that the integral can be expressed in terms of simple functions is incidental, in the sense that we define a solution of a differential equation to be any functional relation, not involving derivatives or integrals of unknown functions, which implies the differential equation.

Similarly, in an equation of the form

$$F(x)G(y)\mathrm{d}x+f(x)g(y)\mathrm{d}y=0.$$

We may separate the variables and obtain a solution by integration in the form

$$\begin{array}{rl}\int \frac{F(x)}{f(x)}\mathrm{d}x+\int \frac{g(y)}{G(y)}\mathrm{d}y& =C\\ ⟹H(x)+I(y)& =C\end{array}$$

Linear Ordinary Differential Equations (ODEs)

Definition: Linear ODEs

Linear differential equation of order $n$, in absence of products or nonlinear functions of the dependent variable $y$ and its derivatives the highest derivative present being of order $n$:

$$a_0(x)\frac{{\mathrm{d}}^{n}y}{\mathrm{d}{x}^n}+a_1(x)\frac{{\mathrm{d}}^{n-1}y}{\mathrm{d}{x}^{n-1}}+\cdots +a_{n-1}(x)\frac{\mathrm{d}y}{\mathrm{d}x}+a_n(x)y=f(x).$$

The coefficients $a$ may be arbitrarily specified functions of the independent variable $x$.

Examples:

• First order:

$$\begin{gathered} \frac{\mathrm{d}y}{\mathrm{d}x}=g_x \\ \frac{\mathrm{d}y}{\mathrm{d}x}+ky=20k,(a_0=1,a_1=k,f(x)=20k) \end{gathered}$$

• Second order:

$$\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}+2\frac{\mathrm{d}y}{\mathrm{d}x}+y=x$$

• Higher orders include:

$$\begin{gathered} (y^{‴}{)}^5+x^4{y}^{″}-x{y}^{\prime }+y^7=1, \\ {y}^{(4)}-4y^{‴}+10y^{″}-12y^{\prime }+5y=\sin 2x \end{gathered}$$

The most general linear differential equation of the nth order can be written in the form (by dividing $a_0(x)$ in both terms):

$$\frac{{\mathrm{d}}^{n}y}{\mathrm{d}{x}^n}+a_1(x)\frac{{\mathrm{d}}^{n-1}y}{\mathrm{d}{x}^{n-1}}+\cdots +a_{n-1}(x)\frac{\mathrm{d}y}{\mathrm{d}x}+a_n(x)y=h(x).$$

This equation is frequently written in the abbreviated form

$$Ly=h(x).$$

where $L$ here represents the linear differential operator:

$$L=\frac{{\mathrm{d}}^n}{\mathrm{d}{x}^n}+a_1(x)\frac{{\mathrm{d}}^{n-1}}{\mathrm{d}{x}^{n-1}}+\cdots +a_{n-1}(x)\frac{\mathrm{d}}{\mathrm{d}x}+a_n(x)$$

General solutions to linear differential equations

• If $n$ linearly independent solutions $u_1(x),u_2(x),...,u_n(x)$ of the associated homogeneous equation $L{y}_H=0$ are known, the general solution is of the form:

$$y_H(x)=c_1{u}_1(x)+c_2{u}_2(x)+\cdots +c_n{u}_n(x)=\sum _{k=1}^n{c}_k{u}_k(x)$$

where $c_i$ are required arbitrary constants.

• Suppose that one particular solution of $L{y}_P=h(x)$ can be obtained by inspection or otherwise, such that $L{y}_P=h(x)$, then the complete solution is of the form:

$$y=y_H(x)+y_P(x)=\sum _{k=1}^n{c}_k{u}_k(x)+y_P(x)$$

The mission is:

1. Find $y_H:L{y}_H=0$;
2. Find $y_P:L{y}_P=h(x)$ need only one solution

Then $y=y_H+y_P:Ly=h(x)$

The question is : how do we know $n$ functions are linearly independent?

Linear Dependence/Independence of functions

The functions $u_1,u_2,\cdots ,u_n$ are said to linear independence over a given interval if over that interval no one of the functions can be expressed as a linear combination of the others (nontrival). Here are more specific:

We assume that each of a set of $n$ functions $u_1,u_2,\cdots ,u_n$ possesses $n$ finite derivatives at all points of an interval $M$. Then, if a set of constants exists such that

$$c_1{u}_1+c_2{u}_2+\cdots +c_n{u}_n=0$$

for all values of $x$ in the interval $M$.

these same constants also satisfy the identities

$$\begin{gathered} c_1\frac{\mathrm{d}{u}_1}{\mathrm{d}x}+c_2\frac{\mathrm{d}{u}_2}{\mathrm{d}x}+\cdots +c_n\frac{\mathrm{d}{u}_n}{\mathrm{d}x}=0 \\ {c}_1\frac{{\mathrm{d}}^2{u}_1}{\mathrm{d}{x}^2}+c_2\frac{{\mathrm{d}}^2{u}_2}{\mathrm{d}{x}^2}+\cdots +c_n\frac{{\mathrm{d}}^2{u}_n}{\mathrm{d}{x}^2}=0 \\ ⋮ \\ {c}_1\frac{{\mathrm{d}}^{n-1}{u}_1}{\mathrm{d}{x}^{n-1}}+c_2\frac{{\mathrm{d}}^{n-1}{u}_2}{\mathrm{d}{x}^{n-1}}+\cdots +c_n\frac{{\mathrm{d}}^{n-1}{u}_n}{\mathrm{d}{x}^{n-1}}=0 \end{gathered}$$

