Initial and Boundary Value Problems Calculator

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Example

Created on 2024-06-20Asked by Aiden Carter (Solvelet student)

Solve the initial value problem

y'' + y = 0

with

y(0) = 1

and

y'(0) = 0

Solution

To solve the initial value problem

y'' + y = 0

with

y(0) = 1

and

y'(0) = 0

: First, solve the homogeneous differential equation

y'' + y = 0

. The characteristic equation is:

r^2 + 1 = 0

r^2 = -1

r = \pm i

The general solution to the differential equation is:

y(x) = C_1 \cos(x) + C_2 \sin(x)

Next, use the initial conditions to find

C_1

and

C_2

. Given

y(0) = 1

y(0) = C_1 \cos(0) + C_2 \sin(0) = C_1 \cdot 1 + C_2 \cdot 0 = C_1

C_1 = 1

Given

y'(0) = 0

y'(x) = -C_1 \sin(x) + C_2 \cos(x)

y'(0) = -C_1 \sin(0) + C_2 \cos(0) = -C_1 \cdot 0 + C_2 \cdot 1 = C_2

C_2 = 0

Therefore, the solution to the initial value problem is:

y(x) = \cos(x)

Solved on Solvelet with Basic AI Model

Some of the related questions asked by Elizabeth Hill on Solvelet

1. Solve the initial value problem for the first-order linear differential equation

y' + 2y = 4

with the initial condition

y(0) = 1

,2. Show that the vector space of polynomials of degree at most

2

with inner product

\langle f, g \rangle = \int_{0}^{1} f(x)g(x) \, dx

is an inner product space.,

DefinitionInitial Value and Boundary Value Problems have another constraint(s) that orders them to be a differential equation as well. Initial value problems (IVPs) consist of finding a solution to some differential equation such that it passes through the prescribed point which satisfies the given initial conditions. Boundary value problems (BVPs) require that the solution has to satisfy some conditions at the boundaries of the domain. However, problems of this kind are fundamental to the modeling of physical systems: the initial conditions describe the state of the system when time t=0, and the boundary conditions describe behavior at the boundaries. The function y is an IVP if, for example, y′′+y=0 with y(0)=1 and y′(0)=0. And it also could be, A BVP could be y′′+y=0 with y(0)=1 and y(π)=0.

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