If a curve y = f(x) passes through the point | vedclass

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Question

(NONE) Given the differential equation: $y(1 + xy)dx = xdy$.
Rearranging the terms: $ydx + xy^2 dx = xdy$.
$xy^2 dx = xdy - ydx$.
Dividing both sid

Vedclass · Accepted Answer

$\frac{4}{11}$

Vedclass · Answer

(NONE) Given the differential equation: $y(1 + xy)dx = xdy$.
Rearranging the terms: $ydx + xy^2 dx = xdy$.
$xy^2 dx = xdy - ydx$.
Dividing both sides by $xy^2$ (assuming $x, y 
eq 0$): $dx = \frac{xdy - ydx}{xy^2} = \frac{1}{x} \cdot \frac{xdy - ydx}{y^2}$.
This can be written as: $dx = \frac{1}{x} d(\frac{x}{y})$.
Integrating both sides: $\int dx = \int \frac{1}{x} d(\frac{x}{y})$ is not the direct path. Let us rewrite: $\frac{xdy - ydx}{y^2} = x dx$.
This is $d(\frac{x}{y}) = x dx$.
Integrating both sides: $\frac{x}{y} = \frac{x^2}{2} + C$.
Since the curve passes through $(1, -1)$,substitute $x=1, y=-1$: $\frac{1}{-1} = \frac{1^2}{2} + C$.
$-1 = \frac{1}{2} + C \Rightarrow C = -\frac{3}{2}$.
So,$\frac{x}{y} = \frac{x^2 - 3}{2} \Rightarrow y = \frac{2x}{x^2 - 3}$.
Now,find $f(-\frac{1}{2})$: $f(-\frac{1}{2}) = \frac{2(-\frac{1}{2})}{(-\frac{1}{2})^2 - 3} = \frac{-1}{\frac{1}{4} - 3} = \frac{-1}{-\frac{11}{4}} = \frac{4}{11}$.

If a curve $y = f(x)$ passes through the point $(1, -1)$ and satisfies the differential equation $y(1 + xy)dx = xdy$,then $f(-\frac{1}{2}) = $ . . . . .

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