JEE MAIN - Mathematics (2017 - 8th April Morning Slot - No. 20)
The curve satisfying the differential equation, ydx $$-$$(x + 3y2)dy = 0 and passing through the point (1, 1), also passes through the point :
$$\left( {{1 \over 4}, - {1 \over 2}} \right)$$
$$\left( { - {1 \over 3},{1 \over 3}} \right)$$
$$\left( {{1 \over 3}, - {1 \over 3}} \right)$$
$$\left( {{1 \over 4}, {1 \over 2}} \right)$$
Explanation
Given,
y dx = $$\left( {x + 3{y^2}} \right)dy$$
$$ \Rightarrow $$$$\,\,\,$$ y $${{dx} \over {dy}}$$ = x + 3y2
$$ \Rightarrow $$$$\,\,\,$$ $${{dx} \over {dy}}$$ $$-$$ $${x \over y} = 3y$$
If = $${e^{ - \int {{1 \over y}dy} }}$$ = $${e^{ - \ln y}}$$ = $${1 \over y}$$
$$\therefore\,\,\,$$ Soluation is ,
x . $${1 \over y}$$ = $$\int {3y.{1 \over y}dy} $$
$$ \Rightarrow $$$$\,\,\,$$ $${x \over y}$$ = 3y + c
This curve passing through (1, 1)
$$\therefore\,\,\,$$ 1 = 3 + c
$$ \Rightarrow $$$$\,\,\,$$ c = $$-$$ 2
$$\therefore\,\,\,$$ Curve is, x = 3y2 $$-$$ 2y
Now put every point in this equation, and see which point satisfy this equation.
Following this method you can see ($$-$$ $${1 \over 3}$$, $${1 \over 3}$$) point satisfy this equation.
y dx = $$\left( {x + 3{y^2}} \right)dy$$
$$ \Rightarrow $$$$\,\,\,$$ y $${{dx} \over {dy}}$$ = x + 3y2
$$ \Rightarrow $$$$\,\,\,$$ $${{dx} \over {dy}}$$ $$-$$ $${x \over y} = 3y$$
If = $${e^{ - \int {{1 \over y}dy} }}$$ = $${e^{ - \ln y}}$$ = $${1 \over y}$$
$$\therefore\,\,\,$$ Soluation is ,
x . $${1 \over y}$$ = $$\int {3y.{1 \over y}dy} $$
$$ \Rightarrow $$$$\,\,\,$$ $${x \over y}$$ = 3y + c
This curve passing through (1, 1)
$$\therefore\,\,\,$$ 1 = 3 + c
$$ \Rightarrow $$$$\,\,\,$$ c = $$-$$ 2
$$\therefore\,\,\,$$ Curve is, x = 3y2 $$-$$ 2y
Now put every point in this equation, and see which point satisfy this equation.
Following this method you can see ($$-$$ $${1 \over 3}$$, $${1 \over 3}$$) point satisfy this equation.
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