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1

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

A differential equation representing the family of parabolas with axis parallel to y-axis and whose length of latus rectum is the distance of the point (2, $$-$$3) from the line 3x + 4y = 5, is given by :
A
$$10{{{d^2}y} \over {d{x^2}}} = 11$$
B
$$11{{{d^2}x} \over {d{y^2}}} = 10$$
C
$$10{{{d^2}x} \over {d{y^2}}} = 11$$
D
$$11{{{d^2}y} \over {d{x^2}}} = 10$$

## Explanation

Length of latus rectum

$$= {{|3(2) + 4( - 3) - 5|} \over 5} = {{11} \over 5}$$

$${(x - h)^2} = {{11} \over 5}(y - k)$$

differentiate w.r.t. 'x' :-

$$2(x - h) = {{11} \over 5}{{dy} \over {dx}}$$

again differentiate

$$2 = {{11} \over 5}{{{d^2}y} \over {d{x^2}}}$$

$${{11{d^2}y} \over {d{x^2}}} = 10$$
2

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

Let us consider a curve, y = f(x) passing through the point ($$-$$2, 2) and the slope of the tangent to the curve at any point (x, f(x)) is given by f(x) + xf'(x) = x2. Then :
A
$${x^2} + 2xf(x) - 12 = 0$$
B
$${x^3} + xf(x) + 12 = 0$$
C
$${x^3} - 3xf(x) - 4 = 0$$
D
$${x^2} + 2xf(x) + 4 = 0$$

## Explanation

$$y + {{xdy} \over {dx}} = {x^2}$$ (given)

$$\Rightarrow {{dy} \over {dx}} + {y \over x} = x$$

If $${e^{\int {{1 \over x}dx} }} = x$$

Solution of DE

$$\Rightarrow y\,.\,x = \int {x\,.\,x\,dx}$$

$$\Rightarrow xy = {{{x^3}} \over 3} + {c \over 3}$$

Passes through ($$-$$2, 2), so

$$-$$12 = $$-$$ 8 + c $$\Rightarrow$$ c = $$-$$ 4

$$\therefore$$ 3xy = x3 $$-$$ 4

i.e. 3x . f(x) = x3 $$-$$ 4
3

### 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$${e^{{x^2}}}$$
4

### 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)

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

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

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

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