Then:

$$\left(\begin{array}{cccc}{u}_1& u_2& \cdots & u_n\\ \frac{\mathrm{d}{u}_1}{\mathrm{d}x}& \frac{\mathrm{d}{u}_2}{\mathrm{d}x}& \cdots & \frac{\mathrm{d}{u}_n}{\mathrm{d}x}\\ ⋮& ⋮& ⋮& ⋮\\ \frac{{\mathrm{d}}^{n-1}{u}_1}{\mathrm{d}{x}^{n-1}}& \frac{{\mathrm{d}}^{n-1}{u}_2}{\mathrm{d}{x}^{n-1}}& \cdots & \frac{{\mathrm{d}}^{n-1}{u}_n}{\mathrm{d}{x}^{n-1}}\end{array}\right)\left(\begin{array}{c}{c}_1\\ c_2\\ ⋮\\ c_n\end{array}\right)=\mathbf{0}.$$

if the functions $u_1,u_2,\cdots ,u_n$ are linearly dependent, then the determinant The following determinant vanishes if the functions are linearly dependent; otherwise, they are linearly independent

Lemma: Wronskian determinant

The vanishing of the Wronskian determinant is necessary but not sufficient for linear dependence. The functions are linearly independent if this determinant does not vanish, i.e.:

$$W(u_1,u_2,\cdots ,u_n)=\left|\begin{array}{cccc}{u}_1& u_2& \cdots & u_n\\ \frac{\mathrm{d}{u}_1}{\mathrm{d}x}& \frac{\mathrm{d}{u}_2}{\mathrm{d}x}& \cdots & \frac{\mathrm{d}{u}_n}{\mathrm{d}x}\\ ⋮& ⋮& ⋮& ⋮\\ \frac{{\mathrm{d}}^{n-1}{u}_1}{\mathrm{d}{x}^{n-1}}& \frac{{\mathrm{d}}^{n-1}{u}_2}{\mathrm{d}{x}^{n-1}}& \cdots & \frac{{\mathrm{d}}^{n-1}{u}_n}{\mathrm{d}{x}^{n-1}}\end{array}\right|\ne 0.$$

First order linear solution

The linear equation of first order is readily solved in general terms, without determining separately homogeneous and particular solutions

we attempt to determine an integrating factor $p(x)$ such that the standard form

$$\frac{\mathrm{d}y}{\mathrm{d}x}+a_1(x)y=h(x).$$

is equivalent to the equation

$$\frac{\mathrm{d}}{\mathrm{d}x}(py)=ph.$$

Here is the procession:

$$\begin{array}{rl}\frac{\mathrm{d}}{\mathrm{d}x}(py)=& p\frac{\mathrm{d}y}{\mathrm{d}x}+y\frac{\mathrm{d}p}{\mathrm{d}x}\\ ⟹\frac{1}{p}\frac{\mathrm{d}}{\mathrm{d}x}(py)=& \frac{\mathrm{d}y}{\mathrm{d}x}+\frac{y}{p}\frac{\mathrm{d}p}{\mathrm{d}x}\triangleq h(x)\\ ⟹\frac{\mathrm{d}}{\mathrm{d}x}(py)=& ph(x).\end{array}$$

• The above equation can be written as $\frac{\mathrm{d}y}{\mathrm{d}x}+\left(\frac{1}{p}\frac{\mathrm{d}p}{\mathrm{d}x}\right)y=h(x)$.
• Therefore, we have $\frac{1}{p}\frac{\mathrm{d}p}{\mathrm{d}x}=a_1(x)$, which leads to $p=\exp (\int a_1(x)\mathrm{d}x)$
• As a result, we obtain the general solution $py=\int ph\mathrm{d}x+C$, $C$ is an arbitrary constant, or$$y=\frac{1}{p}\int ph\mathrm{d}x+\frac{C}{p}.$$

Example1: Integrating factor

To solve the differential equation

$$x\frac{dy}{dx}+(1-x)y=x{e}^x$$

we first rewrite the equation in the standard form,

$$\frac{dy}{dx}+(\frac{1}{x}-1)y=e^x$$

An integrating factor is then

$$p=e^{\int (\frac{1}{x}-1)\mathrm{d}x}=e^{\log x-x}=x{e}^{-x}$$

no constant being added in the integration, since only a factor is needed. The solution is then given by

$$\begin{array}{rl}y& =\frac{e^x}{x}\int x\mathrm{d}x+C\frac{e^x}{x}\\ ⟹y& =\frac{x}{2}{e}^x+C\frac{e^x}{x}\end{array}$$

Linear ODEs with Constant Coefficients

The simplest and perhaps the most important differential equation of higher order is the linear equation (in which the coefficients $a_k$ are constants)

$$Ly=\frac{{\mathrm{d}}^{n}y}{\mathrm{d}{x}^n}+a_1\frac{{\mathrm{d}}^{n-1}y}{\mathrm{d}{x}^{n-1}}+\cdots +a_{n-1}\frac{\mathrm{d}y}{\mathrm{d}x}+a_{n}y=h(x).$$

