1
WB JEE 2008
+1
-0.25

The solution of the differential equation $${{dy} \over {dx}} = {e^{y + x}} + {e^{y - x}}$$ is

A
$${e^{ - y}} = {e^x} - {e^{ - x}} + c,\,c$$ integrating constant
B
$${e^{ - y}} = {e^{ - x}} - {e^x} + c,\,c$$ integrating constant
C
$${e^{ - y}} = {e^x} + {e^{ - x}} + c,\,c$$ integrating constant
D
$${e^{ - y}} + {e^x} + {e^{ - x}} = c,\,c$$ integrating constant
2
WB JEE 2008
+1
-0.25

The order and degree of the following differential equation $${\left[ {1 + {{\left( {{{dy} \over {dx}}} \right)}^2}} \right]^{5/2}} = {{{d^3}y} \over {d{x^3}}}$$ are respectively

A
3, 2
B
3, 10
C
2, 3
D
3, 5
3
WB JEE 2008
+1
-0.25

The differential equation of the family of circles passing through the fixed points (a, 0) and ($$-$$a, 0) is

A
y1(y2 $$-$$ x2) + 2xy + a2 = 0
B
y1y2 + xy + a2x2 = 0
C
y1(y2 $$-$$ x2 + a2) + 2xy = 0
D
y1(y2 + x2) $$-$$ 2xy + a2 = 0
4
WB JEE 2008
+1
-0.25

The differential equation of the family of curves $$y = {e^{2x}}(a\cos x + b\sin x)$$, where a and b are arbitrary constants, is given by

A
y2 $$-$$ 4y1 + 5y = 0
B
2y2 $$-$$ y1 + 5y = 0
C
y2 + 4y1 $$-$$ 5y = 0
D
y2 $$-$$ 2y1 + 5y = 0
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