by barnamah » Tue 29 March, 2011 13:51
to find `(ln(y))^5*dy/dx = x^5*y` with initial condition `y(1) = e^2`
it means to find y in the equation `(ln(y))^5*dy/dx = x^5*y` which means differentiate implicitly.
this means we have to have dx with all xs on one side and dy with ys on the other side.
`(lny)^5*dy/dx = x^5*y` divide by `y` or miltiply by `1/y` so it looks like this: `((lny)^5*dy/dx)*1/y =( x^5*y)*1/y` and we have:
`((lny)^5/7*dy/dx) = x^5`
Now we have to bring dx from the left to right by multiplying both side by dx:
`(lny)^5/7*dy/dx*dx = x^5*dx` and we have `(lny)^5/7dy = x^5dx` which can be written as `(lny)^5*1/7dy = x^5dx`
moving `1/7` out side we have: `1/7int(lny)^5*dy = intx^5dx`
Now we can integrate both sides:
`1/7int(lny)^5dy = intx^5dx`
I can integrate the right side and we have:
`1/7int(lny)^5dy =x^6/6+C`
sorry I am unable to integrate the `1/7int(lny)^5dy` but using Wolfram alpha [url]http://www.wolframalpha.com/input/?i=integrate+%28lny%29^5[/url] I got the following answer
`y (log^5(y)-5 log^4(y)+20 log^3(y)-60 log^2(y)+120 log(y)-120)+C`
so putting them together we have:
`y (log^5(y)-5 log^4(y)+20 log^3(y)-60 log^2(y)+120 log(y)-120)=x^6/6+C`
This far I could go.
Explore and know. That is asked.