Initial value problem

In calculus, an initial value problem (IVP) is an ordinary differential equation together with an initial condition which specifies the value of the unknown function at a given point in the domain. Modeling a system in physics or other sciences frequently amounts to solving an initial value problem. In that context, the IVP is a differential equation which specifies how the system evolves with time plus the initial conditions of the problem.

In calculus, an initial value problem (IVP) is an ordinary differential equation together with an initial condition which specifies the value of the unknown function at a given point in the domain. Modeling a system in physics or other sciences frequently amounts to solving an initial value problem. In that context, the IVP is a differential equation which specifies how the system evolves with time plus the initial conditions of the problem. y'(t) = f(t, y(t)) with f\colon \Omega \subset \mathbb{R} \times \mathbb{R}^n \to \mathbb{R}^n where \Omega is an open set of \mathbb{R} \times \mathbb{R}^n , together with a point in the domain of f (t_0, y_0) \in \Omega, called the initial condition. A solution to an initial value problem is a function y that is a solution to the differential equation and satisfies y(t_0) = y_0. In higher dimensions, the differential equation is replaced with a family of equations y_i'(t)=f_i(t, y_1(t), y_2(t), \dotsc) , and y(t) is viewed as the vector (y_1(t), \dotsc, y_n(t)) , most commonly associated with the position in space. More generally, the unknown function y can take values on infinite dimensional spaces, such as Banach spaces or spaces of distributions. Initial value problems are extended to higher orders by treating the derivatives in the same way as an independent function, e.g. y(t)=f(t,y(t),y'(t)) . For this second-order differential equation, two initial conditions are needed, for example the numerical values of y(0) and y'(0) . ==Examples== A simple example is to solve y'(t) = 0.85 y(t) and y(0) = 19 . We are trying to find a formula for y(t) that satisfies these two equations. Rearrange the equation so that y is on the left hand side \frac{y'(t)}{y(t)} = 0.85 Now integrate both sides with respect to t (this introduces an unknown constant B ). \int \frac{y'(t)}{y(t)}\,dt = \int 0.85\,dt \ln |y(t)| = 0.85t + B Eliminate the logarithm with exponentiation on both sides | y(t) | = e^Be^{0.85t} Let C be a new unknown constant, C = \pm e^B , so y(t) = Ce^{0.85t} Now we need to find a value for C . Use y(0) = 19 as given at the start and substitute 0 for t and 19 for y 19 = C e^{0.85 \cdot 0} C = 19 this gives the final solution of y(t) = 19e^{0.85t} . Second example The solution of y'+3y=6t+5,\qquad y(0)=3 can be found to be y(t)=2e^{-3t}+2t+1. \, Indeed, \begin{align} y'+3y &= \tfrac{d}{dt} (2e^{-3t}+2t+1)+3(2e^{-3t}+2t+1) \\ &= (-6e^{-3t}+2)+(6e^{-3t}+6t+3) \\ &= 6t+5. \end{align} and y(0)=2+0+1=3 , so the given function satisfies the ODE and the initial condition. Third example The solution of y'=y^{\frac 2 3},\qquad y(0)=0 \int \frac{y'}{y^{\frac 2 3}}\,dt = \int y^{-\frac 2 3}\,dy =\int 1\,dt 3 (y(t))^{\frac 1 3}=t+B Applying initial conditions we get B=0 , hence the solution: y(t)= \frac {t^3} {27} . However, the following function is also a solution of the initial value problem: f(t) = \left\{ \begin{array}{lll} \frac{(t-t_1)^3}{27} & \text{if} & t \leq t_1 \\ 0 & \text{if} & t_1 \leq x \leq t_2 \\ \frac{(t-t_2)^3}{27} & \text{if} & t_2 \leq t \\ \end{array} \right. The function is differentiable everywhere and continuous, while satisfying the differential equation as well as the initial value problem. Thus, this is an example of such a problem with infinite number of solutions. ==Notes== ==See also== Boundary value problem Constant of integration Integral curve Norton's dome

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