help me solve this differential equation
6. The normal to the curve $\displaystyle x=2/t$, $\displaystyle y=2t$ at a point has slope $\displaystyle -1/(dy/dx)$
Now:
$\displaystyle
\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
5)
$\displaystyle (x+1)y\frac{dy}{dx} + 1 = 0$
$\displaystyle (x+1)y\frac{dy}{dx} = -1$
$\displaystyle y\frac{dy}{dx} = -\frac{1}{x+1}$
$\displaystyle y dy = -\frac{dx}{x+1}$ 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 $\displaystyle y = 2x+c$
Since (2,5) is on the line, we have $\displaystyle 5 = 2(2) + c \Rightarrow c = 1$
So y = 2x + 1.
$\displaystyle y = 0 \Rightarrow x = -\frac{1}2$
$\displaystyle x = 0 \Rightarrow y = 1$
So the mid point of $\displaystyle (0,1)$ and $\displaystyle \displaystyle (-\frac{1}2, 0)$ is $\displaystyle \displaystyle (-\frac{1}4, \frac12)$