QBManager

Differential Equation

87318 The solution of $25 \frac{d^{2} y}{d x^{2}} - 10 \frac{d y}{d x} + y = 0$ , $y (0) = 1, y (1) = 2 e^{1 / 5}$ is

1

y = e^{5 x} + e^{- 5 x}

2

y = (1 + x) e^{5 x}

3

y = (1 + x) e^{x / 5}

4

y = (1 + x) e^{- x / 5}

Explanation:

(C) : We have given,
$25 \frac{d^{2} y}{d x^{2}} - 10 \frac{d y}{d x} + y = 0$
$y = e^{mx}$
$\frac{d y}{d x} = m e^{m x}$
And, $\frac{d^{2} y}{d x^{2}} = m^{2} e^{m x} \Rightarrow 25 \frac{d^{2} y}{d x^{2}} - 10 \frac{d y}{d x} + y = 0$
$25 m^{2} e^{m x} - 10 m e^{m x} + e^{m x} = 0$
$e^{m x} (25 m^{2} - 10 m + 1) = 0$
For auxiliary equation-
$25 m^{2} - 10 m + 1 = 0$
$(5 m - 1) (5 m - 1) = 0$
$m = \frac{1}{5}, \frac{1}{5}$
Since roots are real and equation for general solution.
Given,
$y = (c_{1} + c_{2} x) e^{x / 5}$
$y (0) = 1$
$c_{1} = 1$
$and, y (1) = 2 e^{1 / 5}$
$(c_{1} + c_{2}) e^{1 / 5} = 2 e^{1 / 5}$
$c_{1} + c_{2} = 2$
$1 + c_{2} = 2$
$c_{2} = 1$
Then general solution,
$y = (1 + 1 \cdot x) e^{x / 5}$
$y = (1 + x) e^{x /}$

WB JEE-2012

Differential Equation

87319 The integrating factor of the differential equation $3 x \log_{e} x \frac{d y}{d x} + y = 2 \log_{e} x$ is given by

1

{(\log_{e} x)}^{3}

2

\log_{e} (\log_{e} x)

3

\log_{e} x

4

{(\log_{e} x)}^{1 / 3}

Explanation:

(D) : We have differential equation,
$3 x \log_{e} x \frac{d y}{d x} + y = 2 \log_{e} x$
It can be written as-
$\frac{d y}{d x} + \frac{1}{3 x \log_{e} x} y = \frac{2}{3 x}$
Which is a linear form $\frac{d y}{d x} + P y = Q$ , where $P$ and $Q$ are function of $x$ and the integrating factor is given by the following formula $e^{\int Pdx}$
$∴ I F = e^{\int \frac{1}{3 x \log_{c} x}} dx$
Put, $\log_{e} x = t$
$\frac{1}{x} dx = dt$
$= e^{1 / 3 \frac{dt}{t}} = e^{1 / 3 \log t} = e^{\log t^{1 / 3}} = t^{1 / 3} = {(\log_{e} x)}^{1 / 3}$

WB JEE-2012

Differential Equation

87320 A solution of the differential equation
${(\frac{d y}{d x})}^{2} - x \frac{d y}{d x} + y = 0$ is :

1

y = 2 x

2

y = - 2 x

3

y = 2 x - 4

4

y = 2 x + 4

Explanation:

(C) : Given,
${(\frac{d y}{d x})}^{2} - x \frac{d y}{d x} + y = 0$
Let, $\frac{d y}{d x} = p$
$p^{2} - p x + y = 0$
$y = x p - p^{2}$
On Differentiating both sides w.r.t $x$ , we get-
$\frac{d y}{d x} = x \cdot \frac{d p}{d x} + p - 2 p \cdot \frac{d p}{d x}$
$\frac{d y}{d x} = (x - 2 p) \frac{d p}{d x} + p$
On putting the value $\frac{d y}{d x}$ in above equation,
$p = (x - 2 p) \frac{d p}{d x} + p \Rightarrow (x - 2 p) \frac{d p}{d x} = 0$
$\frac{d p}{d x} = 0$
On integrating w.r.t. $x$ , we get-
$p = c$
Then, $\frac{d y}{d x} = c \Rightarrow y = x . c - c^{2}$
$c = 2$
Then, $y = 2 x - 4$

