find `(ln(y))^5*dy/dx = x^5*y` with condition `y(1) = e^2`

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find `(ln(y))^5*dy/dx = x^5*y` with condition `y(1) = e^2`

Postby Integr88 » Tue 29 March, 2011 13:44

The question is find `(ln(y))^5*dy/dx = x^5*y` with condition `y(1) = e^2`
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Re: find `(ln(y))^5*dy/dx = x^5*y` with condition `y(1) = e

Postby 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]^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.
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