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Differential Equation

87571 The solution of the differential equation $\frac{d y}{d x} = \frac{1}{x + y^{2}}$ is

1

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

2

y = x^{2} + 2 x + 2 - c e^{x}

3

x = - y^{2} - 2 y + 2 - c e^{y}

4

x = - y^{2} - 2 y - 2 + c e^{y}

5

x = y^{2} + 2 y + 2 - c e^{y}

Explanation:

(D) : Given differential equation,
$\frac{d y}{d x} = \frac{1}{x + y^{2}} \Rightarrow \frac{d x}{d y} = \frac{x + y^{2}}{1} \Rightarrow \frac{d x}{d y} - x = y^{2}$
Assume, $P = - 1, Q = y^{2}$
I.F, $= e \int - 1 d y = e^{- y}$
$x e^{- y} = \int e^{- y} y^{2} d y = - e^{- y} y^{2} + \int 2 e^{- y} y d y$
$= - e^{- y} y^{2} + 2 [- e^{- y} y + \int e^{- y} d y] + c$
$= - e^{- y} y^{2} + 2 [- e^{- y} y - e^{- y}] + c$
$\Rightarrow x e^{- y} = e^{- y} (- y^{2} - 2 y - 2) + c$
$x = - y^{2} - 2 y - 2 + c e^{y}$

Kerala CEE-2009

Differential Equation

87572 The solution of $\frac{d y}{d x} + y \tan x = \sec x$ is :

1

y \sec x = \tan x + c

2

y \tan x = \sec x + c

3

\tan x = y \tan x + c

4

x \sec x = \tan y + c

5

x \tan x = y \tan x + c

Explanation:

(A) : Given,
$\frac{d y}{d x} + y \tan x = \sec x$
It is a first-degree linear D.E. of the form
$\frac{dy}{dx} + py = θ$
Here,
$P = \tan x, θ = \sec x$
I.F. $= e^{\int pdx} = e^{\int \tan x d x} = e^{\log \sec x} = \sec x$
The solution is given by-
$y \cdot I \cdot F \cdot = \int Q \cdot I \cdot F \cdot dx + c$
$y \cdot \sec x = \int \sec^{2} x d x + c$
$y \cdot \sec x = \tan x + c$

Kerala CEE-2004

Differential Equation

87573 Solution of the differential equation
$x = 1 + x y \frac{d y}{d x} + \frac{(x y)^{2}}{2!} {(\frac{d y}{d x})}^{2} + \frac{(x y)^{3}}{3!} {(\frac{d y}{d x})}^{3} + \dots$ is

1

y = \log_{e} (x) + C

2

y = {(\log_{e} x)}^{2} + C

3

y = \pm \sqrt{{(\log_{e} x)}^{2} + 2 C}

4

x y = x^{y} + C

Explanation:

(C) : We have,
$x = e^{x y \frac{d y}{d x}}$
$\Rightarrow \log x = x y \frac{d y}{d x} \Rightarrow y d y = \frac{\log x}{x} d x$
$\Rightarrow ydy = \log xd (\log x)$
On integrating both sides, we get-
$\frac{y^{2}}{2} = \frac{{(\log_{e} x)}^{2}}{2} + C \Rightarrow y^{2} = {(\log_{e} x)}^{2} + 2 C$
$\Rightarrow y = \pm \sqrt{{(\log_{e} x)}^{2} + 2 C}$

Manipal UGET-2015]**#

Differential Equation

87537 Let $y (x)$ be the solution of the differential equation $(x \log x) \frac{d y}{d x} + y = 2 x \log x, (x \geq 1)$ . Then
$y (e)$ is equal to

1 0

2 e

3 2

4

2 e

Explanation:

(C) : Given,
Dividing by $x \log x$
$(x \log x) \frac{d y}{d x} + y = 2 x \log x$
$\frac{d y}{d x} + \frac{y}{x \log x} = 2$
Multiplying by I.F
$d (IF \times y) = 2 \log x d x$
IF $= e^{\int \frac{1}{x \log x} dx}$
$= e^{\log (\log x)}$
$= \log x$
$d \times (I F \times y) = 2 \log x d x$
By integrating
$y \log x = \int 2 \log x d x$
By using product rule on R.W.S
$y \log x = 2 [\log x] 1 - (\int \frac{d}{d x} (\log x)) d x$
$y \log x = 2 [x (\log x - 1)] + C$
Put,
$x = 1$
$0 = 2 [1 (0 - 1)] + C$
$C = 2$
$y \log x = 2 [x (\log x - 1)] + 2$
$at x = e$
$y = 2$

