Simple Differential Equation (Ordinary Differential Equation)

[eskie]

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...

 

[arildno]

Redo the whole thing:

1) What is the order of the ODE?

2) How do you check whether a particular function solves an equation?

 

[HallsofIvy]

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 x^9+ 9x^3- 2x^2+ 3x- 4= 0? 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: y' = -10e^{-2x}+ 4x+ 2 and put it and y= 5e^{-2x}+ 2x^2+ 2x+ 1 into the equation: y'+ 2y= -10e^{-2x}+ 4x+ 2+ 10e^{-2x}+ 4x^2+ 4x+ 2 is that equal to 4(x+ 1)?

 

[eskie]

oh i see... how about integrate the y' of y'+2y=4(x+1)²?

 

[arildno]

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).

 

[eskie]

oh... i see... i just to solve that equation just like this...
y'=4x
dy/dx=4x
dy=4xdx
y=2x²

 

[arildno]

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:
\frac{dg}{dx}=e^{2x}(y'+2y)

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

THIS ODE is separable, and you can proceed to solve for g(x) first, nd then for y(x)..

 

[eskie]

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