• Note that $\frac{{\mathrm{d}}^m}{\mathrm{d}{x}^m}{e}^{rx}=r^m{e}^{rx}$

• Then

  $$L{e}^{rx}=(r^n+a_1{r}^{n-1}+\cdots +a_{n-1}r+a_n)e^{rx}$$

• For a homogeneous equation ($h(x)=0$), we just need (The characteristic equation of $r$):

  $$r^n+a_1{r}^{n-1}+\cdots +a_{n-1}r+a_n=0$$
  • If there are $n$ distinct roots to the characteristic equation:
  $$y_H=\sum _{k=1}^n{c}_k{e}^{r_{k}x}$$
  • if one or more of the roots is repeated, less than $n$ independent solutions are obtained. Suppose that $r=r_1$ is a double root of the characteristic equation:
  $$L{e}^{rx}=(r-r_1{)}^2(r-r_2)\cdots (r-r_n)e^{rx},r_1\ne r_2\ne \cdots \ne r_n$$

  To solve it, note that:

  $$\begin{array}{rl}L[e^{rx}{]}_{r=r_1}=& L[e^{r_{1}x}]=0\\ L{\left[\frac{\partial }{\partial r}{e}^{rx}\right]}_{r=r_1}=& L[x{e}^{r_{1}x}]=0,\end{array}$$

  (Not so accuracy):

  $$L\left[\frac{\partial }{\partial r}{e}^{rx}\right]=\frac{\partial }{\partial r}L\left[e^{rx}\right]$$

  The part of the homogeneous solution corresponding to the double root $r_1$ can be written in the form

  $$c_1{e}^{r_{1}x}+c_{2}x{e}^{r_{1}x}=e^{r_{1}x}(c_1+c_{2}x).$$

  By a simple extension of this argument, it can be shown the part of the homogeneous solution corresponding to an $m$-fold root $r_1$ is of the form

  $$e^{r_{1}x}(c_1+c_{2}x+c_3{x}^2+\cdots +c_m{x}^{m-1}).$$

If the characteristic equation has imaginary roots and if the coefficients of that equation are real, the roots must occur in conjugate pairs. Thus, if $r_1=a+ib$ is one root, a second root must be $r_2=a-ib$. The part of the solution corresponding to these two roots can be written in the form

$$A{e}^{(a+ib)x}+B{e}^{(a-ib)x}=e^{ax}(A{e}^{ibx}+B{e}^{-ibx}).$$

In order that this expression be real, the constants $A$ and $B$ must be imaginary By making use of Euler's formula,*

$$e^{i\theta }=\cos \theta +i\sin \theta ,$$

we find that the solution becomes

$$e^{ax}[A(\cos bx+i\sin bx)+B(\cos bx-i\sin bx)]$$

and hence can be written in the more convenient form,

$$e^{ax}(c_1\cos bx+c_2\sin bx),$$

where $c_1$ and $c_2$ are new arbitrary constants replacing $(A+B)$ and $i(A-B)$, respectively. Thus real values of $c_1$ and $c_2$ correspond to values of $A$ and $B$ which are conjugate complex.

Similarly, if $a\pm ib$ are $m$-fold roots, the corresponding $2m$ terms in the homogeneous solution can be written in the real form,

$$\begin{array}{rl}{e}^{ax}[(c_1+c_{2}x+\cdots & +c_m{x}^{m-1})\cos bx\\ & +(c_{m+1}+c_{m+2}x+\cdots +c_{2m}{x}^{m-1})\sin bx]\end{array}$$

Example2: Linear ODE with constant coefficients [core]

• For the equation

$$\frac{{\mathrm{d}}^{3}y}{\mathrm{d}{x}^3}-\frac{\mathrm{d}y}{\mathrm{d}x}=0,$$

the characteristic equation is $r^3-r=r(r+1)(r-1)=0$, from which there follows $r=0,\pm 1$. The general solution is then

$$y=c_1+c_2{e}^x+c_3{e}^{-x}$$

• For the differential equation

$$\frac{{\mathrm{d}}^{3}y}{\mathrm{d}{x}^3}-5\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}+8\frac{\mathrm{d}y}{\mathrm{d}x}-4y=0$$

the characteristic equation is $(r-1)(r-2{)}^2=0$, from which there follows $r=1,2,2$. The general solution is then

$$y=c_1{e}^x+e^{2x}(c_2+c_{3}x).$$

• The equation

$$\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}+2\frac{\mathrm{d}y}{\mathrm{d}x}+5y=0$$

has the characteristic equation $r^2+2r+5=0$, from which $r=-1\pm 2i$; hence

$$y=e^{-x}(c_1\cos 2x+c_2\sin 2x)$$

Nonhomogeneous Linear ODEs

A shorter method which can be applied in many practical cases is that of undetermined coefficients. This method may be used when the right-hand side of $h(x)$ involves only terms of

$$x^m,\sin (qx),\cos (qx),e^{px},\text{ and/or products of two}$$

or more such functions. The reason for the success of the method is the fact that each of these functions, or any product of a finite number of these functions, has only a finite number of linearly independent derivatives.

