# Math Help - Ordinary Differential Equation (quick question about a step in the process)

1. ## Ordinary Differential Equation (quick question about a step in the process)

Looking through my notes for a maths exam tomorrow I can't figure out the jump from one step to another:

The original question is Solve:

$\frac{dy}{dx}-y=x^2e^2x$

now further down the steps I can't figure out how it jumps from:

$\frac{1}{x}\frac{dy}{dx}-\frac{1}{x^2}y=e^2x$

to

$\frac{d}{dx}(y\frac{1}{x})=e^2x$

What I want to know is how you get from the $\frac{dy}{dx}$ to the $\frac{d}{dx}$ and what happens to the $\frac{-1}{x^2}y$

Thanks

2. I think you have a typo in your problem, the original equation should be $xy'-y=x^2e^2x.$

As for your question, what do you get if you try to differentiate w.r.t. to $x$ the expression $y\cdot\frac1x$ ?

3. I don't see any typo no, that's the notation we use for Ordinary Differential Equations.

Differentiating that will give you the $\frac{-1}{x^2}y$ yes, but what I think I'm meaning to say is, how do you jump from the $\frac{dy}{dx}$ to $\frac{d}{dx}$ what is the difference between them.

And if you do as you say and differentiate w.r.t. to the expression ? How can this still be the same as the line up above it? Surely that would make it: $\frac{-1}{x^2}y=e^2x$ and not

Or am I completely missing something here?

4. Originally Posted by Macca567
I don't see any typo no, that's the notation we use for Ordinary Differential Equations.

Differentiating that will give you the $\frac{-1}{x^2}y$ yes, but what I think I'm meaning to say is, how do you jump from the $\frac{dy}{dx}$ to $\frac{d}{dx}$ what is the difference between them.

And if you do as you say and differentiate w.r.t. to the expression ? How can this still be the same as the line up above it? Surely that would make it: $\frac{-1}{x^2}y=e^2x$ and not

Or am I completely missing something here?
You are missing something.

$
\frac{d}{dx} \left(\frac{y}{x} \right) = \frac{x y' - y}{x^2}
$

by the quotient rule and implicit differentiation.