Simple Differential Equation (Ordinary Differential Equation)

  • Thread starter eskie
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  • #1
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Problem:
y'+2y=4(x+1)2 ----> y=5e-2x+2x2+2x+1

1. What the Order of the ODE?
It's 1st order

2 How do you check whether a particular function solves an equation?
If you solve y'+2y=4(x+1)2 and make it y=5e-2x+2x2+2x+1. I want the whole solution... thanks...
 
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  • #2
arildno
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Redo the whole thing:

1) What is the order of the ODE?

2) How do you check whether a particular function solves an equation?
 
  • #3
HallsofIvy
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Problem:
y'+2y=4(x+1)2 ----> y=5e-2x+2x2+2x+1

1. What the Order of the ODE?
It's 1st order
Yes, that is correct.

2 How do you check whether a particular function solves an equation?
If you solve y'+2y=4(x+1)2 and make it y=5e-2x+2x2+2x+1. I want the whole solution... thanks...
How would you determine whether x= 7 satisifies the equation [itex]x^9+ 9x^3- 2x^2+ 3x- 4= 0[/itex]? Not by actually solving the equation! Just put x= 7 into the equation and see if it makes the equation true or not.

Same thing here. You would NOT need to actually solve the equation. Just calculate the derivative of y: [itex]y' = -10e^{-2x}+ 4x+ 2[/itex] and put it and [itex]y= 5e^{-2x}+ 2x^2+ 2x+ 1[/itex] into the equation: [itex]y'+ 2y= -10e^{-2x}+ 4x+ 2+ 10e^{-2x}+ 4x^2+ 4x+ 2[/itex] is that equal to 4(x+ 1)?
 
  • #4
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oh i see... how about integrate the y' of y'+2y=4(x+1)2?
 
  • #5
arildno
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oh i see... how about integrate the y' of y'+2y=4(x+1)2?
That would be the way to find the solution, rather than verifying that a given function is the solution!

To find a solution is generally a lot harder to do than verifying that a function is a solution (or not).
 
  • #6
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oh... i see... i just to solve that equation just like this...
y'=4x
dy/dx=4x
dy=4xdx
y=2x2
 
  • #7
arildno
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oh... i see... i just to solve that equation just like this...
y'=4x
dy/dx=4x
dy=4xdx
Not quite!

In your original ODE, that approach won't help you much.

You need to use what we call an "integrating factor" here!

Let g(x)=e^{2x}y(x}

Then, we have:
[tex]\frac{dg}{dx}=e^{2x}(y'+2y)[/tex]

Note that the expression in the parenthesis is the left-hand side of your ODE, so that you may write:
[tex]\frac{dg}{dx}=4e^{2x}(x+1)^{2}[/tex]

THIS ODE is separable, and you can proceed to solve for g(x) first, nd then for y(x)..
 
  • #8
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oh i see... thanks anyway... :) our teacher said that we must use bernoulli's eq. to solve the eq... and the 5e-2x is constant.. therefore i can just change it to C
 

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