1. ## Differential Equation

Find ${f{{\left({x}\right)}}}$ if ${y}={f{{\left({x}\right)}}}$ satisfies
${\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={34}{y}{{x}}^{{{16}}}$
and the ${y}$-intercept of the curve ${y}={f{{\left({x}\right)}}}$ is ${4}$.

The f(x) and y and x are all mixing me up. I tried doing...

$1/34 \int 1/y dy = \int x^{16}dx$

and solved for y getting...

$y = e^{2x^{17} + 136 + C}$

But i don't understand how to find C because of all the f(x), y, etc. Then i would have to change it back to f(x) somehow...

2. Originally Posted by twoteenine
But i don't understand how to find C because of all the f(x), y, etc. Then i would have to change it back to f(x) somehow...
This part is telling you to use the point $(0,4)$ in your answer for $y$ to solve for $C$.

3. Originally Posted by twoteenine
Find ${f{{\left({x}\right)}}}$ if ${y}={f{{\left({x}\right)}}}$ satisfies
${\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={34}{y}{{x}}^{{{16}}}$
and the ${y}$-intercept of the curve ${y}={f{{\left({x}\right)}}}$ is ${4}$.

The f(x) and y and x are all mixing me up. I tried doing...

$1/34 \int 1/y dy = \int x^{16}dx$

and solved for y getting...

$y = e^{2x^{17} + 136 + C}$

But i don't understand how to find C because of all the f(x), y, etc. Then i would have to change it back to f(x) somehow...
I've not checked your integration so it may be wrong. To find C use the fact you're given the y intercept.

At the y intercept we get (0,4). Sub this into the equation

To make life easier I'm going to use k such that $k = e^C$

$4 = k \cdot e^{2(0)^{17}+136}$

Solve for k