# Thread: differential equation

1. ## differential equation

help me solve this differential equation

2. Originally Posted by carlasader
help me solve this differential equation
5. The ODE:

$(x+1)y \frac{dy}{dx}+1=0$

is of variables seperable type, so rewrite as:

$
y \frac{dy}{dx}= -~\frac{1}{x+1}
$

so:

$
\frac{d}{dx} \left( \frac{y^2}{2} \right)=-~\frac{1}{x+1}
$

RonL

3. Originally Posted by carlasader
help me solve this differential equation
6. The normal to the curve $x=2/t$, $y=2t$ at a point has slope $-1/(dy/dx)$

Now:

$
\frac{dy}{dx}=\frac{dy}{dt}~\frac{dt}{dx}=(2)\left (-\frac{2}{x^2}\right)
$

So now you know the point, and the slope of the normal at that point, so you
can find the equation of the normal

RonL

4. Originally Posted by carlasader
help me solve this differential equation
5)
$(x+1)y\frac{dy}{dx} + 1 = 0$

$(x+1)y\frac{dy}{dx} = -1$

$y\frac{dy}{dx} = -\frac{1}{x+1}$

$y dy = -\frac{dx}{x+1}$ and so on...

7)
Since the slope is 2 and we know the equation of a line is of the form y = (slope)x+(y-intercept)
So its of the form $y = 2x+c$
Since (2,5) is on the line, we have $5 = 2(2) + c \Rightarrow c = 1$
So y = 2x + 1.

$y = 0 \Rightarrow x = -\frac{1}2$
$x = 0 \Rightarrow y = 1$
So the mid point of $(0,1)$ and $\displaystyle (-\frac{1}2, 0)$ is $\displaystyle (-\frac{1}4, \frac12)$

5. Originally Posted by carlasader
help me solve this differential equation
Don't post multiple questions like this in the same thread. Posting them in a
single thread makes it less likely that you will get help will all of them.

RonL