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Differential Equation

87266 Let the solution curve of the differential equation
$x \frac{d y}{d x} - y = \sqrt{y^{2} + 16 x^{2}}, y (1) = 3 be y = y (x)$
Then $y (2)$ is equal to :

1 15

2 11

3 13

4 17

Explanation:

(A) : Differential equation is,
$x \frac{d y}{d x} - y = \sqrt{y^{2} + 16 x^{2}} \Rightarrow x \frac{d y}{d x} = y + \sqrt{y^{2} + 16 x^{2}}$
$\frac{d y}{d x} = \frac{y + \sqrt{y^{2} + 16 x^{2}}}{x}$
Differentiating w.r.t.x, we get -
$\frac{d y}{d x} = v + x \frac{d v}{d x}$
Then equation (i) becomes-
$v + x \frac{d v}{d x} = \frac{v x + \sqrt{v^{2} x^{2} + 16 x^{2}}}{x} \Rightarrow v + x \frac{d v}{d x} = v + \sqrt{v^{2} + 16}$
$x \frac{d v}{d x} = \sqrt{v^{2} + 16}$
Integration on both side, we get -
$\int \frac{dv}{\sqrt{v^{2} + 16}} = \int \frac{dx}{x} \Rightarrow \log | v + \sqrt{v^{2} + 16} | = \log x + \log c$ $v + \sqrt{v^{2} + 16} = xc$
Put $v = \frac{y}{x}$ , we get
$\frac{y}{x} + \sqrt{\frac{y^{2}}{x^{2}} + 16} = x c \Rightarrow y + \sqrt{y^{2} + 16 x^{2}} = c x^{2}$
Given, $y (1) = 3$ then $x = 1$ they $y = 3$
From equation (ii),
$3 + \sqrt{9 + 16 \times 1^{2}} = c (1)^{2} \Rightarrow 3 + \sqrt{25} = c$
$c = 8$
We find $y (2)$ then, $x = 2$
From equation (ii),
$y + \sqrt{y^{2} + 16 \times 4} = 8 \times 4 \Rightarrow \sqrt{y^{2} + 64} = 32 - y$
Squaring both side we get
$y^{2} + 64 = (32 - y)^{2}$
$y^{2} + 64 = 32 \times 32 + y^{2} - 64 y$
$64 (y + 1) = 32 \times 32$
$y + 1 = 16$
$y = 15$

JEE Main-29.06.2022

Differential Equation

87267 What is the solution of the differential equation $(x + 2 y^{3}) \frac{d y}{d x} = y ?$

1

x = y^{3} + c y

2

x^{2} = y^{3} + c y^{2}

3

x^{2} = y^{3} + c y

4 None of the above

Explanation:

(A) : Given,
$(x + 2 y^{3}) \frac{d y}{d x} = y$
$\frac{d x}{d y} = \frac{x + 2 y^{3}}{y} = \frac{x}{y} + \frac{2 y^{3}}{y}$
$\frac{d x}{d y} - \frac{x}{y} = 2 y^{2}$ (This is linear differential equation)
$\frac{dx}{dy} + Px = Q$
Where, $P = - \frac{1}{y}, Q = 2 y^{2}$
$x ($ I.F. $) = \int Q ($ I.F. $) d y + c$ -------(i)
so, I.F. $= e^{\int pdy} = e^{- \int \frac{dy}{y}}$
I.F. $= e^{- \log y} = e^{\log y^{- 1}}$
I.F. $= \frac{1}{y}$
So, from equation (i) :-
$\frac{x}{y} = \int 2 y^{2}, \frac{1}{y} d y + c = 2 \int y d y + c$
$\frac{x}{y} = 2 [\frac{y^{2}}{2}] + c = y^{2} + c$
$x = y^{3} + cy$

