Unclear concept on simple 2nd order differential eq

In summary, the conversation discusses solving second-order differential equations using integration. It explains the method for solving equations (1) and (2) and then discusses the incorrect approach for solving equation (3). The mistake made in the solution is identified and it is explained why it is not possible to solve equation (3) using direct integration.
  • #1
davon806
148
1

Homework Statement


Hi,I am learning to solve 2nd-order differential eq.
Suppose I have a equation
dy/dx - 3x = 0...(1)
Then dy/dx = 3x -----> x = 3(x^2)/2
Now if I have a 2nd order ODE such that:
d^2y/dx^2 = 3....(2)
Then it could be solved by integrating both sides wrt x twice,which yields
y = 3(x^2)/2 + Ax + B

Now,consider the case:
d^2y/dx^2 -6dy/dx + 9 = 0...(3)
I know it could be solved by using the idea of auxiliary equation(Putting y = Ae^(sx) into the initial eq)

But,why it's not possible to solve (3) by using direct integration on both sides as in (1) and (2)?
I can illustrate it here:
d^2y/dx^2 = 6dy/dx - 9,then I integrate both sides wrt x
dy/dx = 6 -9x + A
y = 6x -9(x^2)/2 + Ax + B
It's obviously incorrect,the (6+A)x term vanishes in the 2nd order derivative.Can someone tell me what's wrong?(I mean on the idea)

Thx

Homework Equations

The Attempt at a Solution


I have illustrated it in the problem statement.
 
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  • #2
davon806 said:
But,why it's not possible to solve (3) by using direct integration on both sides as in (1) and (2)?
I can illustrate it here:
d^2y/dx^2 = 6dy/dx - 9,then I integrate both sides wrt x
dy/dx = 6 -9x + A
Didn't you lose a ##y## here?
As in: ##y'=6y-9x+A##.
 
  • #3
Samy_A said:
Didn't you lose a ##y## here?
As in: ##y'=6y-9x+A##.
Thx so much^^,a silly mistake
 
  • #4
davon806 said:
But,why it's not possible to solve (3) by using direct integration on both sides as in (1) and (2)?
I can illustrate it here:
d^2y/dx^2 = 6dy/dx - 9,then I integrate both sides wrt x
dy/dx = 6 -9x + A
y = 6x -9(x^2)/2 + Ax + B
As already noted by @samy A, you lost a factor of y in the second line of your work.

Starting with y'' = 6y' - 9, if we integrate (w. respect to x), we get
y' = 6y - 9x + A

If you try the same trick again, you will have y on the left side, but will have ##\int y dx## on the right side. Since you don't know what y is in terms of x, the integral can't be calculated.
 

FAQ: Unclear concept on simple 2nd order differential eq

1. What is a 2nd order differential equation?

A 2nd order differential equation is a mathematical equation that involves a function, its derivatives, and independent variables. It is written in the form of y'' + p(x)y' + q(x)y = f(x), where y is the unknown function, p(x) and q(x) are known functions of x, and f(x) is a known function of x.

2. How is a 2nd order differential equation different from a 1st order differential equation?

A 1st order differential equation involves only the first derivative of the unknown function, while a 2nd order differential equation involves the second derivative as well. This means that a 2nd order differential equation is more complex and can have a wider variety of solutions.

3. What are some real-life applications of 2nd order differential equations?

2nd order differential equations are used in many fields of science and engineering, such as physics, chemistry, biology, and economics. They can be used to model motion, heat transfer, population growth, and many other physical phenomena.

4. How do you solve a 2nd order differential equation?

The specific method for solving a 2nd order differential equation depends on the type of equation and its initial conditions. Generally, we can use techniques such as separation of variables, substitution, and variation of parameters to solve these equations.

5. What are some common challenges in solving 2nd order differential equations?

One of the biggest challenges in solving 2nd order differential equations is determining the appropriate method to use. Some equations can be solved analytically, while others require numerical methods. Additionally, finding the correct initial conditions and dealing with complex functions can also be challenging.

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