JEE Main-2015

Differential Equation

87553 If $\frac{d y}{d x} = y + 3 > 0$ and $y (0) = 2$ , then $y (\log 2)$ is equal to

1 5

2 13

3 -2

4 7

Explanation:

(D) : $\frac{dy}{dx} = y + 3$
$\frac{d y}{y + 3} = d x$
$\ln (y + 3) = x + C$
Given, at $x = 0, y = 2$
$\ln 5 = C$
$\ln (y + 3) = x + \ln$
$\ln (\frac{y + 3}{5}) = x$
$y + 3 = 5 e^{x}$
$y = 5 e^{x} - 3$
$y (\ln 2) = 5 e^{\ln 2} - 3 = 7$

AIEEE-2011

Differential Equation

87571 The solution of the differential equation $\frac{d y}{d x} = \frac{1}{x + y^{2}}$ is

1

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

2

y = x^{2} + 2 x + 2 - c e^{x}

3

x = - y^{2} - 2 y + 2 - c e^{y}

4

x = - y^{2} - 2 y - 2 + c e^{y}

5

x = y^{2} + 2 y + 2 - c e^{y}

Explanation:

(D) : Given differential equation,
$\frac{d y}{d x} = \frac{1}{x + y^{2}} \Rightarrow \frac{d x}{d y} = \frac{x + y^{2}}{1} \Rightarrow \frac{d x}{d y} - x = y^{2}$
Assume, $P = - 1, Q = y^{2}$
I.F, $= e \int - 1 d y = e^{- y}$
$x e^{- y} = \int e^{- y} y^{2} d y = - e^{- y} y^{2} + \int 2 e^{- y} y d y$
$= - e^{- y} y^{2} + 2 [- e^{- y} y + \int e^{- y} d y] + c$
$= - e^{- y} y^{2} + 2 [- e^{- y} y - e^{- y}] + c$
$\Rightarrow x e^{- y} = e^{- y} (- y^{2} - 2 y - 2) + c$
$x = - y^{2} - 2 y - 2 + c e^{y}$

Kerala CEE-2009

Differential Equation

87572 The solution of $\frac{d y}{d x} + y \tan x = \sec x$ is :

1

y \sec x = \tan x + c

2

y \tan x = \sec x + c

3

\tan x = y \tan x + c

4

x \sec x = \tan y + c

5

x \tan x = y \tan x + c

Explanation:

(A) : Given,
$\frac{d y}{d x} + y \tan x = \sec x$
It is a first-degree linear D.E. of the form
$\frac{dy}{dx} + py = θ$
Here,
$P = \tan x, θ = \sec x$
I.F. $= e^{\int pdx} = e^{\int \tan x d x} = e^{\log \sec x} = \sec x$
The solution is given by-
$y \cdot I \cdot F \cdot = \int Q \cdot I \cdot F \cdot dx + c$
$y \cdot \sec x = \int \sec^{2} x d x + c$
$y \cdot \sec x = \tan x + c$

Kerala CEE-2004

Differential Equation

87573 Solution of the differential equation
$x = 1 + x y \frac{d y}{d x} + \frac{(x y)^{2}}{2!} {(\frac{d y}{d x})}^{2} + \frac{(x y)^{3}}{3!} {(\frac{d y}{d x})}^{3} + \dots$ is

1

y = \log_{e} (x) + C

2

y = {(\log_{e} x)}^{2} + C

3

y = \pm \sqrt{{(\log_{e} x)}^{2} + 2 C}

4

x y = x^{y} + C

Explanation:

(C) : We have,
$x = e^{x y \frac{d y}{d x}}$
$\Rightarrow \log x = x y \frac{d y}{d x} \Rightarrow y d y = \frac{\log x}{x} d x$
$\Rightarrow ydy = \log xd (\log x)$
On integrating both sides, we get-
$\frac{y^{2}}{2} = \frac{{(\log_{e} x)}^{2}}{2} + C \Rightarrow y^{2} = {(\log_{e} x)}^{2} + 2 C$
$\Rightarrow y = \pm \sqrt{{(\log_{e} x)}^{2} + 2 C}$

Manipal UGET-2015]**#

Differential Equation

87537 Let $y (x)$ be the solution of the differential equation $(x \log x) \frac{d y}{d x} + y = 2 x \log x, (x \geq 1)$ . Then
$y (e)$ is equal to