If we define the family of a function $f(x)$ as the set of linearly independent functions of which the function $f(x)$ and its derivatives with respect to $x$ are linear combinations, the following families may be listed

Term	Family
$x^m$	$x^m,x^{m-1},x^{m-2},\cdots ,x^2,x,1$
$\sin (qx)$	$\sin (qx),\cos (qx)$
$\cos (qx)$	$\sin (qx),\cos (qx)$
$e^{px}$	$e^{px}$

Example3 [core]

Consider the differential equation

$$\frac{{\mathrm{d}}^{3}y}{\mathrm{d}{x}^3}-\frac{\mathrm{d}y}{\mathrm{d}x}=2x+1-4\cos x+2e^x.$$

The general homogeneous solution is

$$y_H=c_1+c_2{e}^x+c_3{e}^{-x}.$$

The families of the terms $x,1,\cos x$, and $e^x$ on the right-hand side of the equation are, respectively,

$$\{x,1\},\{1\},\{\cos x,\sin x\},\left\{e^x\right\}.$$

• The second family is contained in the first, and is discarded.
• Since the first family has the representative $1$ in the homogeneous solution($r_1=0\to e^0=1$), it is replaced by the family $\{x^2,x\}$. Similarly, the last family is replaced by $\left\{x{e}^x\right\}$ ($r_2=1\to e^{1x}$). (?)

A particular solution is then assumed in the form

$$y_P=A{x}^2+Bx+C\cos x+D\sin x+Ex{e}^x.$$

When $y$ is replaced by $y_P$, the differential equation becomes

$$-2Ax-B-2D\cos x+2C\sin x+2E{e}^x=2x+1-4\cos x+2e^x.$$

By equating the coefficients of $x,1,\cos x,\sin x$, and $e^x$, there follows

$$A=-1,B=-1,D=2,C=0,E=1.$$

A particular solution thus is

$$y_P=-x^2-x+2\sin x+x{e}^2,$$

and the general solution is

$$y=c_1+c_2{e}^x+c_3{e}^{-x}-x^2-x+2\sin x+x{e}^x.$$

Equidimensional linear differential equation

An equation of the form

$$Ly=x^n\frac{{\mathrm{d}}^{n}y}{\mathrm{d}{x}^n}+b_1{x}^{n-1}\frac{{\mathrm{d}}^{n-1}y}{\mathrm{d}{x}^{n-1}}+\cdots +b_{n-1}x\frac{\mathrm{d}y}{\mathrm{d}x}+b_{n}y=h(x).$$

where the $b$'s are constants, this equation has the property that each term on the left is unchanged when $x$ is replaced by $cx$, where $c$ is a nonzero constant.

Introducing a new independent variable $z$ by the substitution:

$$x=e^z,z=\log x$$

There then follows

$$\frac{\mathrm{d}}{\mathrm{d}x}=\frac{\mathrm{d}z}{\mathrm{d}x}\frac{\mathrm{d}}{\mathrm{d}z}=\frac{1}{e^z}\frac{\mathrm{d}}{\mathrm{d}z}$$

In particular, we obtain

$$\begin{array}{rl}x\frac{\mathrm{d}y}{\mathrm{d}x}& =\frac{\mathrm{d}y}{\mathrm{d}z}\\ x^2\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}& =\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{z}^2}-\frac{\mathrm{d}y}{\mathrm{d}z}=\frac{\mathrm{d}}{\mathrm{d}z}(\frac{\mathrm{d}}{\mathrm{d}z}-1)y,\\ x^3\frac{{\mathrm{d}}^{3}y}{\mathrm{d}{x}^3}& =\frac{{\mathrm{d}}^{3}y}{\mathrm{d}{z}^3}-3\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{z}^2}+2\frac{\mathrm{d}y}{\mathrm{d}z}\\ & =\frac{\mathrm{d}}{\mathrm{d}z}(\frac{\mathrm{d}}{\mathrm{d}z}-1)(\frac{\mathrm{d}}{\mathrm{d}z}-2)y,\end{array}$$

$$\begin{array}{rl}{x}^2\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}=& x^2\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{\mathrm{d}}{\mathrm{d}x}y\right)=x\frac{\mathrm{d}}{\mathrm{d}z}\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)=x\frac{\mathrm{d}}{\mathrm{d}z}\left(\frac{1}{x}\frac{\mathrm{d}y}{\mathrm{d}z}\right)\\ =& e^z\frac{\mathrm{d}}{\mathrm{d}z}\left(\frac{1}{e^z}\frac{\mathrm{d}y}{\mathrm{d}z}\right)=e^z(-\frac{1}{e^z}\frac{\mathrm{d}y}{\mathrm{d}z}+\frac{1}{e^z}\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{z}^2})\\ =& \frac{{\mathrm{d}}^{2}y}{\mathrm{d}{z}^2}-\frac{\mathrm{d}y}{\mathrm{d}z}\end{array}$$

In general, it is found that

$$x^m\frac{{\mathrm{d}}^{m}y}{\mathrm{d}{x}^m}=\frac{\mathrm{d}}{\mathrm{d}z}(\frac{\mathrm{d}}{\mathrm{d}z}-1)(\frac{\mathrm{d}}{\mathrm{d}z}-2)\cdots (\frac{\mathrm{d}}{\mathrm{d}z}-m+1)y.$$

The transformed equation thus becomes linear with constant coefficients, and $y$ then can be determined in terms of $z$. The final result is obtained by replacing $z$ by $\log x$

Example4: Equidimensional linear differential equation

To solve the differential equation

$$x^2\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}-2x\frac{\mathrm{d}y}{\mathrm{d}x}+2y=x^2+2$$

