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1
AIEEE 2005
MCQ (Single Correct Answer)
+4
-1
If $$x{{dy} \over {dx}} = y\left( {\log y - \log x + 1} \right),$$ then the solution of the equation is :
A
$$y\log \left( {{x \over y}} \right) = cx$$
B
$$x\log \left( {{y \over x}} \right) = cy$$
C
$$\log \left( {{y \over x}} \right) = cx$$
D
$$\log \left( {{x \over y}} \right) = cy$$
2
AIEEE 2004
MCQ (Single Correct Answer)
+4
-1
Out of Syllabus
The differential equation for the family of circle $${x^2} + {y^2} - 2ay = 0,$$ where a is an arbitrary constant is :
A
$$\left( {{x^2} + {y^2}} \right)y' = 2xy$$
B
$$2\left( {{x^2} + {y^2}} \right)y' = xy$$
C
$$\left( {{x^2} - {y^2}} \right)y' =2 xy$$
D
$$2\left( {{x^2} - {y^2}} \right)y' = xy$$
3
AIEEE 2004
MCQ (Single Correct Answer)
+4
-1
Solution of the differential equation $$ydx + \left( {x + {x^2}y} \right)dy = 0$$ is
A
$$log$$ $$y=Cx$$
B
$$ - {1 \over {xy}} + \log y = C$$
C
$${1 \over {xy}} + \log y = C$$
D
$$ - {1 \over {xy}} = C$$
4
AIEEE 2003
MCQ (Single Correct Answer)
+4
-1
The solution of the differential equation

$$\left( {1 + {y^2}} \right) + \left( {x - {e^{{{\tan }^{ - 1}}y}}} \right){{dy} \over {dx}} = 0,$$ is :
A
$$x{e^{2{{\tan }^{ - 1}}y}} = {e^{{{\tan }^{ - 1}}y}} + k$$
B
$$\left( {x - 2} \right) = k{e^{2{{\tan }^{ - 1}}y}}$$
C
$$2x{e^{{{\tan }^{ - 1}}y}} = {e^{2{{\tan }^{ - 1}}y}} + k$$
D
$$x{e^{{{\tan }^{ - 1}}y}} = {\tan ^{ - 1}}y + k$$
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