First Order Differential Equation

In summary: It should always be included in the integrals.In summary, the differential equation is dy/dx = √(4 - y^2) and the general solution is y = 2sin(lnx).
  • #1
143
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Homework Statement


Find the general solution to the following differential equation.

dy/dx = 2x( (y^2) + 1)


Homework Equations





The Attempt at a Solution



I got all x terms on one side and all y terms on the otherside

2x dx = 1/( (y^2) + 1`)dy

integrate

x^2 + c = tan(y)

y = tan-inverse(x^2 + c)

Can somebody tell me if this is right ?

my book says the solution is y = tan(x^2 + c)
 
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  • #2
It's not right. What is
[tex] \int \frac{dy}{1+y^2}[/tex]?
 
  • #3
tan-inverse(1/y)

sorry I see where I went wrong now

it should have been x^2 + c = tan-inverse(y)
 
  • #4
Woolyabyss said:
tan-inverse(1/y)

sorry I see where I went wrong now

it should have been x^2 + c = tan-inverse(y)

∫dy/(1+y2) = tan-1y

So,tan-1y = x2 + c

or , y = tan(x2 + c)
 
  • #5
Tanya Sharma said:
∫dy/(1+y2) = tan-1y

So,tan-1y = x2 + c

or , y = tan(x2 + c)

Could you help me with another one?

Given the differential equation

x(dy/dx) = √(4 - y^2) , find the general solution given that y = 0 when x = 1

I got all the x terms to one side and all of the y terms to the opposite side.

1/√(4 - y^2) dy = 1/x dx

integrating gives

sin-inverse(y/2) = lnx

using the limits and we get

sin-inverse(y/2) = lnx

y/2 = sin(lnx)

y = 2sin(lnx)

my book says the solution is y = sin(lnx)
 
  • #6
Woolyabyss said:
Could you help me with another one?

Given the differential equation

x(dy/dx) = √(4 - y^2) , find the general solution given that y = 0 when x = 1

I got all the x terms to one side and all of the y terms to the opposite side.

1/√(4 - y^2) dy = 1/x dx

integrating gives

sin-inverse(y/2) = lnx

using the limits and we get

sin-inverse(y/2) = lnx

y/2 = sin(lnx)

y = 2sin(lnx)

my book says the solution is y = sin(lnx)

y=2sin(lnx) looks to be the correct answer.You can verify that by putting the solution in the original equation and checking whether it satisfies the eq.

Note : Even though you came up with the correct answer ,please do not ignore the constant of integration C while integrating.
 

1. What is a first order differential equation?

A first order differential equation is a mathematical equation that involves the derivative of an unknown function with respect to a single independent variable. It is called a "first order" equation because it only involves the first derivative, as opposed to higher order equations which involve second, third, or higher derivatives.

2. What is the general form of a first order differential equation?

The general form of a first order differential equation is: dy/dx = f(x,y), where y is the unknown function and f(x,y) is a given function of x and y. This equation represents the relationship between the rate of change of y with respect to x, and the values of x and y.

3. What are some real-world applications of first order differential equations?

First order differential equations are used in many different fields of science, engineering, and economics to model and predict various physical systems and processes. Some common applications include population growth, chemical reactions, electrical circuits, and radioactive decay.

4. How are first order differential equations solved?

There are several methods for solving first order differential equations, including separation of variables, integrating factors, and using specific formulas for certain types of equations. In some cases, numerical methods may also be used to approximate a solution.

5. Can first order differential equations have multiple solutions?

Yes, first order differential equations can have multiple solutions. In fact, there are typically infinitely many solutions to a first order differential equation, as they represent a family of functions that all satisfy the same relationship between the rate of change and the variables. However, a specific solution can be determined by specifying initial conditions or boundary conditions.

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