1 0

2 e

3 2

4

2 e

Explanation:

(C) : Given,
Dividing by $x \log x$
$(x \log x) \frac{d y}{d x} + y = 2 x \log x$
$\frac{d y}{d x} + \frac{y}{x \log x} = 2$
Multiplying by I.F
$d (IF \times y) = 2 \log x d x$
IF $= e^{\int \frac{1}{x \log x} dx}$
$= e^{\log (\log x)}$
$= \log x$
$d \times (I F \times y) = 2 \log x d x$
By integrating
$y \log x = \int 2 \log x d x$
By using product rule on R.W.S
$y \log x = 2 [\log x] 1 - (\int \frac{d}{d x} (\log x)) d x$
$y \log x = 2 [x (\log x - 1)] + C$
Put,
$x = 1$
$0 = 2 [1 (0 - 1)] + C$
$C = 2$
$y \log x = 2 [x (\log x - 1)] + 2$
$at x = e$
$y = 2$

JEE Main-2015

Differential Equation

87553 If $\frac{d y}{d x} = y + 3 > 0$ and $y (0) = 2$ , then $y (\log 2)$ is equal to

1 5

2 13

3 -2

4 7

Explanation:

(D) : $\frac{dy}{dx} = y + 3$
$\frac{d y}{y + 3} = d x$
$\ln (y + 3) = x + C$
Given, at $x = 0, y = 2$
$\ln 5 = C$
$\ln (y + 3) = x + \ln$
$\ln (\frac{y + 3}{5}) = x$
$y + 3 = 5 e^{x}$
$y = 5 e^{x} - 3$
$y (\ln 2) = 5 e^{\ln 2} - 3 = 7$

AIEEE-2011

Differential Equation

87571 The solution of the differential equation $\frac{d y}{d x} = \frac{1}{x + y^{2}}$ is

1

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

2

y = x^{2} + 2 x + 2 - c e^{x}

3

x = - y^{2} - 2 y + 2 - c e^{y}

4

x = - y^{2} - 2 y - 2 + c e^{y}

5

x = y^{2} + 2 y + 2 - c e^{y}

Explanation:

(D) : Given differential equation,
$\frac{d y}{d x} = \frac{1}{x + y^{2}} \Rightarrow \frac{d x}{d y} = \frac{x + y^{2}}{1} \Rightarrow \frac{d x}{d y} - x = y^{2}$
Assume, $P = - 1, Q = y^{2}$
I.F, $= e \int - 1 d y = e^{- y}$
$x e^{- y} = \int e^{- y} y^{2} d y = - e^{- y} y^{2} + \int 2 e^{- y} y d y$
$= - e^{- y} y^{2} + 2 [- e^{- y} y + \int e^{- y} d y] + c$
$= - e^{- y} y^{2} + 2 [- e^{- y} y - e^{- y}] + c$
$\Rightarrow x e^{- y} = e^{- y} (- y^{2} - 2 y - 2) + c$
$x = - y^{2} - 2 y - 2 + c e^{y}$

Kerala CEE-2009

Differential Equation

87572 The solution of $\frac{d y}{d x} + y \tan x = \sec x$ is :

1

y \sec x = \tan x + c

2

y \tan x = \sec x + c

3

\tan x = y \tan x + c

4

x \sec x = \tan y + c

5

x \tan x = y \tan x + c

Explanation:

(A) : Given,
$\frac{d y}{d x} + y \tan x = \sec x$
It is a first-degree linear D.E. of the form
$\frac{dy}{dx} + py = θ$
Here,
$P = \tan x, θ = \sec x$
I.F. $= e^{\int pdx} = e^{\int \tan x d x} = e^{\log \sec x} = \sec x$
The solution is given by-
$y \cdot I \cdot F \cdot = \int Q \cdot I \cdot F \cdot dx + c$
$y \cdot \sec x = \int \sec^{2} x d x + c$
$y \cdot \sec x = \tan x + c$

Kerala CEE-2004

Differential Equation

87573 Solution of the differential equation
$x = 1 + x y \frac{d y}{d x} + \frac{(x y)^{2}}{2!} {(\frac{d y}{d x})}^{2} + \frac{(x y)^{3}}{3!} {(\frac{d y}{d x})}^{3} + \dots$ is

