Finding the solution of differential equations for a given condition

A differential equation has several solutions. Solving the differential equation for a given condition is equivalent to finding the solution that satisfies the condition. We apply the general process described earlier except that we find the constant c that satisfies the condition.

Example: 

Find the solution to the following differential equation: ty' + 2y = t² - t + 1 for the condition y(1) = 1/2

Solution

The general form of a differential equation of first order is : dy/dt + p(t)y = g(t).

Let's put the given equation in the general form by dividing both sides by t:
t/ty' + 2/ty = t²/t - t/t + 1/t
y' + 2/ty = t -1 + 1/t

Let's find μ(t) by substituting p(t) = 2/t in the formula:

Let's multiply the differential equation in its proper form by mu of t = t²

  (t²)y' +  (t²)2/ty = t³ -t² + t² 1/t
t²y' + 2ty = t³ -t² + t
The left side is the derivative of t²y. We have:
(t²y)' =  t³ -t² + t
Integrate both sides:
∫(t²y)'dt = ∫( t³ -t² + t)dt
t²y = t⁴/4 - t³/3 + t²/2 + c
y =  t⁴/4t² - t³/3t² + t²/2t² + c/t²
y = 1/4t² - 1/3t + 1/2 + c/t²
Let's substitute y by 1/2 and t by 1 in order to find c:
1/2 = 1/4(1) - 1/3(1) + 1/2 + c/1
1/2 = 1/4 - 1/3 + 1/2 + c
By reducing the fractions to the same denominator and simplifying:
6 = 3 -4 + 6 + 12c
0 = -1 + 12c
12c = 1 c = 1/12
Let's subsititute c in the expression of y:
y = 1/4t² - 1/3t + 1/2 + 1/12/t²

Practice

Solve the differential equation dv/dt = 9.8 - 0.196v satisfying the condittion v(0) = 48

Interested in reviewing Differential and Integral Calculus visit Center for Integral Development 

Resolution of first order linear differential equations

Let's solve the first order linear differential equation dy/dt + p(t)y = g(t). Solving such an equation is to find y. The first thing we have to do is to find a magical function μ(t) called the integrating factor. We multiply both sides of the equation by this function. A subsequent set of steps will hopefully allow to find y. Rather than solving the general form let's solve a numerical example and then we will generalize the process to the solution of any first linear differential equation.

Let's solve the equation dv/dt = 9.8 - 0.196v
Let's put this equation in its general form: dv/dt + 0.19v = 9.8.
Let's find the magical function  μ(t). The solution of the general form of the first linear differential equation above defines this function as:

In the differential equation p(t) = 0.19 therefore

Let's multiply both sides of the equation by the value of μ(t)

The left side of the equation is the derivative of the product

Therefore we have

Let's integrate both sides of the equation we have:

We generalize the procedure used to solve the differential equation above. In order to solve a first linear differential equation having the form dy/dt + p(t)y = g(t) we follow the following steps:
1) We make sure that the differential equation is written in its proper explicit form
2) We find the integrating factor

3) We multiply both sides of the equation by μ(t)
4) We substitute the left side of the equation by the derivative of a product
5) We integrate both sides of the equation in order to find the unknown

Practice. Solve the following differential equation dy/dx = 3 + 2.5y

Interested in learning about limits, continuity, derivative, integral visit Center for Integral Development

Introduction to Differential Equations

In this post I am going to write about the fundamentals of Differential equations. But before we go any further let's remind the definition of equation.
An equation is a mathematical statement that two things are equal.
Ex: 15 = 9 + 6
Both sides of the sign equal have the same value.
An equation can also involve an unknown quantity
Ex: x + 5 = 9
Solving an equation is to find the value of the unknown. This unknown can have one or more values. It can also have an infinity of values.

Differential equation. Definition

A differential equation is an equation that is differential. Differential is an adjective that derives from the word "derivative". A differential equation is an equation that involves a function and its derivative. Solving a differential equation is to find the value of the unknown of this equation. The unknown is a function. Finding the solution of a differential equation is finding the value of the unknown function.

The solution of a differential equation can be one function, a set of functions or a class of functions.
Example: y + dy/dx = 7
y is the unknown function to find. dy/dx is its derivative.
 This equation can also be written as y + y' = 7 or f(x) + f'(x) = 7.

