1
JEE Main 2021 (Online) 1st September Evening Shift
+4
-1
If y = y(x) is the solution curve of the differential equation $${x^2}dy + \left( {y - {1 \over x}} \right)dx = 0$$ ; x > 0 and y(1) = 1, then $$y\left( {{1 \over 2}} \right)$$ is equal to :
A
$${3 \over 2} - {1 \over {\sqrt e }}$$
B
$$3 + {1 \over {\sqrt e }}$$
C
3 + e
D
3 $$-$$ e
2
JEE Main 2021 (Online) 31st August Evening Shift
+4
-1
If $${{dy} \over {dx}} = {{{2^x}y + {2^y}{{.2}^x}} \over {{2^x} + {2^{x + y}}{{\log }_e}2}}$$, y(0) = 0, then for y = 1, the value of x lies in the interval :
A
(1, 2)
B
$$\left( {{1 \over 2},1} \right]$$
C
(2, 3)
D
$$\left( {0,{1 \over 2}} \right]$$
3
JEE Main 2021 (Online) 31st August Evening Shift
+4
-1
If $$y{{dy} \over {dx}} = x\left[ {{{{y^2}} \over {{x^2}}} + {{\phi \left( {{{{y^2}} \over {{x^2}}}} \right)} \over {\phi '\left( {{{{y^2}} \over {{x^2}}}} \right)}}} \right]$$, x > 0, $$\phi$$ > 0, and y(1) = $$-$$1, then $$\phi \left( {{{{y^2}} \over 4}} \right)$$ is equal to :
A
4 $$\phi$$ (2)
B
4$$\phi$$ (1)
C
2 $$\phi$$ (1)
D
$$\phi$$ (1)
4
JEE Main 2021 (Online) 31st August Morning Shift
+4
-1
If $${{dy} \over {dx}} = {{{2^{x + y}} - {2^x}} \over {{2^y}}}$$, y(0) = 1, then y(1) is equal to :
A
log2(2 + e)
B
log2(1 + e)
C
log2(2e)
D
log2(1 + e2)
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