At the solution of the differential equations the argument is not always obviously available x (or t time in tasks physical). Nevertheless is the simplified special case of a task of the differential equation that often promotes simplification of search of its integral.
Instruction
1. Consider the physical task leading to the differential equation in which there is no t argument. It is a task about fluctuations of the mathematical pendulum the mass of m suspended on thread by length of r located in the vertical plane. It is required to find the equation of the movement of the pendulum if in initial moment pendulum it was not mobile and rejected from equilibrium state on a corner α. Forces of resistance should be neglected (see fig. 1a).
2. Decision. The mathematical pendulum is the material point suspended on weightless and inextensible thread in a point of the Lake. The point is affected by two forces: gravity of G=mg and force of a tension of thread N. Both of these forces lie in the vertical plane. Therefore it is possible to apply the equation of rotary motion of a point around the horizontal axis passing through O. Uravneniye's point of rotary motion of a body to the solution of a task has the appearance given on fig. 1b. At the same time I — the moment of inertia of a material point; j is the thread angle of rotation together with a point counted from a vertical axis counterclockwise; M — the moment of forces applied to a material point.
3. Calculate these sizes. I=mr^2, (M=MG) + M (N). But M (N)=0 as the line of action of force passes through O.'s point of M (G)=-mgrsinj. The sign "-" designates that moment of force is directed opposite to the movement aside. Substitute the moment of inertia and moment of force in the equation of the movement and receive the equation displayed in fig. 1c. Reducing weight, there is a ratio (see fig. 1d). There is no t argument.
4. Generally the differential equation of n-go of an order which does not have x and allowed rather senior derivative y^(n)=f (y, y’, y’’..., y^(n-1)). For the second order it is y’ ’=f (y, y’). Solve its substitution y '=z=z (y). As for the difficult dz/dx= (dz/dy)(dy/dx) function, y’ ’=z’z. It will lead to the equation of the first order of z'z=f (y, z). Solve its any of ways known to you and receive z=φ (y, C1). dy/dx = φ (y, C1), ∫dy/φ (x, C1) =x+C2 is as a result received. Here C1 and C2 are any constants.
5. The concrete decision depends on a type of the arisen differential equation of the first order. So, if this equation with the divided variables, then it is solved directly. If this uniform is relative y the equation, then apply substitution of u(y) =z/y to the decision. For the linear equation of z=u (y) *v(y).