AMU-2013

Differential Equation

87321 Solution of $(x + y)^{2} \frac{d y}{d x} = a^{2}$ (' $a$ ' being a constant) is

1

\frac{(x + y)}{a} = \tan \frac{y + C}{a}, C

is an arbitrary constant

2

x y = a \tan C x, C

is an arbitrary constant

3

\frac{x}{a} = \tan \frac{y}{C}, C

is an arbitrary constant

4

x y = \tan (x + C), C

is an arbitrary constant

Explanation:

(A) : Given,
Let, $x + y = v$
$(x + y)^{2} \frac{d y}{d x} = a^{2}$
$\Rightarrow 1 + \frac{d y}{d x} = \frac{d v}{d x} \Rightarrow \frac{d y}{d x} = \frac{d v}{d x} - 1$
$∴ v^{2} (\frac{dv}{dx} - 1) = a^{2}$
$\Rightarrow \frac{d v}{d x} = \frac{v^{2} + a^{2}}{v^{2}} \Rightarrow \frac{v^{2}}{v^{2} + a^{2}} d v = d x$
On integrating both sides, we get
$\int \frac{v^{2}}{v^{2} + a^{2}} d v = \int d x$
$\Rightarrow \int 1 - \frac{a^{2}}{v^{2} + a^{2}} dv = x + C^{'}$
$\Rightarrow v - \frac{a^{2}}{a} \tan^{- 1} \frac{v}{a} = x + C^{'}$
$\Rightarrow x + y - a \tan^{- 1} \frac{x + y}{a} = x + C^{'}$
$\Rightarrow y = a \tan^{- 1} \frac{x + y}{a} + C^{'}$
$\frac{y - C^{'}}{a} = \tan^{- 1} \frac{x + y}{a}$
$\Rightarrow (\frac{x + y}{a}) = \tan (\frac{y + C}{a})$ .
Hence, $\frac{(x + y)}{a} = \tan \frac{y + C}{a}, C$ is an arbitrary constant

WB JEE-2017

Differential Equation

87322 If $y = e^{{msin}^{- 1} x}$ then $(1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} - k y = 0$ , where $k$ is equal to

1

m^{2}

2 2

3 -1

4

- m^{2}

Explanation:

(A) : Given,
$(1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} - k y = 0$
$y = e^{{msin}^{- 1} x}$
$\Rightarrow \frac{d y}{d x} = e^{m \sin^{- 1} x} \cdot \frac{m}{\sqrt{1 - x^{2}}} \Rightarrow \sqrt{1 - x^{2}} \frac{d y}{d x} = m^{m \sin^{- 1} x}$
$\Rightarrow \sqrt{1 - x^{2}} \frac{d y}{d x} = m y$
$\Rightarrow \sqrt{1 - x^{2}} \frac{d^{2} y}{d x^{2}} + \frac{1}{2 \sqrt{1 - x^{2}}} (- 2 x) \frac{d y}{d x} = m \frac{d y}{d x}$
$\Rightarrow (1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} = m \sqrt{1 - x^{2}} \frac{d y}{d x}$
$\Rightarrow (1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} = m$ .(my)
$\Rightarrow (1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} - m^{2} y = 0$
On comparing equation (i) and (ii), we get-
$∴ k = m^{2}$

WB JEE-2017

Differential Equation

87318 The solution of $25 \frac{d^{2} y}{d x^{2}} - 10 \frac{d y}{d x} + y = 0$ , $y (0) = 1, y (1) = 2 e^{1 / 5}$ is

1

y = e^{5 x} + e^{- 5 x}

2

y = (1 + x) e^{5 x}

3

y = (1 + x) e^{x / 5}

4

y = (1 + x) e^{- x / 5}

Explanation:

(C) : We have given,
$25 \frac{d^{2} y}{d x^{2}} - 10 \frac{d y}{d x} + y = 0$
$y = e^{mx}$
$\frac{d y}{d x} = m e^{m x}$
And, $\frac{d^{2} y}{d x^{2}} = m^{2} e^{m x} \Rightarrow 25 \frac{d^{2} y}{d x^{2}} - 10 \frac{d y}{d x} + y = 0$
$25 m^{2} e^{m x} - 10 m e^{m x} + e^{m x} = 0$
$e^{m x} (25 m^{2} - 10 m + 1) = 0$
For auxiliary equation-
$25 m^{2} - 10 m + 1 = 0$
$(5 m - 1) (5 m - 1) = 0$
$m = \frac{1}{5}, \frac{1}{5}$
Since roots are real and equation for general solution.
Given,
$y = (c_{1} + c_{2} x) e^{x / 5}$
$y (0) = 1$
$c_{1} = 1$
$and, y (1) = 2 e^{1 / 5}$
$(c_{1} + c_{2}) e^{1 / 5} = 2 e^{1 / 5}$
$c_{1} + c_{2} = 2$
$1 + c_{2} = 2$
$c_{2} = 1$
Then general solution,
$y = (1 + 1 \cdot x) e^{x / 5}$
$y = (1 + x) e^{x /}$

WB JEE-2012

Differential Equation

87319 The integrating factor of the differential equation $3 x \log_{e} x \frac{d y}{d x} + y = 2 \log_{e} x$ is given by

1

{(\log_{e} x)}^{3}

2

\log_{e} (\log_{e} x)

3

\log_{e} x

4

{(\log_{e} x)}^{1 / 3}

Explanation:

(D) : We have differential equation,
$3 x \log_{e} x \frac{d y}{d x} + y = 2 \log_{e} x$
It can be written as-
$\frac{d y}{d x} + \frac{1}{3 x \log_{e} x} y = \frac{2}{3 x}$
Which is a linear form $\frac{d y}{d x} + P y = Q$ , where $P$ and $Q$ are function of $x$ and the integrating factor is given by the following formula $e^{\int Pdx}$
$∴ I F = e^{\int \frac{1}{3 x \log_{c} x}} dx$
Put, $\log_{e} x = t$
$\frac{1}{x} dx = dt$
$= e^{1 / 3 \frac{dt}{t}} = e^{1 / 3 \log t} = e^{\log t^{1 / 3}} = t^{1 / 3} = {(\log_{e} x)}^{1 / 3}$

WB JEE-2012

Differential Equation

87320 A solution of the differential equation
${(\frac{d y}{d x})}^{2} - x \frac{d y}{d x} + y = 0$ is :

1

y = 2 x

2

y = - 2 x

3

y = 2 x - 4

4

y = 2 x + 4

Explanation:

(C) : Given,
${(\frac{d y}{d x})}^{2} - x \frac{d y}{d x} + y = 0$
Let, $\frac{d y}{d x} = p$
$p^{2} - p x + y = 0$
$y = x p - p^{2}$
On Differentiating both sides w.r.t $x$ , we get-
$\frac{d y}{d x} = x \cdot \frac{d p}{d x} + p - 2 p \cdot \frac{d p}{d x}$
$\frac{d y}{d x} = (x - 2 p) \frac{d p}{d x} + p$
On putting the value $\frac{d y}{d x}$ in above equation,
$p = (x - 2 p) \frac{d p}{d x} + p \Rightarrow (x - 2 p) \frac{d p}{d x} = 0$
$\frac{d p}{d x} = 0$
On integrating w.r.t. $x$ , we get-
$p = c$
Then, $\frac{d y}{d x} = c \Rightarrow y = x . c - c^{2}$
$c = 2$
Then, $y = 2 x - 4$

AMU-2013

Differential Equation

87321 Solution of $(x + y)^{2} \frac{d y}{d x} = a^{2}$ (' $a$ ' being a constant) is

1

\frac{(x + y)}{a} = \tan \frac{y + C}{a}, C

is an arbitrary constant

2

x y = a \tan C x, C

is an arbitrary constant

3

\frac{x}{a} = \tan \frac{y}{C}, C

is an arbitrary constant

4

x y = \tan (x + C), C

is an arbitrary constant

Explanation:

(A) : Given,
Let, $x + y = v$
$(x + y)^{2} \frac{d y}{d x} = a^{2}$
$\Rightarrow 1 + \frac{d y}{d x} = \frac{d v}{d x} \Rightarrow \frac{d y}{d x} = \frac{d v}{d x} - 1$
$∴ v^{2} (\frac{dv}{dx} - 1) = a^{2}$
$\Rightarrow \frac{d v}{d x} = \frac{v^{2} + a^{2}}{v^{2}} \Rightarrow \frac{v^{2}}{v^{2} + a^{2}} d v = d x$
On integrating both sides, we get
$\int \frac{v^{2}}{v^{2} + a^{2}} d v = \int d x$
$\Rightarrow \int 1 - \frac{a^{2}}{v^{2} + a^{2}} dv = x + C^{'}$
$\Rightarrow v - \frac{a^{2}}{a} \tan^{- 1} \frac{v}{a} = x + C^{'}$
$\Rightarrow x + y - a \tan^{- 1} \frac{x + y}{a} = x + C^{'}$
$\Rightarrow y = a \tan^{- 1} \frac{x + y}{a} + C^{'}$
$\frac{y - C^{'}}{a} = \tan^{- 1} \frac{x + y}{a}$
$\Rightarrow (\frac{x + y}{a}) = \tan (\frac{y + C}{a})$ .
Hence, $\frac{(x + y)}{a} = \tan \frac{y + C}{a}, C$ is an arbitrary constant

WB JEE-2017

Differential Equation

87322 If $y = e^{{msin}^{- 1} x}$ then $(1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} - k y = 0$ , where $k$ is equal to

1

m^{2}

2 2

3 -1

4

- m^{2}

Explanation:

(A) : Given,
$(1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} - k y = 0$
$y = e^{{msin}^{- 1} x}$
$\Rightarrow \frac{d y}{d x} = e^{m \sin^{- 1} x} \cdot \frac{m}{\sqrt{1 - x^{2}}} \Rightarrow \sqrt{1 - x^{2}} \frac{d y}{d x} = m^{m \sin^{- 1} x}$
$\Rightarrow \sqrt{1 - x^{2}} \frac{d y}{d x} = m y$
$\Rightarrow \sqrt{1 - x^{2}} \frac{d^{2} y}{d x^{2}} + \frac{1}{2 \sqrt{1 - x^{2}}} (- 2 x) \frac{d y}{d x} = m \frac{d y}{d x}$
$\Rightarrow (1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} = m \sqrt{1 - x^{2}} \frac{d y}{d x}$
$\Rightarrow (1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} = m$ .(my)
$\Rightarrow (1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} - m^{2} y = 0$
On comparing equation (i) and (ii), we get-
$∴ k = m^{2}$

WB JEE-2017

Differential Equation

87318 The solution of $25 \frac{d^{2} y}{d x^{2}} - 10 \frac{d y}{d x} + y = 0$ , $y (0) = 1, y (1) = 2 e^{1 / 5}$ is

1

y = e^{5 x} + e^{- 5 x}

2

y = (1 + x) e^{5 x}

3

y = (1 + x) e^{x / 5}

4

y = (1 + x) e^{- x / 5}

Explanation:

(C) : We have given,
$25 \frac{d^{2} y}{d x^{2}} - 10 \frac{d y}{d x} + y = 0$
$y = e^{mx}$
$\frac{d y}{d x} = m e^{m x}$
And, $\frac{d^{2} y}{d x^{2}} = m^{2} e^{m x} \Rightarrow 25 \frac{d^{2} y}{d x^{2}} - 10 \frac{d y}{d x} + y = 0$
$25 m^{2} e^{m x} - 10 m e^{m x} + e^{m x} = 0$
$e^{m x} (25 m^{2} - 10 m + 1) = 0$
For auxiliary equation-
$25 m^{2} - 10 m + 1 = 0$
$(5 m - 1) (5 m - 1) = 0$
$m = \frac{1}{5}, \frac{1}{5}$
Since roots are real and equation for general solution.
Given,
$y = (c_{1} + c_{2} x) e^{x / 5}$
$y (0) = 1$
$c_{1} = 1$
$and, y (1) = 2 e^{1 / 5}$
$(c_{1} + c_{2}) e^{1 / 5} = 2 e^{1 / 5}$
$c_{1} + c_{2} = 2$
$1 + c_{2} = 2$
$c_{2} = 1$
Then general solution,
$y = (1 + 1 \cdot x) e^{x / 5}$
$y = (1 + x) e^{x /}$

WB JEE-2012

Differential Equation

87319 The integrating factor of the differential equation $3 x \log_{e} x \frac{d y}{d x} + y = 2 \log_{e} x$ is given by

