# Finding general solution

• May 17th 2008, 02:53 AM
Sweeties
Finding general solution
Hello everyone!

i was asked to use the quotient rule to differentiate
h(x) = ( 1 + ln(x) ) / x where x > 0

I got

k'(x) = ( x + (1/x) - 1 + ln(x) * 1 ) / x^2

k'(x) = ( ln(x) ) / x^2

which i think is right.

it then asks me to use the above answer to find the general soltion of the differentisl equation

dy/dx = -(( ln(x) ) / x^2)) y^1/2 where x>0 and y>0

and give the solution in implicit form which i can't do. i am also not sure how to integrate ln(x), i find it confusing.
• May 17th 2008, 03:40 AM
mr fantastic
Quote:

Originally Posted by Sweeties
Hello everyone!

i was asked to use the quotient rule to differentiate
h(x) = ( 1 + ln(x) ) / x where x > 0

I got

k'(x) = ( x + (1/x) - 1 + ln(x) * 1 ) / x^2

k'(x) = ( ln(x) ) / x^2

which i think is right.

it then asks me to use the above answer to find the general soltion of the differentisl equation

dy/dx = -(( ln(x) ) / x^2)) y^1/2 where x>0 and y>0

and give the solution in implicit form which i can't do. i am also not sure how to integrate ln(x), i find it confusing.

You've found that the derivative of $\frac{1 + \ln x}{x}$ is $\frac{\ln x}{x^2}$.

Therefore $\int \frac{\ln x}{x^2} \, dx = \frac{1 + \ln x}{x} + C$.

$\frac{dy}{dx} = - \frac{\ln x}{x^2} \, y^{1/2} \Rightarrow - \frac{dy}{y^{1/2}} = \frac{\ln x}{x^2} \, dx \Rightarrow - \int \frac{dy}{y^{1/2}} = \int \frac{\ln x}{x^2} \, dx$ ......
• May 17th 2008, 04:15 AM
Sweeties
Quote:

Originally Posted by mr fantastic
You've found that the derivative of $\frac{1 + \ln x}{x}$ is $\frac{\ln x}{x^2}$.

Therefore $\int \frac{\ln x}{x^2} \, dx = \frac{1 + \ln x}{x} + C$.

$\frac{dy}{dx} = - \frac{\ln x}{x^2} \, y^{1/2} \Rightarrow - \frac{dy}{y^{1/2}} = \frac{\ln x}{x^2} \, dx \Rightarrow - \int \frac{dy}{y^{1/2}} = \int \frac{\ln x}{x^2} \, dx$ ......

Sory, its not clear to me why/ how you have got this:
- dy/ y^(1/2)
• May 17th 2008, 04:23 AM
mr fantastic
Quote:

Originally Posted by Sweeties
Sory, its not clear to me why/ how you have got this:
- dy/ y^(1/2)

Re-arrange. Put all the x-stuff on one side and all the y-stuff on the other. The "-" has to go somewhere. I put it with the y-stuff. If you prefer:

$\int \frac{dy}{y^{1/2}} = - \int \frac{\ln x}{x^2} \, dx$

is just as correct .....
• May 17th 2008, 05:06 AM
Sweeties
ok. i'm still not sure about the 'general solution in implicit form' part. thanks.
• May 17th 2008, 06:05 AM
mr fantastic
Quote:

Originally Posted by Sweeties
ok. i'm still not sure about the 'general solution in implicit form' part. thanks.

$2 \sqrt{y} = \frac{1 + \ln x}{x} + C$, perhaps?

Although I don't think it's much of a stretch to get an explicit general solution for y .....
• May 23rd 2008, 01:26 PM
jo holvey
uote=Sweeties;146108]Hello everyone!

i was asked to use the quotient rule to differentiate
h(x) = ( 1 + ln(x) ) / x where x > 0

I got

k'(x) = ( x + (1/x) - 1 + ln(x) * 1 ) / x^2 x + (1/x) this should be x*(1/x)
according to the quotient rule
k'(x) = ( ln(x) ) / x^2 this would then change the
which i think is right.

it then asks me to use the above answer to find the general soltion of the differentisl equation

dy/dx = -(( ln(x) ) / x^2)) y^1/2 where x>0 and y>0

and give the solution in implicit form which i can't do. i am also not sure how to integrate ln(x), i find it confusing.[/quote]
• May 23rd 2008, 03:10 PM
mr fantastic
Quote:

Originally Posted by jo holvey
uote=Sweeties;146108]Hello everyone!

i was asked to use the quotient rule to differentiate
h(x) = ( 1 + ln(x) ) / x where x > 0

I got

k'(x) = ( x + (1/x) - 1 + ln(x) * 1 ) / x^2 x + (1/x) this should be x*(1/x)
according to the quotient rule
k'(x) = ( ln(x) ) / x^2 this would then change the
In fact, the derivative of $y = \frac{1 + \ln x}{x}$ is $\frac{dy}{dx} = {\color{red}-}\frac{\ln x}{x^2}$.