SCRA-2010

Differential Equation

87268 Let $y = y (x)$ be a solution curve of the differential $(1 - x^{2} y^{2}) d x = y d x + x d y$ . If the line $x = 1$ intersects the curve $y = y (x)$ at $y$ $= 2$ and the line $x = 2$ intersects the curve $y =$ $y (x)$ at $y = α$ , then a value of $α$ is

1

\frac{3 e^{2}}{2 (3 e^{2} - 1)}

2

\frac{3 e^{2}}{2 (3 e^{2} + 1)}

3

\frac{1 - 3 e^{2}}{2 (3 e^{2} + 1)}

4

\frac{1 + 3 e^{2}}{2 (3 e^{2} - 1)}

Explanation:

(D) : $(1 - x^{2} y^{2}) d x = y d x + x d y, y (1) = 2$
$y (2) = α = ?$
$d x = \frac{d (x y)}{1 - (x y)^{2}} \Rightarrow \int d x = \int \frac{d (x y)}{1 - (x y)^{2}}$
$x = \frac{1}{2} \ln | \frac{1 + x y}{1 - x y} | + C$
Put $x = 1$ and $y = 2$
$1 = \frac{1}{2} \ln | \frac{1 + 2}{1 - 2} | + C \Rightarrow C = 1 - \frac{1}{2} \ln 3$
Now put $x = 2$ and $y = α$
$2 = \frac{1}{2} \ln | \frac{1 + 2 α}{1 - 2 α} | + 1 - \frac{1}{2} \ln 3$
$1 + \frac{1}{2} \ln 3 = \frac{1}{2} \ln | \frac{1 + 2 α}{1 - 2 α} |$
$2 + \ln 3 = \ln (\frac{1 + 2 α}{1 - 2 α})$
$| \frac{1 + 2 α}{1 - 2 α} | = 3 e^{2} \Rightarrow \frac{1 + 2 α}{1 - 2 α} = 3 e^{2}, - 3 e^{2}$
$\frac{1 + 2 α}{1 - 2 α} = 3 e^{2} \Rightarrow α = \frac{3 e^{2} - 1}{2 (3 e^{2} + 1)}$
And $\frac{1 + 2 α}{1 - 2 α} = - 3 e^{2}$
$α = \frac{3 e^{2} + 1}{2 (3 e^{2} - 1)}$

JEE Main-11.04.2023

Differential Equation

87269 Let $y = y (x)$ be the solution of the differential equation $x (1 - x^{2}) \frac{d y}{d x} + (3 x^{2} y - y - 4 x^{3}) = 0, x >$ 1 , with $y (2) = - 2$ . Then $y (3)$ is equal to

1 -18

2 -12

3 -6

4 -3

Explanation:

(A) : $x (1 - x^{2}) \frac{d y}{d x} + (3 x^{2} y - y - 4 x^{3}) = 0$
$x (1 - x^{2}) \frac{d y}{d x} + (3 x^{2} - 1) y = 4 x^{3}$
$\frac{d y}{d x} + \frac{(3 x^{2} - 1)}{(x - x^{3})} y = \frac{4 x^{3}}{(x - x^{3})} \Rightarrow \frac{d y}{d x} + P y = Q$
$I F = e^{\int P d x} = e^{\int \frac{3 x^{2} - 1}{x - x^{3}} d x} \Rightarrow x - x^{3} = t \Rightarrow I F = e^{\int \frac{- d t}{t}}$
$= e^{- \ln t} = \frac{1}{t}$
$∴ IF = \frac{1}{x - x^{3}}$
$y \times I F = \int Q \times I F d x$
$y (\frac{1}{x - x^{3}}) = \int \frac{4 x^{3}}{x - x^{3}} \times \frac{1}{(x - x^{3})} d x = \int \frac{4 x^{3}}{{(x - x^{3})}^{2}} d x$
$= \int \frac{4 x}{{(1 - x^{2})}^{2}} d x$
$= 2 \int \frac{- d K}{K^{2}} 1 - x^{2} = K$

