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One particular solution is x = 0, y = 0, z = 0. Two other solutions are x = 3, y = 6, z = 1, and x = 8, y = 9, z = 2. There is a unique plane in three-dimensional space which passes through the three points with these coordinates, and this plane is the set of all points whose coordinates are solutions of the equation.
If b = 0, the line is a vertical line (that is a line parallel to the y-axis) of equation =, which is not the graph of a function of x. Similarly, if a ≠ 0, the line is the graph of a function of y, and, if a = 0, one has a horizontal line of equation =.
If two solutions intersect each other, that is, they both go through the same point (x,y), then there is a failure of uniqueness for a first-order ordinary differential equation. Thus, there will be a failure of uniqueness if a solution of the first form intersects the second solution. The condition of intersection is : y s (x) = y c (x). We solve
Graph of x y = y x. The line and curve intersect at (e, e). ... (1, 0), instead of continuing on to infinity. The curved section can be written explicitly as
For example, the second-order equation y′′ = −y can be rewritten as two first-order equations: y′ = z and z′ = −y. In this section, we describe numerical methods for IVPs, and remark that boundary value problems (BVPs) require a different set of tools. In a BVP, one defines values, or components of the solution y at more than one ...
This equation is an equation only of y'' and y', meaning it is reducible to the general form described above and is, therefore, separable. Since it is a second-order separable equation, collect all x variables on one side and all y' variables on the other to get: (′) (′) =.
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The Picard–Lindelöf theorem guarantees a unique solution on some interval containing t 0 if f is continuous on a region containing t 0 and y 0 and satisfies the Lipschitz condition on the variable y. The proof of this theorem proceeds by reformulating the problem as an equivalent integral equation.