Thanks very much!

1) Find the general solution of the differential equation: $xy(\mathrm{dy}/\mathrm{dx})=\ln x$

2) Find the particular solution of: $e^x(\mathrm{dy}/\mathrm{dx})=\sqrt{y}$ given that $y=9$ when $x=0$

  Thanks again.

For (1), we have:

$y(\mathrm{dy}/\mathrm{dx})=(\ln x)/x$

Integrating wrt $x$ (and using the chain rule),

$\int y\mathrm{dy}=\int (\ln x)/x\mathrm{dx}$

Now notice that we've got $\ln x$ and its derivative being multiplied on the RHS, and use the substitution $u=\ln x$.

Thus,

$\int y\mathrm{dy}=\int u\mathrm{du}$ (Can you see why?)

Thus,

$y^2/2=u^2/2+C$

ie. $y^2/2=(\ln x{)}^2/2+C$

[If you're picky you can change it to the equivalent and neater form y² = (ln x)² + A, where A is an arbitrary constant]

You should now be able to do (2) yourself as it follows along similar lines. By all means ask if you get stuck or if you didn't understand any of my post...

Marcos

So, in this case we need to get rid of the $e^x$ on the LHS and the $\sqrt{y}$ on the RHS.

$e^x(\mathrm{dy}/\mathrm{dx})=\sqrt{y}$

So we get: $y^{-1/2}\mathrm{dy}/\mathrm{dx}=e^{-x}$ (Right?)

Integrating wrt $x$,

$\int y^{-1/2}\mathrm{dy}=\int e^{-x}\mathrm{dx}$

That is,

$2y^{1/2}=-e^{-x}+C$

Can you finish it off now using the conditions you're given? (I get $C=7$)

Marcos

let's take a look at a simpler example (this can be done in a couple of ways but we'll do it using substitutions):

$\int x{e}^{x^2}\mathrm{dx}$

We can use the substitution $u=x^2$

Thus $\mathrm{dx}/\mathrm{du}=1/(\mathrm{du}/\mathrm{dx})=1/(2x)$

We can now rewrite our integral:

$\int x{e}^{x^2}\mathrm{dx}=\int x\times e^u\times 1/(2x)\mathrm{du}$ (Make sure you understand this line)

$=\int e^u/2\mathrm{du}$

(You can then finish this off and substitute $u=x^2$ back into the equation)

The important thing is that it is simpler to integrate $\int e^u/2\mathrm{du}$ rather than $\int x{e}^{x^2}\mathrm{dx}$, that was the whole point of the substitution...

Now, in your example we had:

$\int (\ln x)/x\mathrm{dx}$

Say we substitute $u=(\ln x)/x$.

Well, we need to figure out what $\mathrm{dx}/\mathrm{du}$ is (as we are going to multiply by it when changing the integration variable)

Now, it's quitedifficult (I think impossible - someone may care to correct me) to express $\mathrm{dx}/\mathrm{du}$ only in terms of $u$ (which would be ideal as we're going to integrate wrt $u$) but we can do what we did in the previous example (ie. express it in terms of $x$ and hopefully the terms in '' $x$'' will cancel).

Now, as Mark has already said,

$\mathrm{dx}/\mathrm{du}=1/(\mathrm{du}/\mathrm{dx})=1/[(1-\ln x)/x^2]=x^2/(1-\ln x)$

Thus in our integral:

$\int (\ln x)/x\mathrm{dx}=\int u\times x^2/(1-\ln x)\mathrm{du}$

This has definitely not simplified as we've got both $u$ and $x$ (which isn't a constant) to integrate wrt $u$ and this is harder to integrate than the original function $(\ln x)/x$, contrary to what we're trying to do...

That's why in this example it's easier to use $u=\ln x$:

If you notice that $1/x$ is the derivative of $\ln x$, you can immediately substitute $u$ for $\ln x$ and $\mathrm{du}/\mathrm{dx}$ for $1/x$:

$\int (\ln x)/x\mathrm{dx}=\int u\times \mathrm{du}/\mathrm{dx}\mathrm{dx}=\int u\mathrm{du}$

I hope things are clearer now. If you have any questions or don't agree with some point I've made please ask/say so...

Marcos

Minor note:
(1) wrt = with respect to
(2) a^b = a^b
(3) a*b = a multiplied by b