1

y = \log_{e} (x) + C

2

y = {(\log_{e} x)}^{2} + C

3

y = \pm \sqrt{{(\log_{e} x)}^{2} + 2 C}

4

x y = x^{y} + C

Explanation:

(C) : We have,
$x = e^{x y \frac{d y}{d x}}$
$\Rightarrow \log x = x y \frac{d y}{d x} \Rightarrow y d y = \frac{\log x}{x} d x$
$\Rightarrow ydy = \log xd (\log x)$
On integrating both sides, we get-
$\frac{y^{2}}{2} = \frac{{(\log_{e} x)}^{2}}{2} + C \Rightarrow y^{2} = {(\log_{e} x)}^{2} + 2 C$
$\Rightarrow y = \pm \sqrt{{(\log_{e} x)}^{2} + 2 C}$

Manipal UGET-2015]**#

Differential Equation

87537 Let $y (x)$ be the solution of the differential equation $(x \log x) \frac{d y}{d x} + y = 2 x \log x, (x \geq 1)$ . Then
$y (e)$ is equal to

1 0

2 e

3 2

4

2 e

Explanation:

(C) : Given,
Dividing by $x \log x$
$(x \log x) \frac{d y}{d x} + y = 2 x \log x$
$\frac{d y}{d x} + \frac{y}{x \log x} = 2$
Multiplying by I.F
$d (IF \times y) = 2 \log x d x$
IF $= e^{\int \frac{1}{x \log x} dx}$
$= e^{\log (\log x)}$
$= \log x$
$d \times (I F \times y) = 2 \log x d x$
By integrating
$y \log x = \int 2 \log x d x$
By using product rule on R.W.S
$y \log x = 2 [\log x] 1 - (\int \frac{d}{d x} (\log x)) d x$
$y \log x = 2 [x (\log x - 1)] + C$
Put,
$x = 1$
$0 = 2 [1 (0 - 1)] + C$
$C = 2$
$y \log x = 2 [x (\log x - 1)] + 2$
$at x = e$
$y = 2$

JEE Main-2015

Differential Equation

87553 If $\frac{d y}{d x} = y + 3 > 0$ and $y (0) = 2$ , then $y (\log 2)$ is equal to

1 5

2 13

3 -2

4 7

Explanation:

(D) : $\frac{dy}{dx} = y + 3$
$\frac{d y}{y + 3} = d x$
$\ln (y + 3) = x + C$
Given, at $x = 0, y = 2$
$\ln 5 = C$
$\ln (y + 3) = x + \ln$
$\ln (\frac{y + 3}{5}) = x$
$y + 3 = 5 e^{x}$
$y = 5 e^{x} - 3$
$y (\ln 2) = 5 e^{\ln 2} - 3 = 7$

AIEEE-2011

Differential Equation

87571 The solution of the differential equation $\frac{d y}{d x} = \frac{1}{x + y^{2}}$ is

1

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

2

y = x^{2} + 2 x + 2 - c e^{x}

3

x = - y^{2} - 2 y + 2 - c e^{y}

4

x = - y^{2} - 2 y - 2 + c e^{y}

5

x = y^{2} + 2 y + 2 - c e^{y}

Explanation:

(D) : Given differential equation,
$\frac{d y}{d x} = \frac{1}{x + y^{2}} \Rightarrow \frac{d x}{d y} = \frac{x + y^{2}}{1} \Rightarrow \frac{d x}{d y} - x = y^{2}$
Assume, $P = - 1, Q = y^{2}$
I.F, $= e \int - 1 d y = e^{- y}$
$x e^{- y} = \int e^{- y} y^{2} d y = - e^{- y} y^{2} + \int 2 e^{- y} y d y$
$= - e^{- y} y^{2} + 2 [- e^{- y} y + \int e^{- y} d y] + c$
$= - e^{- y} y^{2} + 2 [- e^{- y} y - e^{- y}] + c$
$\Rightarrow x e^{- y} = e^{- y} (- y^{2} - 2 y - 2) + c$
$x = - y^{2} - 2 y - 2 + c e^{y}$

Kerala CEE-2009

Differential Equation

87572 The solution of $\frac{d y}{d x} + y \tan x = \sec x$ is :

1

y \sec x = \tan x + c

2

y \tan x = \sec x + c

3

\tan x = y \tan x + c

4

x \sec x = \tan y + c

5

x \tan x = y \tan x + c

Explanation:

(A) : Given,
$\frac{d y}{d x} + y \tan x = \sec x$
It is a first-degree linear D.E. of the form
$\frac{dy}{dx} + py = θ$
Here,
$P = \tan x, θ = \sec x$
I.F. $= e^{\int pdx} = e^{\int \tan x d x} = e^{\log \sec x} = \sec x$
The solution is given by-
$y \cdot I \cdot F \cdot = \int Q \cdot I \cdot F \cdot dx + c$
$y \cdot \sec x = \int \sec^{2} x d x + c$
$y \cdot \sec x = \tan x + c$

Kerala CEE-2004

Differential Equation

87573 Solution of the differential equation
$x = 1 + x y \frac{d y}{d x} + \frac{(x y)^{2}}{2!} {(\frac{d y}{d x})}^{2} + \frac{(x y)^{3}}{3!} {(\frac{d y}{d x})}^{3} + \dots$ is

1

y = \log_{e} (x) + C

2

y = {(\log_{e} x)}^{2} + C

3

y = \pm \sqrt{{(\log_{e} x)}^{2} + 2 C}

4

x y = x^{y} + C

Explanation:

(C) : We have,
$x = e^{x y \frac{d y}{d x}}$
$\Rightarrow \log x = x y \frac{d y}{d x} \Rightarrow y d y = \frac{\log x}{x} d x$
$\Rightarrow ydy = \log xd (\log x)$
On integrating both sides, we get-
$\frac{y^{2}}{2} = \frac{{(\log_{e} x)}^{2}}{2} + C \Rightarrow y^{2} = {(\log_{e} x)}^{2} + 2 C$
$\Rightarrow y = \pm \sqrt{{(\log_{e} x)}^{2} + 2 C}$

Manipal UGET-2015]**#

Differential Equation

87537 Let $y (x)$ be the solution of the differential equation $(x \log x) \frac{d y}{d x} + y = 2 x \log x, (x \geq 1)$ . Then
$y (e)$ is equal to

1 0

2 e

3 2

4

2 e

Explanation:

(C) : Given,
Dividing by $x \log x$
$(x \log x) \frac{d y}{d x} + y = 2 x \log x$
$\frac{d y}{d x} + \frac{y}{x \log x} = 2$
Multiplying by I.F
$d (IF \times y) = 2 \log x d x$
IF $= e^{\int \frac{1}{x \log x} dx}$
$= e^{\log (\log x)}$
$= \log x$
$d \times (I F \times y) = 2 \log x d x$
By integrating
$y \log x = \int 2 \log x d x$
By using product rule on R.W.S
$y \log x = 2 [\log x] 1 - (\int \frac{d}{d x} (\log x)) d x$
$y \log x = 2 [x (\log x - 1)] + C$
Put,
$x = 1$
$0 = 2 [1 (0 - 1)] + C$
$C = 2$
$y \log x = 2 [x (\log x - 1)] + 2$
$at x = e$
$y = 2$

JEE Main-2015

Differential Equation

87553 If $\frac{d y}{d x} = y + 3 > 0$ and $y (0) = 2$ , then $y (\log 2)$ is equal to

1 5

2 13

3 -2

4 7

Explanation:

(D) : $\frac{dy}{dx} = y + 3$
$\frac{d y}{y + 3} = d x$
$\ln (y + 3) = x + C$
Given, at $x = 0, y = 2$
$\ln 5 = C$
$\ln (y + 3) = x + \ln$
$\ln (\frac{y + 3}{5}) = x$
$y + 3 = 5 e^{x}$
$y = 5 e^{x} - 3$
$y (\ln 2) = 5 e^{\ln 2} - 3 = 7$

AIEEE-2011

Differential Equation

87571 The solution of the differential equation $\frac{d y}{d x} = \frac{1}{x + y^{2}}$ is

1

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

2

y = x^{2} + 2 x + 2 - c e^{x}

3

x = - y^{2} - 2 y + 2 - c e^{y}

4

x = - y^{2} - 2 y - 2 + c e^{y}

5

x = y^{2} + 2 y + 2 - c e^{y}

Explanation:

(D) : Given differential equation,
$\frac{d y}{d x} = \frac{1}{x + y^{2}} \Rightarrow \frac{d x}{d y} = \frac{x + y^{2}}{1} \Rightarrow \frac{d x}{d y} - x = y^{2}$
Assume, $P = - 1, Q = y^{2}$
I.F, $= e \int - 1 d y = e^{- y}$
$x e^{- y} = \int e^{- y} y^{2} d y = - e^{- y} y^{2} + \int 2 e^{- y} y d y$
$= - e^{- y} y^{2} + 2 [- e^{- y} y + \int e^{- y} d y] + c$
$= - e^{- y} y^{2} + 2 [- e^{- y} y - e^{- y}] + c$
$\Rightarrow x e^{- y} = e^{- y} (- y^{2} - 2 y - 2) + c$
$x = - y^{2} - 2 y - 2 + c e^{y}$

