Reduction of order

December 12, 2021

Reduction of order[edit]

Differential equations can usually be solved more easily if the order of the equation can be reduced.

Reduction to a first-order system[edit]

Any explicit differential equation of order n,

F\left(x,y,y',y'',\ \ldots ,\ y^{(n-1)}\right)=y^{(n)}

can be written as a system of n first-order differential equations by defining a new family of unknown functions

y_{i}=y^{(i-1)}.\!

for i = 1, 2,..., n. The n-dimensional system of first-order coupled differential equations is then

{\begin{array}{rcl}y_{1}'&=&y_{2}\\y_{2}'&=&y_{3}\\&\vdots &\\y_{n-1}'&=&y_{n}\\y_{n}'&=&F(x,y_{1},\ldots ,y_{n}).\end{array}}

more compactly in vector notation:

\mathbf {y} '=\mathbf {F} (x,\mathbf {y} )

where

\mathbf {y} =(y_{1},\ldots ,y_{n}),\quad \mathbf {F} (x,y_{1},\ldots ,y_{n})=(y_{2},\ldots ,y_{n},F(x,y_{1},\ldots ,y_{n})).

Summary of exact solutions[edit]

Some differential equations have solutions that can be written in an exact and closed form. Several important classes are given here.

In the table below, $P (x)$ , $Q (x)$ , $P (y)$ , $Q (y)$ , and $M (x, y)$ , $N (x, y)$ are any integrable functions of $x$ , $y$ , and $b$ and $c$ are real given constants, and $C 1, C 2, ...$ are arbitrary constants (complex in general). The differential equations are in their equivalent and alternative forms that lead to the solution through integration.

In the integral solutions, λ and ε are dummy variables of integration (the continuum analogues of indices in summation), and the notation $\int x F (λ) dλ$ just means to integrate $F (λ)$ with respect to $λ$ , then after the integration substitute $λ = x$ , without adding constants (explicitly stated).

Separable equations[edit]

Differential equation	Solution method	General solution
First-order, separable in x and y (general case, see below for special cases)^[27] ${\begin{aligned}P_{1}(x)Q_{1}(y)+P_{2}(x)Q_{2}(y)\,{\frac {dy}{dx}}&=0\\P_{1}(x)Q_{1}(y)\,dx+P_{2}(x)Q_{2}(y)\,dy&=0\end{aligned}}$	Separation of variables (divide by P₂Q₁).	$\int ^{x}{\frac {P_{1}(\lambda )}{P_{2}(\lambda )}}\,d\lambda +\int ^{y}{\frac {Q_{2}(\lambda )}{Q_{1}(\lambda )}}\,d\lambda =C$
First-order, separable in x^[25] ${\begin{aligned}{\frac {dy}{dx}}&=F(x)\\dy&=F(x)\,dx\end{aligned}}$	Direct integration.	$y=\int ^{x}F(\lambda )\,d\lambda +C$
First-order, autonomous, separable in y^[25] ${\begin{aligned}{\frac {dy}{dx}}&=F(y)\\dy&=F(y)\,dx\end{aligned}}$	Separation of variables (divide by F).	$x=\int ^{y}{\frac {d\lambda }{F(\lambda )}}+C$
First-order, separable in x and y^[25] ${\begin{aligned}P(y){\frac {dy}{dx}}+Q(x)&=0\\P(y)\,dy+Q(x)\,dx&=0\end{aligned}}$	Integrate throughout.	$\int ^{y}P(\lambda )\,d\lambda +\int ^{x}Q(\lambda )\,d\lambda =C$

General first-order equations[edit]

Differential equation	Solution method	General solution
First-order, homogeneous^[25] ${\frac {dy}{dx}}=F\left({\frac {y}{x}}\right)$	Set y = ux, then solve by separation of variables in u and x.	$\ln(Cx)=\int ^{y/x}{\frac {d\lambda }{F(\lambda )-\lambda }}$
First-order, separable^[27] ${\begin{aligned}yM(xy)+xN(xy)\,{\frac {dy}{dx}}&=0\\yM(xy)\,dx+xN(xy)\,dy&=0\end{aligned}}$	Separation of variables (divide by xy).	$\ln(Cx)=\int ^{xy}{\frac {N(\lambda )\,d\lambda }{\lambda [N(\lambda )-M(\lambda )]}}$ If N = M, the solution is xy = C.
Exact differential, first-order^[25] ${\begin{aligned}M(x,y){\frac {dy}{dx}}+N(x,y)&=0\\M(x,y)\,dy+N(x,y)\,dx&=0\end{aligned}}$ where ${\frac {\partial M}{\partial y}}={\frac {\partial N}{\partial x}}$	Integrate throughout.	${\begin{aligned}F(x,y)&=\int ^{x}M(\lambda ,y)\,d\lambda +\int ^{y}Y(\lambda )\,d\lambda \\&=\int ^{y}N(x,\lambda )\,d\lambda +\int ^{x}X(\lambda )\,d\lambda =C\end{aligned}}$ where $Y(y)=N(x,y)-{\frac {\partial }{\partial y}}\int ^{x}M(\lambda ,y)\,d\lambda$ and $X(x)=M(x,y)-{\frac {\partial }{\partial x}}\int ^{y}N(x,\lambda )\,d\lambda$
Inexact differential, first-order^[25] ${\begin{aligned}M(x,y){\frac {dy}{dx}}+N(x,y)&=0\\M(x,y)\,dy+N(x,y)\,dx&=0\end{aligned}}$ where ${\frac {\partial M}{\partial x}}\neq {\frac {\partial N}{\partial y}}$	Integration factor $μ (x, y)$ satisfying ${\frac {\partial (\mu M)}{\partial y}}={\frac {\partial (\mu N)}{\partial x}}$	If $μ (x, y)$ can be found in a suitable way, then ${\begin{aligned}F(x,y)=&\int ^{x}\mu (\lambda ,y)M(\lambda ,y)\,d\lambda +\int ^{y}Y(\lambda )\,d\lambda \\=&\int ^{y}\mu (x,\lambda )N(x,\lambda )\,d\lambda +\int ^{x}X(\lambda )\,d\lambda =C\end{aligned}}$ where $Y(y)=N(x,y)-{\frac {\partial }{\partial y}}\int ^{x}\mu (\lambda ,y)M(\lambda ,y)\,d\lambda$ and $X(x)=M(x,y)-{\frac {\partial }{\partial x}}\int ^{y}\mu (x,\lambda )N(x,\lambda )\,d\lambda$

General second-order equations[edit]

Differential equation	Solution method	General solution
Second-order, autonomous^[28] ${\frac {d^{2}y}{dx^{2}}}=F(y)$	Multiply both sides of equation by $2 dy / dx$ , substitute $2{\frac {dy}{dx}}{\frac {d^{2}y}{dx^{2}}}={\frac {d}{dx}}\left({\frac {dy}{dx}}\right)^{2}$ , then integrate twice.	$x=\pm \int ^{y}{\frac {d\lambda }{\sqrt {2\int ^{\lambda }F(\varepsilon )\,d\varepsilon +C_{1}}}}+C_{2}$

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Dr. Balamurugan