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1

### JEE Main 2021 (Online) 27th August Morning Shift

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

$${{dy} \over {dx}} = 2(y + 2\sin x - 5)x - 2\cos x$$ such that y(0) = 7. Then y($$\pi$$) is equal to :
A
$$2{e^{{\pi ^2}}} + 5$$
B
$${e^{{\pi ^2}}} + 5$$
C
$$3{e^{{\pi ^2}}} + 5$$
D
$$7{e^{{\pi ^2}}} + 5$$

## Explanation

$${{dy} \over {dx}} - 2xy = 2(2\sin x - 5)x - 2\cos x$$

IF = $${e^{ - {x^2}}}$$

So, $$y.{e^{ - {x^2}}} = \int {{e^{ - {x^2}}}(2x(2\sin x - 5) - 2\cos x)dx}$$

$$\Rightarrow y.{e^{ - {x^2}}} = {e^{ - {x^2}}}(5 - 2\sin x) + c$$

$$\Rightarrow y = 5 - 2\sin x + c.{e^{{x^2}}}$$

Given at x = 0, y = 7

$$\Rightarrow$$ 7 = 5 + c $$\Rightarrow$$ c = 2

So, $$y = 5 - 2\sin x + 2{e^{{x^2}}}$$

Now, at x = $$\pi$$,

y = 5 + 2$${e^{{x^2}}}$$
2

### JEE Main 2021 (Online) 26th August Evening Shift

Let y(x) be the solution of the differential equation 2x2 dy + (ey $$-$$ 2x)dx = 0, x > 0. If y(e) = 1, then y(1) is equal to :
A
0
B
2
C
loge 2
D
loge (2e)

## Explanation

$$2{x^2}dy + ({e^y} - 2x)dx = 0$$

$${{dy} \over {dx}} + {{{e^y} - 2x} \over {2{x^2}}} = 0 \Rightarrow {{dy} \over {dx}} + {{{e^y}} \over {2{x^2}}} - {1 \over x} = 0$$

$${e^{ - y}}{{dy} \over {dx}} - {{{e^{ - y}}} \over x} = - {1 \over {2{x^2}}} \Rightarrow$$ Put $${e^{ - y}} = z$$

$${{ - dz} \over {dx}} - {z \over x} = - {1 \over {2{x^2}}} \Rightarrow xdz + zdx = {{dx} \over {2x}}$$

$$d(xz) = {{dx} \over {2x}} \Rightarrow xz = {1 \over 2}{\log _e}x + c$$

$$x{e^{ - y}} = {1 \over 2}{\log _e}x + c$$, passes through (e, 1)

$$\Rightarrow C = {1 \over 2}$$

$$x{e^{ - y}} = {{{{\log }_e}ex} \over 2}$$

$${e^{ - y}} = {1 \over 2} \Rightarrow y = {\log _e}2$$
3

### JEE Main 2021 (Online) 26th August Morning Shift

Let y = y(x) be a solution curve of the differential equation $$(y + 1){\tan ^2}x\,dx + \tan x\,dy + y\,dx = 0$$, $$x \in \left( {0,{\pi \over 2}} \right)$$. If $$\mathop {\lim }\limits_{x \to 0 + } xy(x) = 1$$, then the value of $$y\left( {{\pi \over 4}} \right)$$ is :
A
$$- {\pi \over 4}$$
B
$${\pi \over 4} - 1$$
C
$${\pi \over 4} + 1$$
D
$${\pi \over 4}$$

## Explanation

$$(y + 1){\tan ^2}x\,dx + \tan x\,dy + y\,dx = 0$$

or $${{dy} \over {dx}} + {{{{\sec }^2}x} \over {\tan x}}.y = - \tan x$$

$$IF = {e^{\int {{{{{\sec }^2}x} \over {\tan x}}dx} }} = {e^{\ln \tan x}} = \tan x$$

$$\therefore$$ $$y\tan x = - \int {{{\tan }^2}x\,dx}$$

or $$y\tan x = - \tan x + x + C$$

or $$y = - 1 + {x \over {\tan x}} + {C \over {\tan x}}$$

or $$\mathop {\lim }\limits_{x \to 0} xy = - x + {{{x^2}} \over {\tan x}} + {{Cx} \over {\tan x}} = 1$$

or C = 1

$$y(x) = \cot x + x\cot x - 1$$

$$y\left( {{\pi \over 4}} \right) = {\pi \over 4}$$
4

### JEE Main 2021 (Online) 27th July Morning Shift

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

$${\log _{}}\left( {{{dy} \over {dx}}} \right) = 3x + 4y$$, with y(0) = 0.

If $$y\left( { - {2 \over 3}{{\log }_e}2} \right) = \alpha {\log _e}2$$, then the value of $$\alpha$$ is equal to :
A
$$- {1 \over 4}$$
B
$${1 \over 4}$$
C
$$2$$
D
$$- {1 \over 2}$$

## Explanation

$${{dy} \over {dx}} = {e^{3x}}.{e^{4y}} \Rightarrow \int {{e^{ - 4y}}dy = \int {{e^{3x}}dx} }$$

$${{{e^{ - 4y}}} \over { - 4}} = {{{e^{3x}}} \over 3} + C \Rightarrow - {1 \over 4} - {1 \over 3} = C \Rightarrow C = - {7 \over {12}}$$

$${{{e^{ - 4y}}} \over { - 4}} = {{{e^{3x}}} \over 3} - {7 \over {12}} \Rightarrow {e^{ - 4y}} = {{4{e^{3x}} - 7} \over { - 3}}$$

$${e^{4y}} = {3 \over {7 - 4{e^{3x}}}} \Rightarrow 4y = \ln \left( {{3 \over {7 - 4{e^{3x}}}}} \right)$$

$$4y = \ln \left( {{3 \over 6}} \right)$$ when $$x = - {2 \over 3}\ln 2$$

$$y = {1 \over 4}\ln \left( {{1 \over 2}} \right) = - {1 \over 4}\ln 2$$

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