JEE Main-28.06.2022

Differential Equation

87266 Let the solution curve of the differential equation
$x \frac{d y}{d x} - y = \sqrt{y^{2} + 16 x^{2}}, y (1) = 3 be y = y (x)$
Then $y (2)$ is equal to :

1 15

2 11

3 13

4 17

Explanation:

(A) : Differential equation is,
$x \frac{d y}{d x} - y = \sqrt{y^{2} + 16 x^{2}} \Rightarrow x \frac{d y}{d x} = y + \sqrt{y^{2} + 16 x^{2}}$
$\frac{d y}{d x} = \frac{y + \sqrt{y^{2} + 16 x^{2}}}{x}$
Differentiating w.r.t.x, we get -
$\frac{d y}{d x} = v + x \frac{d v}{d x}$
Then equation (i) becomes-
$v + x \frac{d v}{d x} = \frac{v x + \sqrt{v^{2} x^{2} + 16 x^{2}}}{x} \Rightarrow v + x \frac{d v}{d x} = v + \sqrt{v^{2} + 16}$
$x \frac{d v}{d x} = \sqrt{v^{2} + 16}$
Integration on both side, we get -
$\int \frac{dv}{\sqrt{v^{2} + 16}} = \int \frac{dx}{x} \Rightarrow \log | v + \sqrt{v^{2} + 16} | = \log x + \log c$ $v + \sqrt{v^{2} + 16} = xc$
Put $v = \frac{y}{x}$ , we get
$\frac{y}{x} + \sqrt{\frac{y^{2}}{x^{2}} + 16} = x c \Rightarrow y + \sqrt{y^{2} + 16 x^{2}} = c x^{2}$
Given, $y (1) = 3$ then $x = 1$ they $y = 3$
From equation (ii),
$3 + \sqrt{9 + 16 \times 1^{2}} = c (1)^{2} \Rightarrow 3 + \sqrt{25} = c$
$c = 8$
We find $y (2)$ then, $x = 2$
From equation (ii),
$y + \sqrt{y^{2} + 16 \times 4} = 8 \times 4 \Rightarrow \sqrt{y^{2} + 64} = 32 - y$
Squaring both side we get
$y^{2} + 64 = (32 - y)^{2}$
$y^{2} + 64 = 32 \times 32 + y^{2} - 64 y$
$64 (y + 1) = 32 \times 32$
$y + 1 = 16$
$y = 15$

JEE Main-29.06.2022

Differential Equation

87267 What is the solution of the differential equation $(x + 2 y^{3}) \frac{d y}{d x} = y ?$

1

x = y^{3} + c y

2

x^{2} = y^{3} + c y^{2}

3

x^{2} = y^{3} + c y

4 None of the above

Explanation:

(A) : Given,
$(x + 2 y^{3}) \frac{d y}{d x} = y$
$\frac{d x}{d y} = \frac{x + 2 y^{3}}{y} = \frac{x}{y} + \frac{2 y^{3}}{y}$
$\frac{d x}{d y} - \frac{x}{y} = 2 y^{2}$ (This is linear differential equation)
$\frac{dx}{dy} + Px = Q$
Where, $P = - \frac{1}{y}, Q = 2 y^{2}$
$x ($ I.F. $) = \int Q ($ I.F. $) d y + c$ -------(i)
so, I.F. $= e^{\int pdy} = e^{- \int \frac{dy}{y}}$
I.F. $= e^{- \log y} = e^{\log y^{- 1}}$
I.F. $= \frac{1}{y}$
So, from equation (i) :-
$\frac{x}{y} = \int 2 y^{2}, \frac{1}{y} d y + c = 2 \int y d y + c$
$\frac{x}{y} = 2 [\frac{y^{2}}{2}] + c = y^{2} + c$
$x = y^{3} + cy$

