Thread: Help with finding general solution of differential equation

Hi

I think I've managed to work out the first part of a question which is

Use the Quotient Rule to differentiate the function h(x) = (1 + ln(x))/x (x>0)
I get the answer ln(x)/x^2.

Now the second part of the question is

Using your answer to the first part, find the general solution of the differential equation

dy/dx = -(ln(x)/x^2)y^1/2 (x > 0, y > 0) giving the answer in implicit form.


Can you help?

The equation given can be solved using seperation of variables, you can write it as
$\displaystyle
\int \frac{dy}{y^{\frac{1}{2}}}=\int -\frac{\ln x}{x^2}dx$

the integral on the LHS is trivial and can you see how to apply your answer to the first part to the RHS?

Thanks

Ok, then dy/y^1/2 = -ln(x)/x^2 dx

gives 1/(2/3)y^3/2 + a = -((1 + ln(x))/x) + b

which is 1/(2/3)y^3/2 = -((1 + ln(x))/x) + c

Is this right?

Originally Posted by flyingscotsman

Thanks

Ok, then dy/y^1/2 = -ln(x)/x^2 dx

gives 1/(2/3)y^3/2 + a = -((1 + ln(x))/x) + b

which is 1/(2/3)y^3/2 = -((1 + ln(x))/x) + c

Is this right?

Unfortunately not, no, but you're not too far off. You made two errors; one when finding the derivative of h(x), and another when trying to integrate the function of y.

Originally Posted by flyingscotsman

I think I've managed to work out the first part of a question which is

Use the Quotient Rule to differentiate the function h(x) = (1 + ln(x))/x (x>0)
I get the answer ln(x)/x^2.

$\displaystyle h(x)=\frac{1+\ln{x}}{x}$

$\displaystyle \frac{d}{dx}(\frac{u}{v})=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}$

$\displaystyle u=1+\ln{x}$

$\displaystyle \frac{du}{dx}=\frac{1}{x}$

$\displaystyle v=x$

$\displaystyle \frac{dv}{dx}=1$

$\displaystyle \frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}=\frac{x\frac{1}{x}-(1+\ln{x})}{x^2}$

This appears to be where you goofed:

$\displaystyle h'(x)=\frac{1-1-\ln{x}}{x^2}$

$\displaystyle h'(x)=-\frac{\ln{x}}{x^2}$

Now the second part of the question is

Using your answer to the first part, find the general solution of the differential equation

dy/dx = -(ln(x)/x^2)y^1/2 (x > 0, y > 0) giving the answer in implicit form.

$\displaystyle \frac{dy}{dx}=-\frac{\ln{x}}{x^2}\sqrt{y}$

$\displaystyle -\frac{\ln{x}}{x^2}dx=\frac{1}{\sqrt{y}}dy$

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

$\displaystyle h(x)=\int\frac{1}{\sqrt{y}}dy$

The above is fine, but to a beginner it may help to convert the square root to a fraction root:

$\displaystyle h(x)=\int y^{-\frac{1}{2}}dy$

Can you take it from there?

Hi

Thanks for the correction in the first part.

As for h(x) = y^-1/2 dy, is it 2y^1/2 after integration?

Therefore giving as the particular solution

2y^1/2 = -((1 + ln(x))/x) + c

Is this correct?

Originally Posted by flyingscotsman

Hi

Thanks for the correction in the first part.

As for h(x) = y^-1/2 dy, is it 2y^1/2 after integration?

Therefore giving as the particular solution

2y^1/2 = -((1 + ln(x))/x) + c

Is this correct?

Very close, but you're still forgetting your first error. When you find the derivative of h(x), you make a small mistake, which you need to fix.

Yes, I've made the mistake of including the minus.

It should be 2y^1/2 = (1 + lnx)/x + c, without the minus.
I hope that's it.

Originally Posted by flyingscotsman

Yes, I've made the mistake of including the minus.

It should be 2y^1/2 = (1 + lnx)/x + c, without the minus.
I hope that's it.

That is correct, yes.

Many thanks for all your help.

Originally Posted by hatsoff

That is correct, yes.

How did the lnX become 1+lnX again please?

Originally Posted by brainache

How did the lnX become 1+lnX again please?

It was in the original equation:

$\displaystyle
h(x)=\frac{1+\ln{x}}{x}
$

Which is the integral of part of the second equation:

$\displaystyle
\frac{dy}{dx}=-\frac{\ln{x}}{x^2}\sqrt{y}
$

Originally Posted by hatsoff

It was in the original equation:

$\displaystyle
h(x)=\frac{1+\ln{x}}{x}
$

Which is the integral of part of the second equation:

$\displaystyle
\frac{dy}{dx}=-\frac{\ln{x}}{x^2}\sqrt{y}
$

Thank you