1

{(\log_{e} x)}^{3}

2

\log_{e} (\log_{e} x)

3

\log_{e} x

4

{(\log_{e} x)}^{1 / 3}

Explanation:

(D) : We have differential equation,
$3 x \log_{e} x \frac{d y}{d x} + y = 2 \log_{e} x$
It can be written as-
$\frac{d y}{d x} + \frac{1}{3 x \log_{e} x} y = \frac{2}{3 x}$
Which is a linear form $\frac{d y}{d x} + P y = Q$ , where $P$ and $Q$ are function of $x$ and the integrating factor is given by the following formula $e^{\int Pdx}$
$∴ I F = e^{\int \frac{1}{3 x \log_{c} x}} dx$
Put, $\log_{e} x = t$
$\frac{1}{x} dx = dt$
$= e^{1 / 3 \frac{dt}{t}} = e^{1 / 3 \log t} = e^{\log t^{1 / 3}} = t^{1 / 3} = {(\log_{e} x)}^{1 / 3}$

WB JEE-2012

Differential Equation

87320 A solution of the differential equation
${(\frac{d y}{d x})}^{2} - x \frac{d y}{d x} + y = 0$ is :

1

y = 2 x

2

y = - 2 x

3

y = 2 x - 4

4

y = 2 x + 4

Explanation:

(C) : Given,
${(\frac{d y}{d x})}^{2} - x \frac{d y}{d x} + y = 0$
Let, $\frac{d y}{d x} = p$
$p^{2} - p x + y = 0$
$y = x p - p^{2}$
On Differentiating both sides w.r.t $x$ , we get-
$\frac{d y}{d x} = x \cdot \frac{d p}{d x} + p - 2 p \cdot \frac{d p}{d x}$
$\frac{d y}{d x} = (x - 2 p) \frac{d p}{d x} + p$
On putting the value $\frac{d y}{d x}$ in above equation,
$p = (x - 2 p) \frac{d p}{d x} + p \Rightarrow (x - 2 p) \frac{d p}{d x} = 0$
$\frac{d p}{d x} = 0$
On integrating w.r.t. $x$ , we get-
$p = c$
Then, $\frac{d y}{d x} = c \Rightarrow y = x . c - c^{2}$
$c = 2$
Then, $y = 2 x - 4$

AMU-2013

Differential Equation

87321 Solution of $(x + y)^{2} \frac{d y}{d x} = a^{2}$ (' $a$ ' being a constant) is

1

\frac{(x + y)}{a} = \tan \frac{y + C}{a}, C

is an arbitrary constant

2

x y = a \tan C x, C

is an arbitrary constant

3

\frac{x}{a} = \tan \frac{y}{C}, C

is an arbitrary constant

4

x y = \tan (x + C), C

is an arbitrary constant

Explanation:

(A) : Given,
Let, $x + y = v$
$(x + y)^{2} \frac{d y}{d x} = a^{2}$
$\Rightarrow 1 + \frac{d y}{d x} = \frac{d v}{d x} \Rightarrow \frac{d y}{d x} = \frac{d v}{d x} - 1$
$∴ v^{2} (\frac{dv}{dx} - 1) = a^{2}$
$\Rightarrow \frac{d v}{d x} = \frac{v^{2} + a^{2}}{v^{2}} \Rightarrow \frac{v^{2}}{v^{2} + a^{2}} d v = d x$
On integrating both sides, we get
$\int \frac{v^{2}}{v^{2} + a^{2}} d v = \int d x$
$\Rightarrow \int 1 - \frac{a^{2}}{v^{2} + a^{2}} dv = x + C^{'}$
$\Rightarrow v - \frac{a^{2}}{a} \tan^{- 1} \frac{v}{a} = x + C^{'}$
$\Rightarrow x + y - a \tan^{- 1} \frac{x + y}{a} = x + C^{'}$
$\Rightarrow y = a \tan^{- 1} \frac{x + y}{a} + C^{'}$
$\frac{y - C^{'}}{a} = \tan^{- 1} \frac{x + y}{a}$
$\Rightarrow (\frac{x + y}{a}) = \tan (\frac{y + C}{a})$ .
Hence, $\frac{(x + y)}{a} = \tan \frac{y + C}{a}, C$ is an arbitrary constant

