1
BITSAT 2021
+3
-1

The solution of $${x^3}{{dy} \over {dx}} + 4{x^2}\tan y = {e^x}\sec y$$ satisfying y (1) = 0, is

A
$$\tan y = (x - 2){e^x}\log x$$
B
$$\sin y = {e^x}(x - 1){x^{ - 4}}$$
C
$$\tan y = (x - 1){e^x}{x^{ - 3}}$$
D
$$\sin y = {e^x}(x - 1){x^3}$$
2
BITSAT 2020
+3
-1

The solution of the equation $${{dy} \over {dx}} + {1 \over x}\tan y = {1 \over {{x^2}}}\tan y\sin y$$ is

A
$$2y = \sin y(1 - 2c{x^2})$$
B
$$2x = \cot y(1 + 2c{x^2})$$
C
$$2x = \sin y(1 - 2c{x^2})$$
D
$$2x\sin y = 1 - 2c{x^2}$$
3
BITSAT 2020
+3
-1

The solution of differential equation $$(x{y^5} + 2y)dx - xdy = 0$$, is

A
$$9{x^8} + 4{x^9}{y^4} = 9{y^4}C$$
B
$$9{x^8} - 4{x^9}{y^4} - 9{y^4}C = 0$$
C
$${x^8}(9 + 4{y^4}) = 10{y^4}C$$
D
None of these
4
BITSAT 2020
+3
-1

A curve passes through (2, 0) and the slope of the tangent at P(x, y) is equal to $${{{{(x + 1)}^2} + y - 3} \over {x + 1}}$$ then the equation of the curve is

A
y = x2 $$-$$ 2x
B
y = x3 $$-$$ 8
C
y2 = x2 + 2x
D
y2 = 5x2 $$-$$ 6
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