SCRA-2010

Differential Equation

87268 Let $y = y (x)$ be a solution curve of the differential $(1 - x^{2} y^{2}) d x = y d x + x d y$ . If the line $x = 1$ intersects the curve $y = y (x)$ at $y$ $= 2$ and the line $x = 2$ intersects the curve $y =$ $y (x)$ at $y = α$ , then a value of $α$ is

1

\frac{3 e^{2}}{2 (3 e^{2} - 1)}

2

\frac{3 e^{2}}{2 (3 e^{2} + 1)}

3

\frac{1 - 3 e^{2}}{2 (3 e^{2} + 1)}

4

\frac{1 + 3 e^{2}}{2 (3 e^{2} - 1)}

Explanation:

(D) : $(1 - x^{2} y^{2}) d x = y d x + x d y, y (1) = 2$
$y (2) = α = ?$
$d x = \frac{d (x y)}{1 - (x y)^{2}} \Rightarrow \int d x = \int \frac{d (x y)}{1 - (x y)^{2}}$
$x = \frac{1}{2} \ln | \frac{1 + x y}{1 - x y} | + C$
Put $x = 1$ and $y = 2$
$1 = \frac{1}{2} \ln | \frac{1 + 2}{1 - 2} | + C \Rightarrow C = 1 - \frac{1}{2} \ln 3$
Now put $x = 2$ and $y = α$
$2 = \frac{1}{2} \ln | \frac{1 + 2 α}{1 - 2 α} | + 1 - \frac{1}{2} \ln 3$
$1 + \frac{1}{2} \ln 3 = \frac{1}{2} \ln | \frac{1 + 2 α}{1 - 2 α} |$
$2 + \ln 3 = \ln (\frac{1 + 2 α}{1 - 2 α})$
$| \frac{1 + 2 α}{1 - 2 α} | = 3 e^{2} \Rightarrow \frac{1 + 2 α}{1 - 2 α} = 3 e^{2}, - 3 e^{2}$
$\frac{1 + 2 α}{1 - 2 α} = 3 e^{2} \Rightarrow α = \frac{3 e^{2} - 1}{2 (3 e^{2} + 1)}$
And $\frac{1 + 2 α}{1 - 2 α} = - 3 e^{2}$
$α = \frac{3 e^{2} + 1}{2 (3 e^{2} - 1)}$

JEE Main-11.04.2023

Differential Equation

87269 Let $y = y (x)$ be the solution of the differential equation $x (1 - x^{2}) \frac{d y}{d x} + (3 x^{2} y - y - 4 x^{3}) = 0, x >$ 1 , with $y (2) = - 2$ . Then $y (3)$ is equal to

1 -18

2 -12

3 -6

4 -3

Explanation:

(A) : $x (1 - x^{2}) \frac{d y}{d x} + (3 x^{2} y - y - 4 x^{3}) = 0$
$x (1 - x^{2}) \frac{d y}{d x} + (3 x^{2} - 1) y = 4 x^{3}$
$\frac{d y}{d x} + \frac{(3 x^{2} - 1)}{(x - x^{3})} y = \frac{4 x^{3}}{(x - x^{3})} \Rightarrow \frac{d y}{d x} + P y = Q$
$I F = e^{\int P d x} = e^{\int \frac{3 x^{2} - 1}{x - x^{3}} d x} \Rightarrow x - x^{3} = t \Rightarrow I F = e^{\int \frac{- d t}{t}}$
$= e^{- \ln t} = \frac{1}{t}$
$∴ IF = \frac{1}{x - x^{3}}$
$y \times I F = \int Q \times I F d x$
$y (\frac{1}{x - x^{3}}) = \int \frac{4 x^{3}}{x - x^{3}} \times \frac{1}{(x - x^{3})} d x = \int \frac{4 x^{3}}{{(x - x^{3})}^{2}} d x$
$= \int \frac{4 x}{{(1 - x^{2})}^{2}} d x$
$= 2 \int \frac{- d K}{K^{2}} 1 - x^{2} = K$