Making use of the variable transformation, we obtain the transformed equation

$$\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{z}^2}-3\frac{\mathrm{d}y}{\mathrm{d}z}+2y=e^{2z}+2$$

The solution is found to be

$$y=c_1{e}^z+c_2{e}^{2z}+z{e}^{2z}+1$$

or, returning to the variable $x$

$$y=c_{1}x+c_2{x}^2+x^2\log x+1$$

Particular Solutions by Variation of Parameters

Suppose

$$Ly=\frac{{\mathrm{d}}^{n}y}{\mathrm{d}{x}^n}+a_1(x)\frac{{\mathrm{d}}^{n-1}y}{\mathrm{d}{x}^{n-1}}+\cdots +a_{n-1}(x)\frac{\mathrm{d}y}{\mathrm{d}x}+a_n(x)y=h(x).$$

has the homogeneous solution

$$y_H=\sum _{k=1}^n{c}_k{u}_k(x).$$

where the $u$'s are $n$ linearly independent homogeneous solutions and the $c$’s are $n$ arbitrary constants or "parameters."

We assume that the particular solution of the above ODE is of the following form

$$y_P=\sum _{k=1}^n{C}_k{u}_k(x).$$

Given $y_P$, we obtain:

$$\frac{\mathrm{d}}{\mathrm{d}x}{y}_P=\sum _{k=1}^n{C}_k{u}_k^{\prime }+\sum _{k=1}^n{C}_k^{\prime }{u}_k.$$

In order to simplify this expression, we require

$$\sum _{k=1}^n{C}_k^{\prime }{u}_k=0.$$

There then follows

$$\frac{{\mathrm{d}}^2}{\mathrm{d}{x}^2}{y}_P=\sum _{k=1}^n{C}_k{u}_k^{″}+\sum _{k=1}^n{C}_k^{\prime }{u}_k^{\prime }.$$

As the second condition, we require

$$\sum _{k=1}^n{C}_k^{\prime }{u}_k^{\prime }=0.$$

Proceeding in this way through the $(n-1)$th derivative, we have our $(n-1)$th requirement:

$$\sum _{k=1}^n{C}_k^{\prime }{u}_k^{(n-2)}=0.$$

And the $(n-1)$th derivative is

$$\frac{{\mathrm{d}}^{n-1}}{\mathrm{d}{x}^{n-1}}{y}_P=\sum _{k=1}^n{C}_k{u}_k^{(n-1)}.$$

The expression for the $n$th derivative is then

$$\frac{{\mathrm{d}}^n}{\mathrm{d}{x}^n}{y}_P=\sum _{k=1}^n{C}_k^{\prime }{u}_k^{(n-1)}+\sum _{k=1}^n{C}_k{u}_k^{(n)}.$$

Introducing the expressions for $y_P$ and its derivatives into the general linear differential equation:

$$\begin{array}{rl}L{y}_P=\sum _{k=1}^n{C}_k{u}_k^{(n)}& +a_1(x)\sum _{k=1}^n{C}_k{u}_k^{(n-1)}+\cdots \\ & +a_{n-1}(x)\sum _{k=1}^n{C}_k{u}_k^{\prime }+a_n(x)\sum _{k=1}^n{C}_k{u}_k+\sum _{k=1}^n{C}_k^{\prime }{u}_k^{(n-1)}=h(x)\end{array}$$

Combining the first summations, we obtain

$$\begin{array}{rl}& \sum _{k=1}^n{C}_k[\frac{{\mathrm{d}}^n{u}_k}{\mathrm{d}{x}^n}+a_1(x)\frac{{\mathrm{d}}^{n-1}{u}_k}{\mathrm{d}{x}^{n-1}}+\cdots \\ & +a_{n-1}(x)\frac{\mathrm{d}{u}_k}{\mathrm{d}x}+a_n(x)u_k]+\sum _{k=1}^n{C}_k^{\prime }{u}_k^{(n-1)}=h(x)\end{array}$$

Therefore

$$\sum _{k=1}^n{C}_k^{\prime }{u}_k^{(n-1)}=h(x)$$

In summary, the $n$ conditions imposed on the $n$ unknown functions can be written in the expanded form

$$\{\begin{array}{l}{C}_1^{\prime }(x)u_1(x)+C_2^{\prime }(x)u_2(x)+\cdots +C_n^{\prime }(x)u_n(x)=0\\ C_1^{\prime }(x)u_1^{\prime }(x)+C_2^{\prime }(x)u_2^{\prime }(x)+\cdots +C_n^{\prime }(x)u_n^{\prime }(x)=0\\ ⋮\\ C_1^{\prime }(x)u_1^{(n-2)}(x)+C_2^{\prime }(x)u_2^{(n-2)}(x)+\cdots \cdots +C_n^{\prime }(x)u_n^{(n-2)}(x)=0\\ C_1^{\prime }(x)u_1^{(n-1)}(x)+C_2^{\prime }(x)u_2^{(n-1)}(x)+\cdots +C_n^{\prime }(x)u_n^{(n-1)}(x)=h(x)\end{array}$$

If this set of equations is solved for $C_1^{\prime },C_2^{\prime },\dots ,C_n^{\prime }$ by Cramer's rule, the common-denominator determinant is seen to be the Wronskian of $u_1,u_2$, $\dots ,u_n$.

