# Thread: Finding general solution

1. ## 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.

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

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

3. 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)

4. 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 .....

5. ok. i'm still not sure about the 'general solution in implicit form' part. thanks.

6. 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 .....

7. 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]

8. 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
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.
I confess I didn't check the derivative properly. One of my many failings is a total trust in people when it comes to simple things

In fact, the derivative of $y = \frac{1 + \ln x}{x}$ is $\frac{dy}{dx} = {\color{red}-}\frac{\ln x}{x^2}$.

The answer for the solution to the DE changes slightly as a consequence.