JEE Main 2021 (Online) 17th March Evening Shift | Differential Equations Question 130 | Mathematics

If the curve y = y(x) is the solution of the differential equation

$$2({x^2} + {x^{5/4}})dy - y(x + {x^{1/4}})dx = {2x^{9/4}}dx$$, x > 0 which

passes through the point $$\left( {1,1 - {4 \over 3}{{\log }_e}2} \right)$$, then the value of y(16) is equal to :

$$4\left( {{{31} \over 3} - {8 \over 3}{{\log }_e}3} \right)$$

$$\left( {{{31} \over 3} - {8 \over 3}{{\log }_e}3} \right)$$

$$\left( {{{31} \over 3} + {8 \over 3}{{\log }_e}3} \right)$$

$$4\left( {{{31} \over 3} + {8 \over 3}{{\log }_e}3} \right)$$

JEE Main 2021 (Online) 17th March Evening Shift

MCQ (Single Correct Answer)

-1

Let y = y(x) be the solution of the differential equation

$$\cos x(3\sin x + \cos x + 3)dy = (1 + y\sin x(3\sin x + \cos x + 3))dx,0 \le x \le {\pi \over 2},y(0) = 0$$. Then, $$y\left( {{\pi \over 3}} \right)$$ is equal to :

$$2{\log _e}\left( {{{\sqrt 3 + 7} \over 2}} \right)$$

$$2{\log _e}\left( {{{3\sqrt 3 - 8} \over 4}} \right)$$

$$2{\log _e}\left( {{{2\sqrt 3 + 10} \over {11}}} \right)$$

$$2{\log _e}\left( {{{2\sqrt 3 + 9} \over 6}} \right)$$

JEE Main 2021 (Online) 17th March Morning Shift

MCQ (Single Correct Answer)

-1

Which of the following is true for y(x) that satisfies the differential equation

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

y(1) = 1

y(1) = e^{$$-$$$${1 \over 2}$$} $$-$$ 1

y(1) = e^{$${1 \over 2}$$} $$-$$ e^{$$-$$$${1 \over 2}$$}

y(1) = e^{$${1 \over 2}$$} $$-$$ 1

JEE Main 2021 (Online) 16th March Evening Shift

MCQ (Single Correct Answer)

-1

If y = y(x) is the solution of the differential equation

$${{dy} \over {dx}}$$ + (tan x) y = sin x, $$0 \le x \le {\pi \over 3}$$, with y(0) = 0, then $$y\left( {{\pi \over 4}} \right)$$ equal to :

$${1 \over 2}$$log_e 2

$$\left( {{1 \over {2\sqrt 2 }}} \right)$$ log_e 2

log_e 2

$${1 \over 4}$$ log_e 2

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