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Onto Function

Suppose y = f...

Question

Suppose y = f(x) satisfies the differential equation $y d x + y^{2} d x = x d y$ . If y(x) > 0 for all x $ϵ$ R and y(1) = 1, then y (-3) =

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Solution

The correct option is C 3
$y d x + y^{2} d y = x d y y d x - x d y + y^{2} d y = 0 \Rightarrow \frac{y d x - x d y}{y^{2}} + d y = 0 \Rightarrow d (\frac{x}{y}) + d y = 0 \Rightarrow \frac{x}{y} + y = C W h e n x = 1, y = 1, 1 + 1 = C \Rightarrow c = 2 ∴ \frac{x}{y} + y = 2, w h e n x = 3, \frac{- 3}{y} + y = 2 \Rightarrow y^{2} - 2 y - 3 = 0 \Rightarrow y = 3$

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Similar questions

y d x + y^{2} d y = x d y; x \in R, y > 0, y (x), y (1) = - 1

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f : [0, 1] \to R

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x, y \in [0, 1]

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y (0) = 1, then y (\frac{1}{4}) + y (\frac{3}{4})

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