Ordinary differential equation

An ordinary differential equation is an equation where the unknown function has one independent variable.

Examples:

Partial differential equation

A partial differential equation is an equation where the unknown function has more than one independent variable.

Examples:

-
Order

The order of a differential equation is the highest derivative.

Examples:

dy/dx + y = 2x + 1 is a first order differential equation

dy/dx + d²y/dx² is a second order differential equation.

Linear differential equations

An ordinary differential equation has the form:

In a linear differential equation, there are no products of the function and its derivatives. The function and its derivatives have only the power of 1.

Examples:

2 y''' + 5y'' + 3y' + 6y = sint

cosydy/dx = y²dy/dx + y³ is non linear. The function has the power 2.

Interested in reviewing in a simple manner the notions of limit, continuity, derivatives and integral visit
Center for Integral Development

Volume of a solid of revolution: method of cylindrical shells

In this post we are going to compute the volume of a solid rotating around the y-axis by utilizing the method of cylindrical shells.

Let's consider a function f. positive over an interval [a,b]. Let R be the region bounded by the curve of the function, two verticals x = a and x = b and the x-axis..

A solid of revolution is obtained by revolving the region around the y-axis. Our goal is to determine the volume of this solid. We are going to use the method of cylindrical shells for that purpose.

Method.

Let's divide the region R in small rectangles of width dx and consider one rectangle.

Let's revolve the rectangle around the y-axis. We obtain a cylindrical shell



Let's calculate the volume of this cylindrical shell. The volume of this cylindrical shell is obtained by multiplying the volume of the outer cylinder by the thickness of the cylindrical shell. We can write:

Volume cylindrical shell = Volume outer shell * thickness.

The volume of a cylinder is equal to the surface of the base by the height. Therefore the volume of the outer shell can be written as:

Volume of outer shell = Area of base * height. The base is a circle. Its area is 2𝜋r. Then
Volume outer shell  = 2𝜋rh .

Let's substitute in Volume cylindrical shell

Volume cylindrical shell = 2𝜋rhdx (dx is the thickness)

The radius r is variable. We call it x. The height is a function of the radius. We represent it by f(x)

V cylindrical shell =  2𝜋xf(x)dx
Adding the volume of all these small cylindrical shells is going to give the volume of the entire solid of revolution. This is equivalent to integrate the volume over the interval  [a,b]. Therefore the volume of the solid of revolution is given by:

Example

Use the cylindrical shells to find the volume generated when the region bounded by the curves is revolved about  the y-axis.

y = 1/x  y = 0 x = 1 x = 3

Solution 

The volume of a cylindrical shell generated by a region around the y-axis is given by 

Practice

Find the volume generated by the lines y = 2x-1 y = -2x + 3 and 
x  = 2 about the y-axis.

Interested in Calculus and other math subjects tutoring visit 
New Direction Education Services for contact information 
Interested in online Calculus courses visit Center for Integral Development

Volume of a solid of revolution: method of washer

According to the Merriam-Webster dictionary a washer is a flat ring or a perforated plate used in joints or assemblies to ensure tightness, prevent linkage or release friction.



The automobile mechanic technician use this type of ring to relate two pieces in fixing a car..

Washer method to compute the volume of a solid of revolution 

Let's consider two continuous and positive functions f and g over an interval [a,b]. Let R be the region bounded by the curves of these functions.



Let's have the shaded region revolve around the x-axis.


This is the solid we obtain. It has the shape of a hollow pot. In order to find its volume we have to integrate the cross-section, which is a washer or a flat ring as defined previously. The cross-section is shaded in the figure above.

The area of the cross-section is the difference between the area the of the circle of radius f(x) and the circle of radius g(x).

The volume of the solid of revolution is found by integrating the area:

Example

Find the volume generated when the region between the graphs f(x) = x² + 1 and g(x) = x over the interval [0,3] is revolved around the x-axis.


Solution


When the green region revolves around the x-axis it creates a solid of revolution like this:



The cross-section is a ring or washer. The volume of the solid it generates is given by:

Exercise

Find the volume generated when the region between the graphs f(x) = 2x² + 3 and g(x) = 2x over the interval [0,2]

Interested in Calculus tutoring visit New Direction Education Services for contact information
Interested in learning Calculus online visit Center for Integral Development