Kerala CEE-2009

Differential Equation

87572 The solution of $\frac{d y}{d x} + y \tan x = \sec x$ is :

1

y \sec x = \tan x + c

2

y \tan x = \sec x + c

3

\tan x = y \tan x + c

4

x \sec x = \tan y + c

5

x \tan x = y \tan x + c

Explanation:

(A) : Given,
$\frac{d y}{d x} + y \tan x = \sec x$
It is a first-degree linear D.E. of the form
$\frac{dy}{dx} + py = θ$
Here,
$P = \tan x, θ = \sec x$
I.F. $= e^{\int pdx} = e^{\int \tan x d x} = e^{\log \sec x} = \sec x$
The solution is given by-
$y \cdot I \cdot F \cdot = \int Q \cdot I \cdot F \cdot dx + c$
$y \cdot \sec x = \int \sec^{2} x d x + c$
$y \cdot \sec x = \tan x + c$

Kerala CEE-2004

Differential Equation

87573 Solution of the differential equation
$x = 1 + x y \frac{d y}{d x} + \frac{(x y)^{2}}{2!} {(\frac{d y}{d x})}^{2} + \frac{(x y)^{3}}{3!} {(\frac{d y}{d x})}^{3} + \dots$ is

1

y = \log_{e} (x) + C

2

y = {(\log_{e} x)}^{2} + C

3

y = \pm \sqrt{{(\log_{e} x)}^{2} + 2 C}

4

x y = x^{y} + C

Explanation:

(C) : We have,
$x = e^{x y \frac{d y}{d x}}$
$\Rightarrow \log x = x y \frac{d y}{d x} \Rightarrow y d y = \frac{\log x}{x} d x$
$\Rightarrow ydy = \log xd (\log x)$
On integrating both sides, we get-
$\frac{y^{2}}{2} = \frac{{(\log_{e} x)}^{2}}{2} + C \Rightarrow y^{2} = {(\log_{e} x)}^{2} + 2 C$
$\Rightarrow y = \pm \sqrt{{(\log_{e} x)}^{2} + 2 C}$

Manipal UGET-2015]**#

Differential Equation

87537 Let $y (x)$ be the solution of the differential equation $(x \log x) \frac{d y}{d x} + y = 2 x \log x, (x \geq 1)$ . Then
$y (e)$ is equal to

1 0

2 e

3 2

4

2 e

Explanation:

(C) : Given,
Dividing by $x \log x$
$(x \log x) \frac{d y}{d x} + y = 2 x \log x$
$\frac{d y}{d x} + \frac{y}{x \log x} = 2$
Multiplying by I.F
$d (IF \times y) = 2 \log x d x$
IF $= e^{\int \frac{1}{x \log x} dx}$
$= e^{\log (\log x)}$
$= \log x$
$d \times (I F \times y) = 2 \log x d x$
By integrating
$y \log x = \int 2 \log x d x$
By using product rule on R.W.S
$y \log x = 2 [\log x] 1 - (\int \frac{d}{d x} (\log x)) d x$
$y \log x = 2 [x (\log x - 1)] + C$
Put,
$x = 1$
$0 = 2 [1 (0 - 1)] + C$
$C = 2$
$y \log x = 2 [x (\log x - 1)] + 2$
$at x = e$
$y = 2$

JEE Main-2015

Differential Equation

87553 If $\frac{d y}{d x} = y + 3 > 0$ and $y (0) = 2$ , then $y (\log 2)$ is equal to

1 5

2 13

3 -2

4 7

Explanation:

(D) : $\frac{dy}{dx} = y + 3$
$\frac{d y}{y + 3} = d x$
$\ln (y + 3) = x + C$
Given, at $x = 0, y = 2$
$\ln 5 = C$
$\ln (y + 3) = x + \ln$
$\ln (\frac{y + 3}{5}) = x$
$y + 3 = 5 e^{x}$
$y = 5 e^{x} - 3$
$y (\ln 2) = 5 e^{\ln 2} - 3 = 7$

AIEEE-2011

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Question Bank Series

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