JEE Main-28.06.2022

Differential Equation

87266 Let the solution curve of the differential equation
$x \frac{d y}{d x} - y = \sqrt{y^{2} + 16 x^{2}}, y (1) = 3 be y = y (x)$
Then $y (2)$ is equal to :

1 15

2 11

3 13

4 17

Explanation:

(A) : Differential equation is,
$x \frac{d y}{d x} - y = \sqrt{y^{2} + 16 x^{2}} \Rightarrow x \frac{d y}{d x} = y + \sqrt{y^{2} + 16 x^{2}}$
$\frac{d y}{d x} = \frac{y + \sqrt{y^{2} + 16 x^{2}}}{x}$
Differentiating w.r.t.x, we get -
$\frac{d y}{d x} = v + x \frac{d v}{d x}$
Then equation (i) becomes-
$v + x \frac{d v}{d x} = \frac{v x + \sqrt{v^{2} x^{2} + 16 x^{2}}}{x} \Rightarrow v + x \frac{d v}{d x} = v + \sqrt{v^{2} + 16}$
$x \frac{d v}{d x} = \sqrt{v^{2} + 16}$
Integration on both side, we get -
$\int \frac{dv}{\sqrt{v^{2} + 16}} = \int \frac{dx}{x} \Rightarrow \log | v + \sqrt{v^{2} + 16} | = \log x + \log c$ $v + \sqrt{v^{2} + 16} = xc$
Put $v = \frac{y}{x}$ , we get
$\frac{y}{x} + \sqrt{\frac{y^{2}}{x^{2}} + 16} = x c \Rightarrow y + \sqrt{y^{2} + 16 x^{2}} = c x^{2}$
Given, $y (1) = 3$ then $x = 1$ they $y = 3$
From equation (ii),
$3 + \sqrt{9 + 16 \times 1^{2}} = c (1)^{2} \Rightarrow 3 + \sqrt{25} = c$
$c = 8$
We find $y (2)$ then, $x = 2$
From equation (ii),
$y + \sqrt{y^{2} + 16 \times 4} = 8 \times 4 \Rightarrow \sqrt{y^{2} + 64} = 32 - y$
Squaring both side we get
$y^{2} + 64 = (32 - y)^{2}$
$y^{2} + 64 = 32 \times 32 + y^{2} - 64 y$
$64 (y + 1) = 32 \times 32$
$y + 1 = 16$
$y = 15$

JEE Main-29.06.2022

Differential Equation

87267 What is the solution of the differential equation $(x + 2 y^{3}) \frac{d y}{d x} = y ?$

1

x = y^{3} + c y

2

x^{2} = y^{3} + c y^{2}

3

x^{2} = y^{3} + c y

4 None of the above

Explanation:

(A) : Given,
$(x + 2 y^{3}) \frac{d y}{d x} = y$
$\frac{d x}{d y} = \frac{x + 2 y^{3}}{y} = \frac{x}{y} + \frac{2 y^{3}}{y}$
$\frac{d x}{d y} - \frac{x}{y} = 2 y^{2}$ (This is linear differential equation)
$\frac{dx}{dy} + Px = Q$
Where, $P = - \frac{1}{y}, Q = 2 y^{2}$
$x ($ I.F. $) = \int Q ($ I.F. $) d y + c$ -------(i)
so, I.F. $= e^{\int pdy} = e^{- \int \frac{dy}{y}}$
I.F. $= e^{- \log y} = e^{\log y^{- 1}}$
I.F. $= \frac{1}{y}$
So, from equation (i) :-
$\frac{x}{y} = \int 2 y^{2}, \frac{1}{y} d y + c = 2 \int y d y + c$
$\frac{x}{y} = 2 [\frac{y^{2}}{2}] + c = y^{2} + c$
$x = y^{3} + cy$