Example6: the general solution to the second order linear differential equation

$$\frac{d^{2}y}{d{x}^2}+a_1(x)\frac{dy}{dx}+a_2(x)y=h(x),$$

there follows

$$y=C_1(x)u_1(x)+C_2(x)u_2(x)$$

where

$$C_1^{\prime }=\frac{\left|\begin{array}{ll}0& u_2\\ h& u_2^{\prime }\end{array}\right|}{\left|\begin{array}{ll}{u}_1& u_2\\ u_1^{\prime }& u_2^{\prime }\end{array}\right|}=-\frac{h(x)u_2(x)}{W[u_1(x),u_2(x)]}$$

and, similarly,

$$C_2^{\prime }=\frac{h(x)u_1(x)}{W[u_1(x),u_2(x)]}.$$

Thus we can write

$$\begin{array}{rl}& C_1=-{\int }^x\frac{h(x)u_2(x)}{W[u_1(x),u_2(x)]}\mathrm{d}x+c_1,\\ & C_2={\int }^x\frac{h(x)u_1(x)}{W[u_1(x),u_2(x)]}\mathrm{d}x+c_2\end{array}$$

Example7: 2nd order linear differential equation

$$\frac{d^{2}y}{d{x}^2}+y=f(x)$$

two linearly independent homogeneous solutions are $u_1=\cos x,u_2=\sin x$. The Wronskian is

$$W(\cos x,\sin x)=\left|\begin{array}{rr}\cos x& \sin x\\ -\sin x& \cos x\end{array}\right|={\sin }^{2}x+{\cos }^{2}x=1.$$

Therefore

$$y=-\cos x[{\int }^{x}f(x)\sin x\mathrm{d}x+c_1]+\sin x[{\int }^{x}f(x)\cos x\mathrm{d}x+c_2]$$

or

$$y={\int }^{x}f(\xi )[\cos \xi \sin x-\sin \xi \cos x]\mathrm{d}\xi +c_1\cos x+c_2\sin x$$

or

$$y={\int }^{x}f(\xi )\sin (x-\xi )\mathrm{d}\xi +c_1\cos x+c_2\sin x$$

Example8: 3rd order linear differential equation

$$\frac{d^{3}y}{d{x}^3}-3\frac{d^{2}y}{d{x}^2}+2\frac{dy}{dx}=f(x),$$

we may take $u_1=1,u_2=e^x,u_3=e^{2x}$. Equations (52) then become

$$\begin{array}{rlrl}{C}_1^{\prime }+& C_2^{\prime }{e}^x+C_3^{\prime }{e}^{2x}& & =0\\ & C_2^{\prime }{e}^x+2C_3^{\prime }{e}^{2x}& & =0\\ & C_2^{\prime }{e}^x+4C_3^{\prime }{e}^{2x}& & =f(x)\end{array}$$

Solving the three simultaneous equations, we obtain

$$C_1^{\prime }=\frac{1}{2}f(x),C_2^{\prime }=-e^{-x}f(x),C_3^{\prime }=\frac{1}{2}{e}^{-2x}f(x)$$

The solution of the differential equation is then

$$\begin{array}{rl}y=& 1[\frac{1}{2}{\int }^{x}f(\xi )\mathrm{d}\xi +c_1]+e^x[-{\int }^x{e}^{-\xi }f(\xi )\mathrm{d}\xi +c_2]\\ & +e^{2x}[\frac{1}{2}{\int }^x{e}^{-2\xi }f(\xi )\mathrm{d}\xi +c_3]\end{array}$$

or, equivalently,

$$y=\frac{1}{2}{\int }^{x}f(\xi )[1-2e^{x-\xi }+e^{2(x-\xi )}]\mathrm{d}\xi +c_1+c_2{e}^x+c_3{e}^{2x}$$

Reduction of order

One of the important properties of linear differential equations is the fact that:

if one homogeneous solution of an equation of order $n$ is known, a new linear differential equation of order $n-1$,determining the remainder of the solution, can be obtained.

This procedure is in a sense analogous to the reduction of the degree of an algebraic equation when one solution is known.

Suppose that one homogeneous solution $u_1(x)$ is known.

1. We next write $y=v(x)u_1(x)$ and attempt to determine the function $v(x)$.
2. Substituting $v{u}_1$ for $y$ in the left-hand side of the differential equation, we obtain a new linear differential equation of order $n$ to determine $v$. But since $y=c{u}_1(x)$ is a homogeneous solution of the original equation, $v=c$ must be a homogeneous solution of the new equation.
3. Hence the new equation must lack the term of zero order in $v$; that is, the coefficient of $v$ must be zero in equations. Thus the new equation is of order $n-1$ in the variable $\mathrm{d}v/\mathrm{d}x$

Example9: general 2nd-order linear equation

$$\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}+a_1(x)\frac{\mathrm{d}y}{\mathrm{d}x}+a_2(x)y=h(x)$$

