Such a curve is called sinusoidal; an example is shown in Figure 3. FLAP material © copyright 1996 Open University, MATH 6.3: Solving second order differential equations, Methods of solution for various second–order differential equations, Classifying second–order differential equations, A worked example: damped, driven harmonic motion. Let the speed of the particle be v0 when it is at position p (at a distance no from O), At t = 0 the particle at P(moving towards the right), At t = t the particle is at Q(at a distance x from O), The restoring force F→\overrightarrow{F}F at Q is given by, ⇒ F→=−Kx→\overrightarrow{F}=-K\overrightarrow{x}F=−Kx K – is positive constant, ⇒ F→=ma→\overrightarrow{F}=m\overrightarrow{a}F=ma a→\overrightarrow{a}a- acceleration at Q, ⇒ ma→=−Kx→m\overrightarrow{a}=-K\overrightarrow{x}ma=−Kx, ⇒ a→=−(Km)x→\overrightarrow{a}=-\left( \frac{K}{m} \right)\overrightarrow{x}a=−(mK​)x, Put, Km=ω2\frac{K}{m}={{\omega }^{2}}mK​=ω2, ⇒ a→=−(Km)m→=−ω2x→\overrightarrow{a}=-\left( \frac{K}{m} \right)\overrightarrow{m}=-{{\omega }^{2}}\overrightarrow{x}a=−(mK​)m=−ω2x Since, [a→=d2xdt2]\left[ \overrightarrow{a}=\frac{{{d}^{2}}x}{d{{t}^{2}}} \right][a=dt2d2x​] Equations which can be directly integrated often arise when considering objects that are moving along a line. So we have found a particular solution; it is y = t/5 + 6/25. and the particular solution if y = 6 and dy/dx = 0 at x = 0. This is a linear ordinary differential equation with constant coefficients. where a, b, c are constants and a ≠ 0; it is equations of this type that will be discussed in this module. However, in some of the examples, drawn from specific problems in physics, we will use the notation for the variables that is appropriate to the situation. Any solution of this equation must be a function such that when we differentiate it twice, we recover the function itself, multiplied by −ω02. You can probably guess what will happen if we choose a value for b somewhere between these two extremes. is the sum of two decreasing exponentials, and as t becomes very large and positive, y tends to zero. Remember, this range of ϕ values is non–standard and will not be given automatically by your calculator. See the Glossary for details. ⇒v2A2+v2A2ω2=1\frac{{{v}^{2}}}{{{A}^{2}}}+\frac{{{v}^{2}}}{{{A}^{2}}{{\omega }^{2}}}=1A2v2​+A2ω2v2​=1 this is an equation of an ellipse. Comment If you managed to do the question this way, congratulations! The idea is that, guided by the form of f (t), we should assume a particular form for yp(t) which contains some undetermined constants, and simply substitute this into the differential equation. The differential equation for the Simple harmonic motion has the following solutions: These solutions can be verified by substituting this x values in the above differential equation for the linear simple harmonic motion. If it is slightly pushed from its mean position and released, it makes angular oscillations. Convert between a particular solution of the form y = B cos(ωt) + C sin(ωt) and one of the form y = A sin(ωt + ϕ) and vice versa. Let us suppose that we have somehow managed to find a particular solution to Equation 5; we will call it yp(t). If b = 0 the motion is dampingundamped and we have oscillations corresponding to SHM (as in Equation 15). Since π < ϕ < 3π/2, cos ϕ must be negative; so $\cos\varphi = \frac12\sqrt{3\os}$. Consider about a particle P describing uniform circular motion in anticlockwise direction in a circle of radius A and centre O. An equation of the form given in Equation 27 also arises in the theory of a.c. circuits. You will also need to be familiar with various trigonometric identities (although we will repeat them here); to be able to solve first–order differential equations by direct integration (which requires a fair knowledge of integration methods); to know how to check a proposed solution_mathematicalsolution to a differential equation by substitution (which requires good differentiation skills); and to be able to use initial conditions to obtain a particular solution from a general solution. for the case b2 < 4ac. As we mentioned at the end of Subsection 2.4, we can obtain the general solution to Equation 29 by adding these two solutions together: The general solution of Equation 29 in the case b2 > 4ac is, where$p_1 = \dfrac{-b+\sqrt{b^2-4ac}}{2a}\quad\text{and}\quad p_2 = \dfrac{-b-\sqrt{b^2-4ac}}{2a}$, $y = B\exp\left(\dfrac{-b+\sqrt{b^2-4ac}}{2a}\right) + C\exp\left(\dfrac{-b-\sqrt{b^2-4ac}}{2a}\right)$(34). It looks as though this solution will give complex values to y. ), The current in the circuit shown in Figure 1 satisfies the differential equation. Here, ω is the angular velocity of the particle. The second treatment does not require as much knowledge of complex numbers, we will just give you the solution of the differential equation and ask you to verify that what we say is correct. Based upon initial conditions of particle, general equations for simple harmonic motion can be expressed in different forms –, (1) In previous discussion, in equation (1) above, we have considered the following initial conditions –, Then general equation of SHM is x = A \cos \left ( \omega t + \phi _ { 0 } \right ) …… (1a), (2) If we consider the following initial conditions –, Then general equation of SHM becomes x = - A \cos \left ( \omega t + \phi _ { 0 } \right ) …… (1b), (3) If we consider the following initial conditions –, Then general equation of SHM becomes x = A \sin \left ( \omega t + \phi _ { 0 } \right ) …… (1c), (4) If we consider the following initial conditions –, Then general equation of SHM becomes x = - A \sin \left ( \omega t + \phi _ { 0 } \right ) …… (1d). Since f (t) is a polynomial of degree 1, we try a solution of the form y = Ht + K, where H and K are undetermined coefficients. State whether the following differential equations are linear with constant coefficients and, if so, whether they are homogeneous or inhomogeneous: (a) $\dfrac{d^2y}{dx^2} = x-3y$,   (b) $x\dfrac{d^2y}{dx^2} + 2\dfrac{dy}{dx}+ y = 0$,   (c) $\dfrac{d^2x}{dt^2} - 5\dfrac{dx}{dt} = 6x$. We will now return to the case of negative h in Equation 14. Ok. If the form we have chosen is correct, we will be able to determine the constants appearing in our trial solution from the requirement that we must obtain an identity on making the substitution. Solve an equation of the form $\dfrac{d^2y}{dt^2} - \lambda^2 y = 0$. Since a ≠ 0, we can divide both sides of the equation by a, and write h = c/a to give: You will see shortly that the form of solution of Equation 14 depends on whether the constant h is positive or negative. Here, $\omega_0= 1/\sqrt{LC\os}$ = 1.25 × 107 s−1. In equation (1), multiplying by 2( dx/dt),we get i. All that remains after a long time, therefore, is the particular solution, which represents the response of the oscillating mass to the applied force.