Already have an account? Get multiple benefits of using own account!
Login in your account..!
Remember me
Don't have an account? Create your account in less than a minutes,
Forgot password? how can I recover my password now!
Enter right registered email to receive password!
Initial Conditions and Boundary Conditions
In many problems on integration, an initial condition (y = y0 when x = 0) or a boundary condition (y = y0 when x = x0 ) is given which uniquely determines the constant of integration. As a result of the unique determination of 'c', a specific curve can be singled out from a family of curves.
Example
For the boundary condition, y = 15 when x = 2, the integral y = ∫4dx is evaluated as follows:
y = ∫4dx = 4x + c
Substituting, y = 15, when x = 2
15 = 4(2) + c or c = 7
y = 4x + 7
Note that even though 'c' is specified, ∫4dx remains an indefinite integral because xn is unspecified. Thus, the integral 4x + 7 can assume an infinite number of possible values.
Complex Numbers In the radicals section we noted that we won't get a real number out of a square root of a negative number. For example √-9 isn't a real number as there is no
how do you do it
Find the sum of all natural numbers amongst first one thousand numbers which are neither divisible 2 or by 5 Ans: Sum of all natural numbers in first 1000 integers which ar
10 statements must be shown to be logically equivalent to the Statement the nxn matrix is invertible.
This topic is specified its own section for a couple of purposes. Firstly, understanding direction fields and what they tell us regarding a differential equation as well as its sol
i need help in writing about a magic car?..
We want to find the integral of a function at an arbitrary location x from the origin. Thus, where I(x=0) is the value of the integral for all times less than 0. (Essenti
Vector Arithmetic In this part we need to have a brief discussion of vector arithmetic. Addition We will begin with addition of two vectors. Thus, given the vectors a
Reason for why limits not existing : In the previous section we saw two limits that did not. We saw that did not exist since the function did not settle down to a sing
The value of K for (k+1)x^2-2(k-1)x+1 = 0 has real and equal roots.
Get guaranteed satisfaction & time on delivery in every assignment order you paid with us! We ensure premium quality solution document along with free turntin report!
whatsapp: +91-977-207-8620
Phone: +91-977-207-8620
Email: [email protected]
All rights reserved! Copyrights ©2019-2020 ExpertsMind IT Educational Pvt Ltd