SCRA-2010

Differential Equation

87268 Let $y = y (x)$ be a solution curve of the differential $(1 - x^{2} y^{2}) d x = y d x + x d y$ . If the line $x = 1$ intersects the curve $y = y (x)$ at $y$ $= 2$ and the line $x = 2$ intersects the curve $y =$ $y (x)$ at $y = α$ , then a value of $α$ is

1

\frac{3 e^{2}}{2 (3 e^{2} - 1)}

2

\frac{3 e^{2}}{2 (3 e^{2} + 1)}

3

\frac{1 - 3 e^{2}}{2 (3 e^{2} + 1)}

4

\frac{1 + 3 e^{2}}{2 (3 e^{2} - 1)}

Explanation:

(D) : $(1 - x^{2} y^{2}) d x = y d x + x d y, y (1) = 2$
$y (2) = α = ?$
$d x = \frac{d (x y)}{1 - (x y)^{2}} \Rightarrow \int d x = \int \frac{d (x y)}{1 - (x y)^{2}}$
$x = \frac{1}{2} \ln | \frac{1 + x y}{1 - x y} | + C$
Put $x = 1$ and $y = 2$
$1 = \frac{1}{2} \ln | \frac{1 + 2}{1 - 2} | + C \Rightarrow C = 1 - \frac{1}{2} \ln 3$
Now put $x = 2$ and $y = α$
$2 = \frac{1}{2} \ln | \frac{1 + 2 α}{1 - 2 α} | + 1 - \frac{1}{2} \ln 3$
$1 + \frac{1}{2} \ln 3 = \frac{1}{2} \ln | \frac{1 + 2 α}{1 - 2 α} |$
$2 + \ln 3 = \ln (\frac{1 + 2 α}{1 - 2 α})$
$| \frac{1 + 2 α}{1 - 2 α} | = 3 e^{2} \Rightarrow \frac{1 + 2 α}{1 - 2 α} = 3 e^{2}, - 3 e^{2}$
$\frac{1 + 2 α}{1 - 2 α} = 3 e^{2} \Rightarrow α = \frac{3 e^{2} - 1}{2 (3 e^{2} + 1)}$
And $\frac{1 + 2 α}{1 - 2 α} = - 3 e^{2}$
$α = \frac{3 e^{2} + 1}{2 (3 e^{2} - 1)}$

JEE Main-11.04.2023

Differential Equation

87269 Let $y = y (x)$ be the solution of the differential equation $x (1 - x^{2}) \frac{d y}{d x} + (3 x^{2} y - y - 4 x^{3}) = 0, x >$ 1 , with $y (2) = - 2$ . Then $y (3)$ is equal to

1 -18

2 -12

3 -6

4 -3

Explanation:

(A) : $x (1 - x^{2}) \frac{d y}{d x} + (3 x^{2} y - y - 4 x^{3}) = 0$
$x (1 - x^{2}) \frac{d y}{d x} + (3 x^{2} - 1) y = 4 x^{3}$
$\frac{d y}{d x} + \frac{(3 x^{2} - 1)}{(x - x^{3})} y = \frac{4 x^{3}}{(x - x^{3})} \Rightarrow \frac{d y}{d x} + P y = Q$
$I F = e^{\int P d x} = e^{\int \frac{3 x^{2} - 1}{x - x^{3}} d x} \Rightarrow x - x^{3} = t \Rightarrow I F = e^{\int \frac{- d t}{t}}$
$= e^{- \ln t} = \frac{1}{t}$
$∴ IF = \frac{1}{x - x^{3}}$
$y \times I F = \int Q \times I F d x$
$y (\frac{1}{x - x^{3}}) = \int \frac{4 x^{3}}{x - x^{3}} \times \frac{1}{(x - x^{3})} d x = \int \frac{4 x^{3}}{{(x - x^{3})}^{2}} d x$
$= \int \frac{4 x}{{(1 - x^{2})}^{2}} d x$
$= 2 \int \frac{- d K}{K^{2}} 1 - x^{2} = K$

