JEE MAIN - Mathematics (2021 - 20th July Morning Shift - No. 11)
Let y = y(x) be the solution of the differential equation $${e^x}\sqrt {1 - {y^2}} dx + \left( {{y \over x}} \right)dy = 0$$, y(1) = $$-$$1. Then the value of (y(3))2 is equal to :
1 $$-$$ 4e3
1 $$-$$ 4e6
1 + 4e3
1 + 4e6
Explanation
$${e^x}\sqrt {1 - {y^2}} dx + {y \over x}dy = 0$$
$$ \Rightarrow {e^x}\sqrt {1 - {y^2}} dx + {{ - y} \over x}dy$$
$$ \Rightarrow \int {{{ - y} \over {\sqrt {1 - {y^2}} }}} dy = \int_{II}^{{e^x}} {_1^xdx} $$
$$ \Rightarrow \sqrt {1 - {y^2}} = {e^x}(x - 1) + c$$
Given : At x = 1, y = $$-$$1
$$\Rightarrow$$ 0 = 0 + c $$\Rightarrow$$ c = 0
$$\therefore$$ $$\sqrt {1 - {y^2}} = {e^x}(x - 1)$$
At x = 3
$$1 - {y^2} = {({e^3}2)^2} \Rightarrow {y^2} = 1 - 4{e^6}$$
$$ \Rightarrow {e^x}\sqrt {1 - {y^2}} dx + {{ - y} \over x}dy$$
$$ \Rightarrow \int {{{ - y} \over {\sqrt {1 - {y^2}} }}} dy = \int_{II}^{{e^x}} {_1^xdx} $$
$$ \Rightarrow \sqrt {1 - {y^2}} = {e^x}(x - 1) + c$$
Given : At x = 1, y = $$-$$1
$$\Rightarrow$$ 0 = 0 + c $$\Rightarrow$$ c = 0
$$\therefore$$ $$\sqrt {1 - {y^2}} = {e^x}(x - 1)$$
At x = 3
$$1 - {y^2} = {({e^3}2)^2} \Rightarrow {y^2} = 1 - 4{e^6}$$
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