Assuming $y=v(x)u_1(x)$, then introduce into $2^{\text{nd }}$-order linear equation

$$v^{\prime \prime }{u}_1+2v^{\prime }{u}_1^{\prime }+a_1{v}^{\prime }{u}_1+v\left(\underset{=0}{\underbrace{u_1^{\prime \prime }+a_1{u}_1^{\prime }+a_2{u}_1}}\right)=h.$$

since $u_1$ is a homogeneous solution, the expression in parentheses vanishes and the differential equation determining $v$ becomes

$$v^{\prime \prime }{u}_1+2v^{\prime }{u}_1^{\prime }+a_1{v}^{\prime }{u}_1=h$$

or

$${\left(v^{\prime }\right)}^{\prime }+(2\frac{u_1^{\prime }}{u_1}+a_1)v^{\prime }=\frac{h}{u_1}.$$

This equation is of first order in $v^{\prime }$, with an integrating factor (See first order linear solution) in the form

$$e^{2\log u_1+\int a_1\mathrm{d}x}=p{u}_1^2,\text{ where }p=e^{\int a_1\mathrm{d}x}$$

Hence

$$\begin{array}{rl}& v^{\prime }=\frac{1}{p{u}_1^2}{\int }^{x}ph{u}_1\mathrm{d}x+\frac{c_1}{p{u}_1^2}\\ & v={\int }^x\frac{{\int }^{x}ph{u}_1\mathrm{d}x}{p{u}_1^2}\mathrm{d}x+c_1{\int }^x\frac{\mathrm{d}x}{p{u}_1^2}+c_2\end{array}$$

With $y={vu}_1(x)$,

$$y=u_1(x)\int \frac{{\int }^{x}ph{u}_1\mathrm{d}x}{p{u}_1^2}\mathrm{d}x+c_1{u}_1(x){\int }^x\frac{\mathrm{d}x}{p{u}_1^2}+c_2{u}_1(x)$$

Determination of Constants

The $n$ arbitrary constants present in the general solution of a linear differential equation of order $n$ are to be determined by $n$ suitably prescribed supplementary conditions

Frequently these conditions consist of the requirement that the function and its first $n-1$ derivatives take on prescribed values at a given point $x=a$:

$$y(a)=y_0,y^{\prime }(a)=y_0^{\prime },\dots ,y^{(n-1)}(a)=y_0^{(n-1)}$$

When such conditions are prescribed, the problem is known as an initial-value problem. In this case it can be shown that if the point $x=a$ is included in an interval where the coefficients $a_1(x),\cdots ,a_n(x)$ and the right-hand side $h(x)$ of the differential equation, in standard form, are continuous, there exists a unique solution satisfying above Equations. Here, if the complete solution is written in the form

$$\begin{array}{rl}& y=\sum _{k=1}^n{c}_k{u}_k(x)+y_p(x)\\ & ⟹\sum _{k=1}^n{c}_k{u}_k^{(m)}(a)=y_0^{(m)}-y_P^{(m)}(a)(m=0,1,2,\dots ,n-1)\\ & ⟹\left(\begin{array}{cccc}{u}_1^{(0)}& u_2^{(0)}& \cdots & u_n^{(0)}\\ u_1^{(1)}& u_2^{(1)}& \cdots & u_n^{(1)}\\ ⋮& ⋮& \ddots & ⋮\\ u_1^{(n-1)}& u_2^{(n-1)}& \cdots & u_n^{(n-1)}\end{array}\right)\left(\begin{array}{c}{c}_1\\ c_2\\ ⋮\\ c_n\end{array}\right)=\left(\begin{array}{c}{y}_0^{(0)}-y_p^{(0)}(a)\\ y_0^{(1)}-y_p^{(1)}(a)\\ ⋮\\ y_0^{(n-1)}-y_p^{(n-1)}(a)\end{array}\right)\end{array}$$

determinant of the coefficients of the constants $c_k$ is the value of the Wronskian of the linearly independent solutions $u_k$ at the point $x=a$

Special solvable types of nonlinear equations

1. Separable equations

Q:

$$(1+x^2)\mathrm{d}y+(1+y^2)\mathrm{d}x=0$$

A:

$$\begin{array}{rl}\frac{\mathrm{d}x}{1+x^2}+\frac{\mathrm{d}y}{1+y^2}& =0\\ {\tan }^{-1}x+{\tan }^{-1}y& ={\tan }^{-1}c\\ x+y& =c(1-xy)\end{array}$$

Q:

$${\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)}^2-4y+4=0$$

A: yields two separable equations when solved algebraically for $\mathrm{d}y/\mathrm{d}x$

$$\frac{\mathrm{d}y}{\mathrm{d}x}=\pm 2\sqrt{y-1},$$

provided that the division by $\sqrt{y-1}$ is legitimate, and hence there follows $\pm \sqrt{y-1}=x-c$, or

$$y=1+(x-c{)}^2.$$

2. Exact first-order equations

A first-order equation, written in the form

$$P(x,y)\mathrm{d}x+Q(x,y)\mathrm{d}y=0.$$

where $P$ and $Q$ are assumed to have continuous first partial derivatives, is said to be exact when $P$ and $Q$ satisfy the condition

$$\frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x}.$$

In this case, and in this case only, there exists a function $u(x,y)$ such that,

$$\begin{array}{rl}& \mathrm{d}u=P\mathrm{d}x+Q\mathrm{d}y=0\\ ⟹& u(x,y)=c.\end{array}$$

since

$$\frac{\partial u}{\partial x}=P,\frac{\partial u}{\partial y}=Q,$$

the necessity follows from the fact that

$$\frac{\partial }{\partial x}\left(\frac{\partial u}{\partial y}\right)=\frac{\partial }{\partial y}\left(\frac{\partial u}{\partial x}\right)$$

