differentiation

**lbj2322** · May 12th 2009, 09:27 PM

if the graph of y=f(x) contains the point (0,2), dy/dx=-x/ye^(x^2) and f(x)>0 for all x then f(x) = ?

**redsoxfan325** · May 12th 2009, 09:49 PM

Originally Posted by lbj2322

if the graph of y=f(x) contains the point (0,2), dy/dx=-x/ye^(x^2) and f(x)>0 for all x then f(x) = ?

$y'=-\frac{x}{ye^{x^2}} \implies yy' = -\frac{x}{e^{x^2}}$

Integrate both sides. (Use u-substitution with $u=x^2$ and $du=2x\,dx$ on the righthand integral.)

$\int yy'\,dx = -\int\frac{x}{e^{x^2}}\,dx \implies \frac{1}{2}y^2 = \frac{1}{2}e^{-x^2}+c \implies y = \sqrt{e^{-x^2}+C}$

Plug in $(0,2)$ : $2=\sqrt{1+C} \implies C=3$

So $y=f(x)=\sqrt{e^{-x^2}+3}$

**lbj2322** · May 12th 2009, 09:52 PM

thanks alot

i was confused on what to do with the coordinates but now i understand. Thanks again

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