How Do I Solve This Initial-Value Problem Correctly?

[morbello]

Have been looking at this problem .A few times nowI've come up with different answers.Its getting close to needing to know or wether, to just put in what i have.so I've placed a question on here to see if I am missing something.



The equations that are said to help are


\int\f'(x)/f(x) dx = In (f(x))+c (f(x)>0)

dy/dx = x^4+1/x^5+5x+6 (x>-1) is the equation to work off.


my work so far is

1/5 In (x^5+5)+c

 

[rock.freak667]

your answer is nearly correct...you are just missing something...check back to what f(x) is and what your answer is...

 

[morbello]

so the x^5+5x+6 does not differeniate to become x^5+5 ok

there are a few activitys in my books that are like it.

 

[rock.freak667]

so the x^5+5x+6 does not differeniate to become x^5+5 ok

there are a few activitys in my books that are like it.

\int \frac{x^4+1}{x^5+5x+6} dx

Let t=x^5+5x+6 \Rightarrow \frac{dt}{dx}=5(x^4+1)

\frac{dt}{5}=(x^4+1)dx

\int \frac {x^4+1}{x^5+5x+6} dx \equiv \int \frac{1}{5} \frac{1}{t} dt

and \int \frac{1}{x} dx = ln(x)+C

\int \frac{1}{5} \frac{1}{t} dt = \frac{1}{5} lnt + C

 

[morbello]

that blew me a little bit there.ive sat and looked at it a little and i think i know what you have done just i would not have thought off it. i think we did one function that was worked out that way.ill have to look it up tonight.