# Thread: Begginer DE student with a question or 2...

1. ## Begginer DE student with a question or 2...

When asked to prove that x = -1 is a solution to the equation

x2 + 3x + 2 = 0

my lecturer told us that substituting the x = -1 into the equation was not enough as we are assuming we have a solution rather than proving x = -1 was. This confused me.

He went on to show us how to do it which involved subbing in another variable for x say x = x0 and letting the RHS=LHS. To me this was exactly the same as what he had told us not to do which further confused me. So question 1 how can I prove x = -1 is a solution to the above equation.

Also on a tutorial sheet Im working through we have to give examples of ODE's with various properties, namely what order they are, linearity, whether the are homogeneous, autunomous.

in one question Im asked to give example of an ODE of third order, linear, homogeneous and non autonomous..

Is this ok??

d3y/dx3 + d2y/dx2 + dy/dx + xy = 0

the term im not sure of is the xy, would this term effect the linearity of equation? Can an equation be homogeneous and non autonomous??

Cheers

Nappy

2. When determining linearity of a differential equation, you effectively treat the independent variable like a constant. So the xy term does not make the ODE nonlinear. In fact, you could replace x with any nonlinear function of x and the ODE would remain linear. Replacing y with y^2, on the other hand, would make the ODE nonlinear.

For your first query, I'm not sure where this lecturer is coming from. For any practical purposes, when you plug values into an equation you will either prove that the value correspond to a solution or prove that they don't. There's no assumptions necessary going in. Maybe this is more informal than what he's talking about.

3. Originally Posted by Nappy
When asked to prove that x = -1 is a solution to the equation

x2 + 3x + 2 = 0

my lecturer told us that substituting the x = -1 into the equation was not enough as we are assuming we have a solution rather than proving x = -1 was. This confused me.

He went on to show us how to do it which involved subbing in another variable for x say x = x0 and letting the RHS=LHS. To me this was exactly the same as what he had told us not to do which further confused me. So question 1 how can I prove x = -1 is a solution to the above equation.

Cheers

Nappy
If you are asked to prove that x= -1 is a solution to the equation, Nappy,
then if f(-1)=0, it's proven.

If you are asked to find the solutions of f(x)=0, without being asked to test any particular value of x, then you just use the other methods of writing f(x) as a multiplication [factors of 2 that add to give 3] or using the "quadratic formula".

$\displaystyle x^2+3x+2=0\ \Rightarrow\ (x+2)(x+1)=0$

This is just an alternative way to write it which makes the solutions obvious.
Any number multiplied by zero is zero, hence x+2=0 and x+1=0 are the solutions.

Or, $\displaystyle f(x)=ax^2+bx+c=0\ \Rightarrow\ x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$