1
JEE Main 2022 (Online) 29th June Evening Shift
Numerical
+4
-1
Change Language

Let y = y(x), x > 1, be the solution of the differential equation $$(x - 1){{dy} \over {dx}} + 2xy = {1 \over {x - 1}}$$, with $$y(2) = {{1 + {e^4}} \over {2{e^4}}}$$. If $$y(3) = {{{e^\alpha } + 1} \over {\beta {e^\alpha }}}$$, then the value of $$\alpha + \beta $$ is equal to _________.

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2
JEE Main 2022 (Online) 29th June Morning Shift
Numerical
+4
-1
Change Language

Let y = y(x) be the solution of the differential equation $${{dy} \over {dx}} + {{\sqrt 2 y} \over {2{{\cos }^4}x - {{\cos }^2}x}} = x{e^{{{\tan }^{ - 1}}(\sqrt 2 \cot 2x)}},\,0 < x < {\pi \over 2}$$ with $$y\left( {{\pi \over 4}} \right) = {{{\pi ^2}} \over {32}}$$. If $$y\left( {{\pi \over 3}} \right) = {{{\pi ^2}} \over {18}}{e^{ - {{\tan }^{ - 1}}(\alpha )}}$$, then the value of 3$$\alpha$$2 is equal to ___________.

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3
JEE Main 2022 (Online) 27th June Evening Shift
Numerical
+4
-1
Change Language

Let $$y = y(x)$$ be the solution of the differential equation $$(1 - {x^2})dy = \left( {xy + ({x^3} + 2)\sqrt {1 - {x^2}} } \right)dx, - 1 < x < 1$$, and $$y(0) = 0$$. If $$\int_{{{ - 1} \over 2}}^{{1 \over 2}} {\sqrt {1 - {x^2}} y(x)dx = k} $$, then k$$-$$1 is equal to _____________.

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4
JEE Main 2022 (Online) 26th June Morning Shift
Numerical
+4
-1
Change Language

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

$$(4 + {x^2})dy - 2x({x^2} + 3y + 4)dx = 0$$ pass through the origin. Then y(2) is equal to _____________.

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