# Problem 01 | Separation of Variables

**Problem 01**

$\dfrac{dr}{dt} = -4rt$, when $t = 0$, $r = r_o$

**Solution 01**

$\dfrac{dr}{dt} = -4rt$

$\dfrac{dr}{r} = -4t\,dt$

**Problem 01**

$\dfrac{dr}{dt} = -4rt$, when $t = 0$, $r = r_o$

**Solution 01**

$\dfrac{dr}{dt} = -4rt$

$\dfrac{dr}{r} = -4t\,dt$

**Linear Equations of Order One**

Linear equation of order one is in the form

$\dfrac{dy}{dx} + P(x) \, y = Q(x).$

The general solution of equation in this form is

$\displaystyle ye^{\int P\,dx} = \int Qe^{\int P\,dx}\,dx + C$

The differential equation

$M(x, y) \, dx + N(x, y) \, dy = 0$

is an exact equation if

$\dfrac{\partial M}{\partial y} = \dfrac{\partial N}{\partial x}$

If the function f(x, y) remains unchanged after replacing x by kx and y by ky, where k is a constant term, then f(x, y) is called a **homogeneous function**. A differential equation

$M \, dx + N \, dy = 0 \,\, \to \,\,$ Equation (1)

is homogeneous in x and y if M and N are homogeneous functions of the same degree in x and y.

The depreciation charge and the total depreciation at any time m using the sum-of-the-years-digit method is given by the following formulas:

**Depreciation Charge:**

$d_m = (FC - SV) \dfrac{n - m + 1}{SYD}$

**Total depreciation at any time m**

$D_m = (FC - SV) \dfrac{m(2n - m + 1)}{2 \times SYD}$

Where:

FC = first cost

SV = salvage value

n = economic life (in years)

m = any time before n (in years)

SYD = sum of years digit = 1 + 2 + ... + n = n(1 + n)/2

Given the differential equation

$M(x, y)\,dx + N(x, y)\,dy = 0 \,\, \to \,\,$ Equation (1)

where $\,M\,$ and $\,N\,$ may be functions of both $\,x\,$ and $\,y\,$. If the above equation can be transformed into the form

$f(x)\,dx + f(y)\,dy = 0\,\, \to \,\,$ Equation (2)

where $\,f(x)\,$ is a function of $\,x\,$ alone and $\,f(y)\,$ is a function of $\,y\,$ alone, equation (1) is called **variables separable**.

**Problem 532**

A beam simply supported at the ends of a 25-ft span carries a uniformly distributed load of 1000 lb/ft over its entire length. Select the lightest S section that can be used if the allowable stress is 20 ksi. What is the actual maximum stress in the beam selected?

**Problem 531**

A 15-ft beam simply supported at the ends carries a concentrated load of 9000 lb at midspan. Select the lightest S section that can be employed using an allowable stress of 18 ksi. What is the actual maximum stress in the beam selected?

**Problem 530**

Repeat Prob. 529 if the distributed load is 12 kN/m and the length of the beam is 8 m.

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- Pa answer nga po tnx
- Differential Equation
- Physics
- Application of Differential Equation: mixture problem
- Arbitrary constant
- Homogeneous equations- general solution
- What am I missing when I rotate a tripod leg 45 degrees to solve for loads
- Families of Curves: family of circles with center on the line y= -x and passing through the origin
- Differential Equation: Eliminate C1, C2, and C3 from y=C1e^x+C2e^2x+C3e^3x
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