DIFFERENTIAL( HELP REQUIRED )

1) f'(x) = f(x) - ln x + 1/x
2) ( 1 + xâˆš(x² + y²))dx + ( -1 + âˆš(x² + y²) )y dy = 0
3) 2x³y dy +( 1 - y²)(x²y² + y² -1) dx = 0

[find the soln. of the above differential eqn.]

[ PLEAZ DON'T ASK ME ANS. BECOZ I DON'T HAVE THEM ]

5 Answers

f'(x) = f(x) - ln x + 1/x

multiply by e^-x

Why this came to my mind is because 1/x is the derivative of ln x and f'(x) of f(x)
moreover there was a - sign involved...

\\1/x-\ln x=f(x)-f'(x) \\e^{-x}(1/x-\ln x)=e^{-x}(f(x)-f'(x)) \\\frac{d}{dx}e^{-x}(\ln x)=\frac{d}{dx}e^{-x}(f(x)) \\e^{-x}\ln x=e^{-x}f(x)+c

341

Hari Shankar ·2010-03-10 06:34:01

Or write it as f'(x) - \frac{1}{x} = f(x) - \ln x

So if g(x) = f(x) - \ln x, then the equation is just g'(x) = g(x)

what abt the others sir.

2nd- Its an exact differential equation.Now solve.

wat abt third

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