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Ordinary Differential Equation--First-Order

Given a first-order Ordinary Differential Equation

{dy\over dx} = F(x,y),
\end{displaymath} (1)

if $F(x,y)$ can be expressed using Separation of Variables as
\end{displaymath} (2)

then the equation can be expressed as
{dy\over Y(y)} = X(x)\,dx
\end{displaymath} (3)

and the equation can be solved by integrating both sides to obtain
\int{dy\over Y(y)} = \int X(x)\,dx.
\end{displaymath} (4)

Any first-order ODE of the form

{dy\over dx} + p(x)y = q(x)
\end{displaymath} (5)

can be solved by finding an Integrating Factor $\mu=\mu(x)$ such that
{d\over dx} (\mu y) = \mu {dy\over dx}+y{d\mu\over dx} = \mu q(x).
\end{displaymath} (6)

Dividing through by $\mu y$ yields
{1\over y} {dy\over dx} + {1\over\mu}{d\mu\over dx} = {q(x)\over y}.
\end{displaymath} (7)

However, this condition enables us to explicitly determine the appropriate $\mu$ for arbitrary $p$ and $q$. To accomplish this, take
p(x) = {1\over\mu} {d\mu\over dx}
\end{displaymath} (8)

in the above equation, from which we recover the original equation (5), as required, in the form
{1\over y}{dy\over dx}+p(x)={q(x)\over y}.
\end{displaymath} (9)

But we can integrate both sides of (8) to obtain
\int p(x)\,dx = \int {d\mu\over \mu} = \ln \mu+c
\end{displaymath} (10)

\mu = e^{\int p(x)\,dx}.
\end{displaymath} (11)

Now integrating both sides of (6) gives
\mu y=\int \mu q(x)\,dx+c
\end{displaymath} (12)

(with $\mu$ now a known function), which can be solved for $y$ to obtain
y = {\int \mu q(x)\,dx+c\over\mu} = {\int e^{\int^x p(x')\,dx'} q(x)\,dx+c\over e^{\int^x p(x')\,dx'}},
\end{displaymath} (13)

where $c$ is an arbitrary constant of integration.

Given an $n$th-order linear ODE with constant Coefficients

{d^ny\over dx^n} + a_{n-1} {d^{n-1}y\over dx^{n-1}} +\ldots + a_1 {dy\over dx} + a_0y = Q(x),
\end{displaymath} (14)

first solve the characteristic equation obtained by writing
y\equiv e^{rx}
\end{displaymath} (15)

and setting $Q(x)= 0$ to obtain the $n$ Complex Roots.
r^ne^{rx}+a_{n-1}r^{n-1}e^{rx}+\ldots+a_1 r e^{rx} + a_0 e^{rx}=0
\end{displaymath} (16)

r^n+a_{n-1}r^{n-1}+\ldots+a_1 r + a_0=0.
\end{displaymath} (17)

Factoring gives the Roots $r_i$,
\end{displaymath} (18)

For a nonrepeated Real Root $r$, the corresponding solution is
\end{displaymath} (19)

If a Real Root $r$ is repeated $k$ times, the solutions are degenerate and the linearly independent solutions are
y=e^{rx}, y=x e^{rx}, \ldots, y=x^{k-1}e^{rx}.
\end{displaymath} (20)

Complex Roots always come in Complex Conjugate pairs, $r_\pm = a\pm i b$. For nonrepeated Complex Roots, the solutions are
y=e^{ax}\cos(bx), y=e^{ax}\sin(bx).
\end{displaymath} (21)

If the Complex Roots are repeated $k$ times, the linearly independent solutions are

y=e^{ax}\cos(bx), y=e^{ax}\sin(bx), \ldots, y=x^{k-1}e^{ax}\cos(bx), y=x^{k-1}e^{ax}\sin(bx).
\end{displaymath} (22)

Linearly combining solutions of the appropriate types with arbitrary multiplicative constants then gives the complete solution. If initial conditions are specified, the constants can be explicitly determined. For example, consider the sixth-order linear ODE

(\tilde D-1)(\tilde D-2)^3(\tilde D^2+\tilde D+1)y=0,
\end{displaymath} (23)

which has the characteristic equation
\end{displaymath} (24)

The roots are 1, 2 (three times), and $(-1\pm\sqrt{3}\,i)/2$, so the solution is

...}}\sqrt{3}\,x)+Fe^{-x}\sin({\textstyle{1\over 2}}\sqrt{3}\,x).
\end{displaymath} (25)

If the original equation is nonhomogeneous ($Q(x)\not = 0$), now find the particular solution $y^*$ by the method of Variation of Parameters. The general solution is then

y(x) = \sum_{i=1}^n c_iy_i(x)+y^*(x),
\end{displaymath} (26)

where the solutions to the linear equations are $y_1(x)$, $y_2(x)$, ..., $y_n(x)$, and $y^*(x)$ is the particular solution.

See also Integrating Factor, Ordinary Differential Equation--First-Order Exact, Separation of Variables, Variation of Parameters


Arfken, G. Mathematical Methods for Physicists, 3rd ed. Orlando, FL: Academic Press, pp. 440-445, 1985.

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© 1996-9 Eric W. Weisstein