COMEDK 2021 | Differential Equations Question 12 | Mathematics

The solution of the differential equation $$y{{dy} \over {dx}} = x\left[ {{{{y^2}} \over {{x^2}}} + {{\phi \left( {{{{y^2}} \over {{x^2}}}} \right)} \over {\phi '\left( {{{{y^2}} \over {{x^2}}}} \right)}}} \right]$$ is (where, C is a constant)

$$\phi \left( {{{{y^2}} \over {{x^2}}}} \right) = Cx$$

$$x\phi \left( {{{{y^2}} \over {{x^2}}}} \right) = C$$

$$\phi \left( {{{{y^2}} \over {{x^2}}}} \right) = C{x^2}$$

$${x^2}\phi \left( {{{{y^2}} \over {{x^2}}}} \right) = C$$

COMEDK 2021

MCQ (Single Correct Answer)

-0

The solution of the differential equation $$(1 + {y^2}) + (x - {e^{{{\tan }^{ - 1}}y}}){{dy} \over {dx}} = 0$$ is

$$2x{e^{{{\tan }^{ - 1}}y}} = {e^{2{{\tan }^{ - 1}}y}} + C$$

$$x{e^{{{\tan }^{ - 1}}y}} = {\tan ^{ - 1}}y + C$$

$$x{e^{2{{\tan }^{ - 1}}y}} = {e^{{{\tan }^{ - 1}}y}} + C$$

$$(x - 2) = C{e^{ - {{\tan }^{ - 1}}y}}$$

COMEDK 2020

MCQ (Single Correct Answer)

-0

The differential equation of the family of straight lines whose slope is equal to y-intercept is

$$(x + 1){{dy} \over {dx}} + y = 0$$

$$(x + 1){{dy} \over {dx}} - y = 0$$

$${{dy} \over {dx}} = {{x + 1} \over {y + 1}}$$

$${{dy} \over {dx}} = {{x - 1} \over {y - 1}}$$

COMEDK 2020

MCQ (Single Correct Answer)

-0

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

2, 1

1, 5

2, 3

2, 5

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