Question

Solve linear diff. Equation

(Xlogx)dy/dx +y = 3x^3

Answer 1

I am assuming that log x is the natural log here?

Answer 2

\[xln(x) \frac{dy}{dx}+y=3x^3\]
\[xln(x) \frac{dy}{dx} f(x)+yf(x)=f(x)3x^3\]
Define
\[xln(x) f(x)=\frac{df(x)}{dx}\]

Use this to find explicit formula for f(x)
Then,
\[\frac{dy}{dx} f'(x)+yf(x)=f(x)3x^3\]
turns into
\[\frac{d(y*f(x))}{dx}=f(x)3x^3\]

Answer 3

\[yf(x)=\int\limits f(x)3x^3dx\]Divide by f(x)

Answer 4

All you have to do now is
(1) Understand what I did- ask if confused
(2) Find explicitly the formula for f(x) and solve for y

Answer 5

Sorry bout my weakness but i dont understand what is f(x)

Answer 6

I use integrating factor to solve this so i am not able to get e^integral(1/xlogx)

Answer 7

I believe with the aforementioned method by henpen, \(f(x)\) is acting as our integrating factor.

If we want to figure out the integrating factor using your formula:
\(\displaystyle \large{ e^{\int \frac{1}{x \log x} \; \text{d}x} } \)
We just need to figure out this integral:
\(\int \frac{1}{x \log x} \; \text{d}x \)
Which we should notice the log x and a 1/x in the integral. That looks good for u-substitution with u=log x to figure out.

We can also solve henpen's equation, which should also work out. :)

Answer 8

Thnx

Answer 9

You're welcome!

Answer 10

Is ther an ipad app for this amazing site?

Answer 11

Hm, I'm not sure there is, although I have heard talk about some people wanting one.