JEE Main-28.06.2022

Differential Equation

87266 Let the solution curve of the differential equation
$x \frac{d y}{d x} - y = \sqrt{y^{2} + 16 x^{2}}, y (1) = 3 be y = y (x)$
Then $y (2)$ is equal to :

1 15

2 11

3 13

4 17

Explanation:

(A) : Differential equation is,
$x \frac{d y}{d x} - y = \sqrt{y^{2} + 16 x^{2}} \Rightarrow x \frac{d y}{d x} = y + \sqrt{y^{2} + 16 x^{2}}$
$\frac{d y}{d x} = \frac{y + \sqrt{y^{2} + 16 x^{2}}}{x}$
Differentiating w.r.t.x, we get -
$\frac{d y}{d x} = v + x \frac{d v}{d x}$
Then equation (i) becomes-
$v + x \frac{d v}{d x} = \frac{v x + \sqrt{v^{2} x^{2} + 16 x^{2}}}{x} \Rightarrow v + x \frac{d v}{d x} = v + \sqrt{v^{2} + 16}$
$x \frac{d v}{d x} = \sqrt{v^{2} + 16}$
Integration on both side, we get -
$\int \frac{dv}{\sqrt{v^{2} + 16}} = \int \frac{dx}{x} \Rightarrow \log | v + \sqrt{v^{2} + 16} | = \log x + \log c$ $v + \sqrt{v^{2} + 16} = xc$
Put $v = \frac{y}{x}$ , we get
$\frac{y}{x} + \sqrt{\frac{y^{2}}{x^{2}} + 16} = x c \Rightarrow y + \sqrt{y^{2} + 16 x^{2}} = c x^{2}$
Given, $y (1) = 3$ then $x = 1$ they $y = 3$
From equation (ii),
$3 + \sqrt{9 + 16 \times 1^{2}} = c (1)^{2} \Rightarrow 3 + \sqrt{25} = c$
$c = 8$
We find $y (2)$ then, $x = 2$
From equation (ii),
$y + \sqrt{y^{2} + 16 \times 4} = 8 \times 4 \Rightarrow \sqrt{y^{2} + 64} = 32 - y$
Squaring both side we get
$y^{2} + 64 = (32 - y)^{2}$
$y^{2} + 64 = 32 \times 32 + y^{2} - 64 y$
$64 (y + 1) = 32 \times 32$
$y + 1 = 16$
$y = 15$

JEE Main-29.06.2022

Differential Equation

87267 What is the solution of the differential equation $(x + 2 y^{3}) \frac{d y}{d x} = y ?$

1

x = y^{3} + c y

2

x^{2} = y^{3} + c y^{2}

3

x^{2} = y^{3} + c y

4 None of the above

Explanation:

(A) : Given,
$(x + 2 y^{3}) \frac{d y}{d x} = y$
$\frac{d x}{d y} = \frac{x + 2 y^{3}}{y} = \frac{x}{y} + \frac{2 y^{3}}{y}$
$\frac{d x}{d y} - \frac{x}{y} = 2 y^{2}$ (This is linear differential equation)
$\frac{dx}{dy} + Px = Q$
Where, $P = - \frac{1}{y}, Q = 2 y^{2}$
$x ($ I.F. $) = \int Q ($ I.F. $) d y + c$ -------(i)
so, I.F. $= e^{\int pdy} = e^{- \int \frac{dy}{y}}$
I.F. $= e^{- \log y} = e^{\log y^{- 1}}$
I.F. $= \frac{1}{y}$
So, from equation (i) :-
$\frac{x}{y} = \int 2 y^{2}, \frac{1}{y} d y + c = 2 \int y d y + c$
$\frac{x}{y} = 2 [\frac{y^{2}}{2}] + c = y^{2} + c$
$x = y^{3} + cy$

