# Simple Differential Equation (Ordinary Differential Equation)

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...

Last edited:

Related Differential Equations News on Phys.org
arildno
Homework Helper
Gold Member
Dearly Missed
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
Homework Helper
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)?

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

arildno
Homework Helper
Gold Member
Dearly Missed
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).

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

arildno
Homework Helper
Gold Member
Dearly Missed
oh... i see... i just to solve that equation just like this...
y'=4x
dy/dx=4x
dy=4xdx
Not quite!

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)..

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