The following techniques apply to one-dimensional autonomous differential equations. Any one-dimensional equation of order n is equivalent to an n-dimensional first-order system (as described in
reduction to a first-order system), but not necessarily vice versa.
First order The first-order autonomous equation \frac{dx}{dt} = f(x) is
separable, so it can be solved by rearranging it into the integral form t + C = \int \frac{dx}{f(x)}
Second order The second-order autonomous equation \frac{d^2x}{dt^2} = f(x, x') is more difficult, but it can be solved by introducing the new variable v = \frac{dx}{dt} and expressing the
second derivative of x via the
chain rule as \frac{d^2x}{dt^2} = \frac{dv}{dt} = \frac{dx}{dt}\frac{dv}{dx} = v\frac{dv}{dx} so that the original equation becomes v\frac{dv}{dx} = f(x, v) which is a first order equation containing no reference to the independent variable t. Solving provides v as a function of x. Then, recalling the definition of v: \frac{dx}{dt} = v(x) \quad \Rightarrow \quad t + C = \int \frac{d x}{v(x)} which is an implicit solution.
Special case: The special case where f is independent of x' \frac{d^2 x}{d t^2} = f(x) benefits from separate treatment. These types of equations are very common in
classical mechanics because they are always
Hamiltonian systems. The idea is to make use of the identity \frac{d x}{d t} = \left(\frac{d t}{d x}\right)^{-1} which follows from the
chain rule, barring any issues due to
division by zero. By inverting both sides of a first order autonomous system, one can immediately integrate with respect to x: \frac{d x}{d t} = f(x) \quad \Rightarrow \quad \frac{d t}{d x} = \frac{1}{f(x)} \quad \Rightarrow \quad t + C = \int \frac{dx}{f(x)} which is another way to view the separation of variables technique. The second derivative must be expressed as a derivative with respect to x instead of t: \begin{align} \frac{d^2 x}{d t^2} &= \frac{d}{d t}\left(\frac{d x}{d t}\right) = \frac{d}{d x}\left(\frac{d x}{d t}\right) \frac{d x}{d t} \\[4pt] &= \frac{d}{d x}\left(\left(\frac{d t}{d x}\right)^{-1}\right) \left(\frac{d t}{d x}\right)^{-1} \\[4pt] &= - \left(\frac{d t}{d x}\right)^{-2} \frac{d^2 t}{d x^2} \left(\frac{d t}{d x}\right)^{-1} = - \left(\frac{d t}{d x}\right)^{-3} \frac{d^2 t}{d x^2} \\[4pt] &= \frac{d}{d x}\left(\frac{1}{2}\left(\frac{d t}{d x}\right)^{-2}\right) \end{align} To reemphasize: what's been accomplished is that the second derivative with respect to t has been expressed as a derivative of x. The original second order equation can now be integrated: \begin{align} \frac{d^2 x}{d t^2} &= f(x) \\ \frac{d}{d x}\left(\frac{1}{2}\left(\frac{d t}{d x}\right)^{-2}\right) &= f(x) \\ \left(\frac{d t}{d x}\right)^{-2} &= 2 \int f(x) dx + C_1 \\ \frac{d t}{d x} &= \pm \frac{1}{\sqrt{2 \int f(x) dx + C_1}} \\ t + C_2 &= \pm \int \frac{dx}{\sqrt{2 \int f(x) dx + C_1}} \end{align} This is an implicit solution. The greatest potential problem is inability to simplify the integrals, which implies difficulty or impossibility in evaluating the integration constants.
Special case: Using the above approach, the technique can extend to the more general equation \frac{d^2 x}{d t^2} = \left(\frac{d x}{d t}\right)^n f(x) where n is some parameter not equal to two. This will work since the second derivative can be written in a form involving a power of x'. Rewriting the second derivative, rearranging, and expressing the left side as a derivative: \begin{align} &- \left(\frac{d t}{d x}\right)^{-3} \frac{d^2 t}{d x^2} = \left(\frac{d t}{d x}\right)^{-n} f(x) \\[4pt] &- \left(\frac{d t}{d x}\right)^{n - 3} \frac{d^2 t}{d x^2} = f(x) \\[4pt] &\frac{d}{d x}\left(\frac{1}{2 - n}\left(\frac{d t}{d x}\right)^{n - 2}\right) = f(x) \\[4pt] &\left(\frac{d t}{d x}\right)^{n - 2} = (2 - n) \int f(x) dx + C_1 \\[2pt] &t + C_2 = \int \left((2 - n) \int f(x) dx + C_1\right)^{\frac{1}{n - 2}} dx \end{align} The right will carry +/− if n is even. The treatment must be different if n = 2: \begin{align} - \left(\frac{d t}{d x}\right)^{-1} \frac{d^2 t}{d x^2} &= f(x) \\ -\frac{d}{d x}\left(\ln\left(\frac{d t}{d x}\right)\right) &= f(x) \\ \frac{d t}{d x} &= C_1 e^{-\int f(x) dx} \\ t + C_2 &= C_1 \int e^{-\int f(x) dx} dx \end{align}
Higher orders There is no analogous method for solving third- or higher-order autonomous equations. Such equations can only be solved exactly if they happen to have some other simplifying property, for instance
linearity or dependence of the right side of the equation on the dependent variable only (i.e., not its derivatives). This should not be surprising, considering that nonlinear autonomous systems in three dimensions can produce truly
chaotic behavior such as the
Lorenz attractor and the
Rössler attractor. Likewise, general non-autonomous equations of second order are unsolvable explicitly, since these can also be chaotic, as in a periodically forced pendulum.
Multivariate case In \mathbf x'(t) = A \mathbf x(t), where \mathbf x(t) is an n-dimensional column vector dependent on t. The solution is \mathbf x(t) = e^{A t} \mathbf c where \mathbf c is an n \times 1 constant vector.
Finite durations For non-linear autonomous ODEs it is possible under some conditions to develop solutions of finite duration, meaning here that from its own dynamics, the system will reach the value zero at an ending time and stay there in zero forever after. These finite-duration solutions cannot be
analytical functions on the whole real line, and because they will be non-
Lipschitz functions at the ending time, they don't stand uniqueness of solutions of Lipschitz differential equations. As example, the equation: :y'= -\text{sgn}(y)\sqrt,\,\,y(0)=1 Admits the finite duration solution: :y(x)=\frac{1}{4}\left(1-\frac{x}{2}+\left|1-\frac{x}{2}\right|\right)^2 == See also ==