1
JEE Main 2023 (Online) 25th January Morning Shift
+4
-1

Let $$y = y(x)$$ be the solution curve of the differential equation $${{dy} \over {dx}} = {y \over x}\left( {1 + x{y^2}(1 + {{\log }_e}x)} \right),x > 0,y(1) = 3$$. Then $${{{y^2}(x)} \over 9}$$ is equal to :

A
$${{{x^2}} \over {5 - 2{x^3}(2 + {{\log }_e}{x^3})}}$$
B
$${{{x^2}} \over {3{x^3}(1 + {{\log }_e}{x^2}) - 2}}$$
C
$${{{x^2}} \over {7 - 3{x^3}(2 + {{\log }_e}{x^2})}}$$
D
$${{{x^2}} \over {2{x^3}(2 + {{\log }_e}{x^3}) - 3}}$$
2
JEE Main 2023 (Online) 24th January Evening Shift
+4
-1

Let $$y=y(x)$$ be the solution of the differential equation $$(x^2-3y^2)dx+3xy~dy=0,y(1)=1$$. Then $$6y^2(e)$$ is equal to

A
$$\frac{3}{2}\mathrm{e}^2$$
B
$$3\mathrm{e}^2$$
C
$$\mathrm{e}^2$$
D
$$2\mathrm{e}^2$$
3
JEE Main 2023 (Online) 24th January Morning Shift
+4
-1

Let $$y = y(x)$$ be the solution of the differential equation $${x^3}dy + (xy - 1)dx = 0,x > 0,y\left( {{1 \over 2}} \right) = 3 - \mathrm{e}$$. Then y (1) is equal to

A
2 $$-$$ e
B
3
C
1
D
e
4
JEE Main 2022 (Online) 29th July Evening Shift
+4
-1

If the solution curve of the differential equation $$\frac{d y}{d x}=\frac{x+y-2}{x-y}$$ passes through the points $$(2,1)$$ and $$(\mathrm{k}+1,2), \mathrm{k}>0$$, then

A
$$2 \tan ^{-1}\left(\frac{1}{k}\right)=\log _{e}\left(k^{2}+1\right)$$
B
$$\tan ^{-1}\left(\frac{1}{k}\right)=\log _{e}\left(k^{2}+1\right)$$
C
$$2 \tan ^{-1}\left(\frac{1}{k+1}\right)=\log _{e}\left(k^{2}+2 k+2\right)$$
D
$$2 \tan ^{-1}\left(\frac{1}{k}\right)=\log _{e}\left(\frac{k^{2}+1}{k^{2}}\right)$$
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