BITSAT 2021 | Differential Equations Question 2 | Mathematics

Solution of $$\left( {{{x + y - 1} \over {x + y - 2}}} \right){{dy} \over {dx}} = \left( {{{x + y + 1} \over {x + y + 2}}} \right)$$, given that y = 1 when x = 1 is

$$\ln \left| {{{{{(x - y)}^2} - 2} \over 2}} \right| = 2(x + y)$$

$$\ln \left| {{{{{(x + y)}^2} - 2} \over 2}} \right| = 2(x - y)$$

$$\ln \left| {{{{{(x - y)}^2} + 2} \over 2}} \right| = 2(x + y)$$

$$\ln \left| {{{{{(x + y)}^2} + 2} \over 2}} \right| = 2(x + y)$$

BITSAT 2021

MCQ (Single Correct Answer)

-1

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

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

$$\sin y = {e^x}(x - 1){x^{ - 4}}$$

$$\tan y = (x - 1){e^x}{x^{ - 3}}$$

$$\sin y = {e^x}(x - 1){x^3}$$

BITSAT 2020

MCQ (Single Correct Answer)

-1

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

$$2y = \sin y(1 - 2c{x^2})$$

$$2x = \cot y(1 + 2c{x^2})$$

$$2x = \sin y(1 - 2c{x^2})$$

$$2x\sin y = 1 - 2c{x^2}$$

BITSAT 2020

MCQ (Single Correct Answer)

-1

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

$$9{x^8} + 4{x^9}{y^4} = 9{y^4}C$$

$$9{x^8} - 4{x^9}{y^4} - 9{y^4}C = 0$$

$${x^8}(9 + 4{y^4}) = 10{y^4}C$$

None of these

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