Solving Applied Math Problems

Introduction to solving applied math problems:

Mostly college level students are using the applied mathematics problems. Applied mathematics is very helpful for students for learning about math. Applied math is a division of math that concerns itself with the mathematical techniques typically used in the application of math knowledge to other domains. The following topics are covered in applied mathematics; they are functions of a complex variable, calculus, ordinary differential equations, partial differential equations and the calculus of variations.

Example Problems for Solving Applied Math Problems:

Solving applied math problems – Example: 1

Transform the BVP to an integral equation: `(d^2 y)/(dx^2) + y = x, y(0) = 0,` `y'(1) = 0.`

Solution:
First, integrate both sides from x to 1 (so that the boundary condition `y'(1) = 0 ` can be applied). This gives

`y'(1) - y'(x) + int_x^1 y(u) du = int_x^1 y(u)du = int_x^1 udu = (u^2)/(2)|_x^1 = 1/2 – (x^2)/(2)`
Applying the condition` y'(1) = 0 ` and multiplying both sides by (-1) results in

`y'(x) - int_x^1 y(u)du = (x^2)/(2)- 1/2`

Next, integrate both sides from 0 to x:

`(y(x) - y(0)) - int_0^x int_z^1 y(u)dudz = int_0^x (u^2)/(2) - 1/2 du = ((u^3)/(6) - (u)/(2))|_0^x = (x^3)/(6) - (x)/(2)`

or, applying the condition y(0) = 0,

`y(x) - int_0^x int_z^1 y(u) dudz = (x^3)/(6) - (x)/(2)`

We see that the region must be split into two pieces, and a separate integral written for each piece:

`int_0^1 int_z^1 y(u)dudz = int_0^x int_0^u dz y(u) du + int_x^1 int_0^x dz y(u) du`

or after doing the trivial z-integration,

`int_0^1 int_z^1 y(u)dudz = int_0^x u y(u) du + int_x^1 x y(u) du`

The desired integral equation, then, is

`y(x) - int_0^x u y(u) du - int_1^x xy(u) du = (x^3)/(6) - (x)/(2)`

Such an integral equation is often written in the form

`y(x) - int_0^1 k(x, u) y(u) du = (x^3)/(6) - (x)/(2)`

where

k(x, u) = `{ u, 0 < u < x`

`{ x, x< u < 1`

`Answer: k(x, u) = { u, 0 < u < x`

`{ x, x< u < 1`

Solving applied math problems – Example: 2

If `f(t) = ti + (t^2 - 2t)j + (3t^2 + 3t^3)k` , find `int_0^1 f(t)dt.`

Solution:

`f(t) = ti + (t^2 - 2t)j + (3t^2 + 3t^3)k`

` = i int_0^1 tdt + j int_0^1 (t^2 - 2t)dt + k int_0^1 (3T2 + 3t^3)dt`

`= i[(t^2)/(2)] + j[(t^3)/(3) - t^2] + k[t^3 + (3t^4)/(4) at (0,1)`

`= i[1/2 + j[1/3 - 1] + k[1 + 3/4]`

`= 1/2 i - 2/3 j + 7/4 k`

`Answer: 1/2 i - 2/3 j + 7/4 k`

Practice Problems for Solving Applied Math Problems:

1. Solve:  `g(s) = f(s) + lambda int_0^(2 pi) sin(s) cos(t)g(t) dt`

`Answer: g(s) = f(s) + lambda sin(s) int_0^(2 pi) cos(t)f(t) dt`

2. If  `f(t) = (t - t^2)I + 2t^3 j - 3k,` Find i) `int f(t) dt ` ii) `int_1^2 f(t) dt`

`Answer: (i) i[(t^2)/(2) - (t^3)/(3)] + (t^4)/(2) j - 3tk + c`

` (ii) - 5/6 I + 15/2 j - 3k`