Equation of curve passing through 3-9 which satisfies the differential equation dy - dx - x -1- x 2- isA- 6 x y-3 x2-6 x-29B- 6 x y-3 x3-29 x-6C- 6 xy -3 x 3-29 x -6D- None of these

The correct option is C 6xy=3x^3 +29 x 6dy/dx=x+1/x^2 ⇒∫ dy =∫(x+1/x^2 )dx ⇒ y =x^2/2 1/x+c Since it passes through (3, 9), therefore 9=9/2 1/3+c ⇒ c =29/6 ∴ ...

Equation of c...

Question

Equation of curve passing through (3, 9) which satisfies the differential equation $\frac{d y}{d x} = x + \frac{1}{x^{2}}$ , is

6 x y = 3 x^{2} - 6 x + 29

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6 x y = 3 x^{3} - 29 x + 6

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6 x y = 3 x^{3} + 29 x - 6

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Solution

The correct option is C $6 x y = 3 x^{3} + 29 x - 6$
$\frac{d y}{d x} = x + \frac{1}{x^{2}} \Rightarrow \int d y = \int (x + \frac{1}{x^{2}}) d x \Rightarrow y = \frac{x^{2}}{2} - \frac{1}{x} + c$
Since it passes through (3, 9), therefore
$9 = \frac{9}{2} - \frac{1}{3} + c \Rightarrow c = \frac{29}{6} ∴ y = \frac{x^{2}}{2} - \frac{1}{x} + \frac{29}{6} \Rightarrow 6 x y = 3 x^{3} + 29 x - 6$ .

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The correct option is C 6xy=3x3+29x−6dydx=x+1x2⇒∫dy=∫(x+1x2)dx⇒y=x22−1x+c Since it passes through (3, 9), therefore 9=92−13+c⇒c=296∴y=x22−1x+296⇒6xy=3x3+29x−6.

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