SolveODE Command

SolveODE( <f'(x, y)> )

Attempts to find the exact solution of the first order ordinary differential equation (ODE) \frac{dy}{dx}(x)=f'(x, y(x)).

Example: SolveODE(2x / y) yields \sqrt{2} \sqrt{-c_{1}+x^{2}}, where c_{1} is a constant.

Note: c_{1} will be created as an auxiliary object with a corresponding slider.

SolveODE( <f'(x, y)>, <Point on f> )

Attempts to find the exact solution of the first order ODE \frac{dy}{dx}(x)=f'(x, y(x)) and use the solution which goes through the given point.

Example: SolveODE(y / x, (1, 2)) yields y = 2x.

SolveODE( <f'(x, y)>, <Start x>, <Start y>, <End x>, <Step> )

Solves first order ODE \frac{dy}{dx}=f'(x, y) numerically with given start point, end and step for x.

Example: SolveODE(-x*y, x(A), y(A), 5, 0.1) solves \frac{dy}{dx}=-xy using previously defined A as a starting point.

Note:

• Length( <Locus> ) allows you to find out how many points are in the computed locus.
• First( <Locus>, <Number> ) allows you to extract the points as a list.
• To find the "reverse" solution, just enter a negative value for End x, for example SolveODE(-x*y, x(A), y(A), -5, 0.1)

SolveODE( <y'>, <x'>, <Start x>, <Start y>, <End t>, <Step> )

Solves first order ODE \frac{dy}{dx}=\frac{f(x, y)}{g(x, y)} with given start point, maximal value of an internal parameter t and step for t. This version of the command may work where the first one fails e.g. when the solution curve has vertical points.

Example: SolveODE(-x, y, x(A), y(A), 5, 0.1) solves \frac{dy}{dx}=- \frac{x}{y} using previously defined A as a starting point.

Note: To find the "reverse" solution, just enter a negative value for End t, for example SolveODE(-x, y, x(A), y(A), -5, 0.1).

SolveODE( <b(x)>, <c(x)>, <f(x)>, <Start x>, <Start y>, <Start y'>, <End x>, <Step> )

Solves second order ODE y'' + b(x) y' + c(x) y = f(x).

Example: SolveODE(x^2, 2x, 2x^2 + x, x(A), y(A), 0, 5, 0.1) solves the second order ODE using previously defined A as a starting point.

Note: Always returns the result as locus. The algorithms are currently based on Runge-Kutta numeric methods.

CAS Syntax

SolveODE( <Equation> )

Attempts to find the exact solution of the first or second order ODE. For first and second derivative of y you can use y' and y'' respectively.

Example: SolveODE(y' = y / x) yields y = c₁ x.

SolveODE( <Equation>, <Point(s) on f> )

Attempts to find the exact solution of the given first or second order ODE which goes through the given point(s).

Example: SolveODE(y' = y / x, (1, 2)) yields y = 2x.

SolveODE( <Equation>, <Point(s) on f>, <Point(s) on f'> )

Attempts to find the exact solution of the given first or second order ODE and goes through the given point(s) on f and f' goes through the given point(s) on f' .

Example: SolveODE(y'' - 3y' + 2 = x, (2, 3), (1, 2)) yields y = \frac{-9 x^2 e^3 + 30 x e^3 - 32 {(e^3)}^2 + 138 e^3 + 32 e^{3 x} }{54 e^3} .

SolveODE( <Equation>, <Dependent Variable>, <Independent Variable>, <Point(s) on f> )

Attempts to find the exact solution of the given first or second order ODE which goes through the given point(s).

Example: SolveODE(v' = v / w, v, w, (1, 2)) yields v = 2w.

SolveODE( <Equation>, <Dependent Variable>, <Independent Variable>, <Point(s) on f>, <Point(s) on f'> )

Attempts to find the exact solution of the given first or second order ODE which goes through the given point(s) on f and f' goes through the given point(s) on f' .

Example: SolveODE(v' = v / w, v, w, (1, 2), (0, 2)) yields v = 2w.