1
JEE Main 2022 (Online) 25th July Morning Shift
+4
-1

The general solution of the differential equation $$\left(x-y^{2}\right) \mathrm{d} x+y\left(5 x+y^{2}\right) \mathrm{d} y=0$$ is :

A
$$\left(y^{2}+x\right)^{4}=\mathrm{C}\left|\left(y^{2}+2 x\right)^{3}\right|$$
B
$$\left(y^{2}+2 x\right)^{4}=C\left|\left(y^{2}+x\right)^{3}\right|$$
C
$$\left|\left(y^{2}+x\right)^{3}\right|=\mathrm{C}\left(2 y^{2}+x\right)^{4}$$
D
$$\left|\left(y^{2}+2 x\right)^{3}\right|=C\left(2 y^{2}+x\right)^{4}$$
2
JEE Main 2022 (Online) 30th June Morning Shift
+4
-1
Out of Syllabus

Let $${{dy} \over {dx}} = {{ax - by + a} \over {bx + cy + a}},\,a,b,c \in R$$, represents a circle with center ($$\alpha$$, $$\beta$$). Then, $$\alpha$$ + 2$$\beta$$ is equal to :

A
$$-$$1
B
0
C
1
D
2
3
JEE Main 2022 (Online) 29th June Evening Shift
+4
-1

If y = y(x) is the solution of the differential equation $$\left( {1 + {e^{2x}}} \right){{dy} \over {dx}} + 2\left( {1 + {y^2}} \right){e^x} = 0$$ and y (0) = 0, then $$6\left( {y'(0) + {{\left( {y\left( {{{\log }_e}\sqrt 3 } \right)} \right)}^2}} \right)$$ is equal to

A
2
B
$$-$$2
C
$$-$$4
D
$$-$$1
4
JEE Main 2022 (Online) 29th June Morning Shift
+4
-1

Let the solution curve of the differential equation

$$x{{dy} \over {dx}} - y = \sqrt {{y^2} + 16{x^2}}$$, $$y(1) = 3$$ be $$y = y(x)$$. Then y(2) is equal to:

A
15
B
11
C
13
D
17
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