JEE Main 2020 (Online) 4th September Morning Slot | Differential Equations Question 149 | Mathematics

Let y = y(x) be the solution of the differential equation,
xy'- y = x²(xcosx + sinx), x > 0. if y ($$\pi $$) = $$\pi $$ then
$$y''\left( {{\pi \over 2}} \right) + y\left( {{\pi \over 2}} \right)$$ is equal to :

$$1 + {\pi \over 2}$$

$$2 + {\pi \over 2} + {{{\pi ^2}} \over 4}$$

$$2 + {\pi \over 2}$$

$$1 + {\pi \over 2} + {{{\pi ^2}} \over 4}$$

JEE Main 2020 (Online) 3rd September Evening Slot

MCQ (Single Correct Answer)

-1

If x³dy + xy dx = x²dy + 2y dx; y(2) = e and
x > 1, then y(4) is equal to :

$${{\sqrt e } \over 2}$$

$${1 \over 2} + \sqrt e $$

$${3 \over 2} + \sqrt e $$

$${3 \over 2}\sqrt e $$

JEE Main 2020 (Online) 3rd September Morning Slot

MCQ (Single Correct Answer)

-1

The solution curve of the differential equation,

(1 + e^-x)(1 + y²)$${{dy} \over {dx}}$$ = y²,

which passes through the point (0, 1), is :

y² + 1 = y$$\left( {{{\log }_e}\left( {{{1 + {e^{ - x}}} \over 2}} \right) + 2} \right)$$

y² + 1 = y$$\left( {{{\log }_e}\left( {{{1 + {e^{ x}}} \over 2}} \right) + 2} \right)$$

y² = 1 + $${y{{\log }_e}\left( {{{1 + {e^{ - x}}} \over 2}} \right)}$$

y² = 1 + $${y{{\log }_e}\left( {{{1 + {e^{ x}}} \over 2}} \right)}$$

JEE Main 2020 (Online) 2nd September Evening Slot

MCQ (Single Correct Answer)

-1

If a curve y = f(x), passing through the point (1, 2), is the solution of the differential equation,
2x²dy= (2xy + y²)dx, then $$f\left( {{1 \over 2}} \right)$$ is equal to :

$${1 \over {1 - {{\log }_e}2}}$$

$${1 \over {1 + {{\log }_e}2}}$$

$${{ - 1} \over {1 + {{\log }_e}2}}$$

$${1 + {{\log }_e}2}$$

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