SCRA-2010

Differential Equation

87268 Let $y = y (x)$ be a solution curve of the differential $(1 - x^{2} y^{2}) d x = y d x + x d y$ . If the line $x = 1$ intersects the curve $y = y (x)$ at $y$ $= 2$ and the line $x = 2$ intersects the curve $y =$ $y (x)$ at $y = α$ , then a value of $α$ is

1

\frac{3 e^{2}}{2 (3 e^{2} - 1)}

2

\frac{3 e^{2}}{2 (3 e^{2} + 1)}

3

\frac{1 - 3 e^{2}}{2 (3 e^{2} + 1)}

4

\frac{1 + 3 e^{2}}{2 (3 e^{2} - 1)}

Explanation:

(D) : $(1 - x^{2} y^{2}) d x = y d x + x d y, y (1) = 2$
$y (2) = α = ?$
$d x = \frac{d (x y)}{1 - (x y)^{2}} \Rightarrow \int d x = \int \frac{d (x y)}{1 - (x y)^{2}}$
$x = \frac{1}{2} \ln | \frac{1 + x y}{1 - x y} | + C$
Put $x = 1$ and $y = 2$
$1 = \frac{1}{2} \ln | \frac{1 + 2}{1 - 2} | + C \Rightarrow C = 1 - \frac{1}{2} \ln 3$
Now put $x = 2$ and $y = α$
$2 = \frac{1}{2} \ln | \frac{1 + 2 α}{1 - 2 α} | + 1 - \frac{1}{2} \ln 3$
$1 + \frac{1}{2} \ln 3 = \frac{1}{2} \ln | \frac{1 + 2 α}{1 - 2 α} |$
$2 + \ln 3 = \ln (\frac{1 + 2 α}{1 - 2 α})$
$| \frac{1 + 2 α}{1 - 2 α} | = 3 e^{2} \Rightarrow \frac{1 + 2 α}{1 - 2 α} = 3 e^{2}, - 3 e^{2}$
$\frac{1 + 2 α}{1 - 2 α} = 3 e^{2} \Rightarrow α = \frac{3 e^{2} - 1}{2 (3 e^{2} + 1)}$
And $\frac{1 + 2 α}{1 - 2 α} = - 3 e^{2}$
$α = \frac{3 e^{2} + 1}{2 (3 e^{2} - 1)}$

JEE Main-11.04.2023

Differential Equation

87269 Let $y = y (x)$ be the solution of the differential equation $x (1 - x^{2}) \frac{d y}{d x} + (3 x^{2} y - y - 4 x^{3}) = 0, x >$ 1 , with $y (2) = - 2$ . Then $y (3)$ is equal to

1 -18

2 -12

3 -6

4 -3

Explanation:

(A) : $x (1 - x^{2}) \frac{d y}{d x} + (3 x^{2} y - y - 4 x^{3}) = 0$
$x (1 - x^{2}) \frac{d y}{d x} + (3 x^{2} - 1) y = 4 x^{3}$
$\frac{d y}{d x} + \frac{(3 x^{2} - 1)}{(x - x^{3})} y = \frac{4 x^{3}}{(x - x^{3})} \Rightarrow \frac{d y}{d x} + P y = Q$
$I F = e^{\int P d x} = e^{\int \frac{3 x^{2} - 1}{x - x^{3}} d x} \Rightarrow x - x^{3} = t \Rightarrow I F = e^{\int \frac{- d t}{t}}$
$= e^{- \ln t} = \frac{1}{t}$
$∴ IF = \frac{1}{x - x^{3}}$
$y \times I F = \int Q \times I F d x$
$y (\frac{1}{x - x^{3}}) = \int \frac{4 x^{3}}{x - x^{3}} \times \frac{1}{(x - x^{3})} d x = \int \frac{4 x^{3}}{{(x - x^{3})}^{2}} d x$
$= \int \frac{4 x}{{(1 - x^{2})}^{2}} d x$
$= 2 \int \frac{- d K}{K^{2}} 1 - x^{2} = K$

JEE Main-28.06.2022