In order to obtain a function u satisfying the two relations, for example, we start with the first relation $\partial u/\partial x=P$

$$u(x,y)={\int }^{x}P(x,y)dx+f(y)$$

$f(y)$ is the added constant of integration to be determined by second relation $\partial u/\partial y=Q$

$${\int }^x\frac{\partial P}{\partial y}dx+f^{\prime }(y)=Q$$

hence

$$f^{\prime }(y)=Q-{\int }^x\frac{\partial P}{\partial y}dx$$

the right-hand member is indeed only a function of $y$, so that $f(y)$ can be determined (with an irrelevant arbitrary additive constant) by direct integration $f^{\prime }(y)$ has no relation with $x$, because

$$\frac{\partial }{\partial x}(Q-{\int }^x\frac{\partial P}{\partial y}dx)=\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=0$$

Example 10

$$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{1+y^2+3x^{2}y}{1-2xy-x^3}$$

can be written in the form

$$(3x^{2}y+y^2+1)\mathrm{d}x+(x^3+2xy-1)\mathrm{d}y=0$$

and the condition of exactness is satisfied. From the relation

$$\frac{\partial u}{\partial x}=3x^{2}y+y^2+1$$

there follows $u=x^{3}y+x{y}^2+x+f(y)$. The relation

$$\frac{\partial u}{\partial y}=x^3+2xy-1$$

then gives $x^3+2xy+f^{\prime }(y)=x^3+2xy-1$ or $f^{\prime }(y)=-1$, from which $f(y)=-y$, apart from an irrelevant arbitrary additive constant. Hence the required solution $u=c$ is

$$x^{3}y+x{y}^2+x-y=c.$$

3. Homogeneous first-order equations

A function $f(x,y)$ is said to be homogeneous of degree $n$ if there exists a constant $n$ such that, for every number $\lambda$

$$f(\lambda x,\lambda y)={\lambda }^{n}f(x,y).$$

The first-order differential equation, $P(x,y)\mathrm{d}x+Q(x,y)\mathrm{d}y=0$, is said to be homogeneous if $P$ and $Q$ are both homogeneous of degree $n$ , for some constant $n$ . Differential equation becomes separable upon the change of variables $y=vx⟹\mathrm{d}y=v\mathrm{d}x+x\mathrm{d}v$,

$$\begin{array}{rl}& x^{n}P(1,v)\mathrm{d}x+x^{n}Q(1,v)(v\mathrm{d}x+x\mathrm{d}v)=0\\ \text{ or }& [P(1,v)+vQ(1,v)]\mathrm{d}x+xQ(1,v)\mathrm{d}v=0\end{array}$$

Example 11

$$(3y^2-x^2)\mathrm{d}x=2xy\mathrm{d}y$$

is homogeneous (with $n=2$ ) and, with $y=vx$, it becomes

$$(3v^2{x}^2-x^2)\mathrm{d}x=2v{x}^2(v\mathrm{d}x+x\mathrm{d}v)$$

$$⟹(v^2-1)\mathrm{d}x+2xv\mathrm{d}v=0$$

from which there follows

$$\frac{2v\mathrm{d}v}{v^2-1}=\frac{\mathrm{d}x}{x}(x\ne 0,v^2\ne 1)$$

and hence

$$\log |v^2-1|=\log |x|+\log |c|=\log |cx|$$

or

$$v^2-1=cx$$

or

$$y^2-x^2=c{x}^3.$$

4. Second-order equations lacking one variable

The general equation of second order is of the form

$$F(x,y,\frac{\mathrm{d}y}{\mathrm{d}x},\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2})=0.$$

Assume

$$\frac{\mathrm{d}y}{\mathrm{d}x}=p⟹\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}=\frac{\mathrm{d}p}{\mathrm{d}x}=\frac{\mathrm{d}p}{\mathrm{d}y}p$$

1. Lacking $y$ or else:$$\{\begin{array}{l}F(x,y,p,\frac{\mathrm{d}p}{\mathrm{d}x})=0\\ \frac{\mathrm{d}y}{\mathrm{d}x}=p\end{array}.$$
2. Lacking $x$ or else:$$\{\begin{array}{l}F(x,y,p,p\frac{\mathrm{d}p}{\mathrm{d}y})=0\\ \frac{\mathrm{d}y}{\mathrm{d}x}=p\end{array}.$$

Example 12

$$\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}=x{\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)}^3$$

lacks the variable $y$. With $\frac{\mathrm{d}y}{\mathrm{d}x}=p$ and $\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}=\frac{\mathrm{d}p}{\mathrm{d}x}$, there follows

$$\frac{\mathrm{d}p}{\mathrm{d}x}=x{p}^3$$

which separates to give

Hence

$$\begin{array}{rl}p& =\frac{\pm 1}{\sqrt{c_1^2-x^2}}\\ \mathrm{d}y& =\pm \frac{\mathrm{d}x}{\sqrt{c_1^2-x^2}}\end{array}$$

from which there follows

$$y=\pm {\sin }^{-1}\frac{x}{c_1}+c_2\text{ or }x=c_1^{\prime }\sin (y-c_2),$$

where $c_1^{\prime }=\pm c_1$.