# Simple System of Differential Equations

• Jul 31st 2013, 03:40 AM
Educated
Simple System of Differential Equations
$x' = x + y$
$y' = 2x$

Find an expression for $\frac{dy}{dx}$ in terms of x,y
Hint: Note $\frac{d}{dt}y(t) = \frac{d}{dt}y(x(t))=...$

Find an implicit solution of the differential equation in $x \mapsto y$ for y(x)

So, I don't know if I'm interpreting this right or not, but rewriting the equations in fraction notation:
$\frac{dx}{dt} = x + y$
$\frac{dy}{dt} = 2x$

I divide the second one by the first one to get:

$\frac{dy}{dt} \frac{dt}{dx} = \dfrac{2x}{x + y}$

$\frac{dy}{dx}= \dfrac{2x}{x + y}$

So I get an equation for dy/dx

And then I let y(x) = x v(x)
Which means y'(x) = v(x) + x v'(x)

$v(x) + x v'(x) = \dfrac{2x}{x + x v(x)}$

$v(x) + x v'(x) = \dfrac{2}{1 + v(x)}$

Have I done everything right so far?
Now what do I do here? I get this non-linear equation and I don't know how to solve it... any hints?
• Jul 31st 2013, 04:06 AM
Prove It
Re: Simple System of Differential Equations
Quote:

Originally Posted by Educated
$x' = x + y$
$y' = 2x$

Find an expression for $\frac{dy}{dx}$ in terms of x,y
Hint: Note $\frac{d}{dt}y(t) = \frac{d}{dt}y(x(t))=...$

Find an implicit solution of the differential equation in $x \mapsto y$ for y(x)

So, I don't know if I'm interpreting this right or not, but rewriting the equations in fraction notation:
$\frac{dx}{dt} = x + y$
$\frac{dy}{dt} = 2x$

I divide the second one by the first one to get:

$\frac{dy}{dt} \frac{dt}{dx} = \dfrac{2x}{x + y}$

$\frac{dy}{dx}= \dfrac{2x}{x + y}$

So I get an equation for dy/dx

And then I let y(x) = x v(x)
Which means y'(x) = v(x) + x v'(x)

$v(x) + x v'(x) = \dfrac{2x}{x + x v(x)}$

$v(x) + x v'(x) = \dfrac{2}{1 + v(x)}$

Have I done everything right so far?
Now what do I do here? I get this non-linear equation and I don't know how to solve it... any hints?

Yes it's correct, though it is VERY sloppy to continually switch between Leibnitz and Newtonian notation. Choose one and stick to it (I prefer Leibnitz for DEs).

When you get to \displaystyle \begin{align*} v + x\,\frac{dv}{dx} = \frac{2}{1 + v} \end{align*}, this is a separable equation.

\displaystyle \begin{align*} v + x\,\frac{dv}{dx} &= \frac{2}{1 + v} \\ x\,\frac{dv}{dx} &= \frac{2}{1 + v} - v \\ x\,\frac{dv}{dx} &= \frac{2 - v - v^2}{1 + v} \\ \frac{1 + v}{2 - v - v^2}\,\frac{dv}{dx} &= \frac{1}{x} \end{align*}

Now that the variables have been separated, you can solve the DE using integration.