For y > 0 and x ∈ R , y dx + y 2 dy = xdy where y = f ( x ) . If f ( 1 ) = 1 , then find f ( − 3 ) .

y 2 xdy − ydx = dy − ∫ d ( y x ) = ∫ d y ⇒ − y x = y + c At x = 1 , y − 1 , c = − 2 At x = − 3 , y 3 = y − 2 ⇒ y 2 − 2 y − 3 = 0 ∴ y = 3

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MathematicsDifferential EquationMiscellaneous/MixedMedium2 min

MathematicsMediuminteger

For $y > 0$ and $x \in R, y dx + y^{2} dy = xdy$ where $y = f (x)$ . If $f (1) = 1$ , then find $f (- 3)$ .

Answer:

Solution:

$\frac{xdy - ydx}{y ^{2}} = dy$

$- \int d (\frac{x}{y}) = \int d y$

$\Rightarrow - \frac{x}{y} = y + c$

At $x = 1, y - 1, c = - 2$

At $x = - 3, \frac{3}{y} = y - 2$

$\Rightarrow y^{2} - 2 y - 3 = 0$

$∴ y = 3$

Stream:JEESubject:MathematicsTopic:Differential EquationSubtopic:Miscellaneous/Mixed

⏱ 2m

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