WB JEE-2017

Differential Equation

87322 If $y = e^{{msin}^{- 1} x}$ then $(1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} - k y = 0$ , where $k$ is equal to

1

m^{2}

2 2

3 -1

4

- m^{2}

Explanation:

(A) : Given,
$(1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} - k y = 0$
$y = e^{{msin}^{- 1} x}$
$\Rightarrow \frac{d y}{d x} = e^{m \sin^{- 1} x} \cdot \frac{m}{\sqrt{1 - x^{2}}} \Rightarrow \sqrt{1 - x^{2}} \frac{d y}{d x} = m^{m \sin^{- 1} x}$
$\Rightarrow \sqrt{1 - x^{2}} \frac{d y}{d x} = m y$
$\Rightarrow \sqrt{1 - x^{2}} \frac{d^{2} y}{d x^{2}} + \frac{1}{2 \sqrt{1 - x^{2}}} (- 2 x) \frac{d y}{d x} = m \frac{d y}{d x}$
$\Rightarrow (1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} = m \sqrt{1 - x^{2}} \frac{d y}{d x}$
$\Rightarrow (1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} = m$ .(my)
$\Rightarrow (1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} - m^{2} y = 0$
On comparing equation (i) and (ii), we get-
$∴ k = m^{2}$

WB JEE-2017

Differential Equation

87318 The solution of $25 \frac{d^{2} y}{d x^{2}} - 10 \frac{d y}{d x} + y = 0$ , $y (0) = 1, y (1) = 2 e^{1 / 5}$ is

1

y = e^{5 x} + e^{- 5 x}

2

y = (1 + x) e^{5 x}

3

y = (1 + x) e^{x / 5}

4

y = (1 + x) e^{- x / 5}

Explanation:

(C) : We have given,
$25 \frac{d^{2} y}{d x^{2}} - 10 \frac{d y}{d x} + y = 0$
$y = e^{mx}$
$\frac{d y}{d x} = m e^{m x}$
And, $\frac{d^{2} y}{d x^{2}} = m^{2} e^{m x} \Rightarrow 25 \frac{d^{2} y}{d x^{2}} - 10 \frac{d y}{d x} + y = 0$
$25 m^{2} e^{m x} - 10 m e^{m x} + e^{m x} = 0$
$e^{m x} (25 m^{2} - 10 m + 1) = 0$
For auxiliary equation-
$25 m^{2} - 10 m + 1 = 0$
$(5 m - 1) (5 m - 1) = 0$
$m = \frac{1}{5}, \frac{1}{5}$
Since roots are real and equation for general solution.
Given,
$y = (c_{1} + c_{2} x) e^{x / 5}$
$y (0) = 1$
$c_{1} = 1$
$and, y (1) = 2 e^{1 / 5}$
$(c_{1} + c_{2}) e^{1 / 5} = 2 e^{1 / 5}$
$c_{1} + c_{2} = 2$
$1 + c_{2} = 2$
$c_{2} = 1$
Then general solution,
$y = (1 + 1 \cdot x) e^{x / 5}$
$y = (1 + x) e^{x /}$

WB JEE-2012

Differential Equation

87319 The integrating factor of the differential equation $3 x \log_{e} x \frac{d y}{d x} + y = 2 \log_{e} x$ is given by

1

{(\log_{e} x)}^{3}

2

\log_{e} (\log_{e} x)

3

\log_{e} x

4

{(\log_{e} x)}^{1 / 3}

Explanation:

(D) : We have differential equation,
$3 x \log_{e} x \frac{d y}{d x} + y = 2 \log_{e} x$
It can be written as-
$\frac{d y}{d x} + \frac{1}{3 x \log_{e} x} y = \frac{2}{3 x}$
Which is a linear form $\frac{d y}{d x} + P y = Q$ , where $P$ and $Q$ are function of $x$ and the integrating factor is given by the following formula $e^{\int Pdx}$
$∴ I F = e^{\int \frac{1}{3 x \log_{c} x}} dx$
Put, $\log_{e} x = t$
$\frac{1}{x} dx = dt$
$= e^{1 / 3 \frac{dt}{t}} = e^{1 / 3 \log t} = e^{\log t^{1 / 3}} = t^{1 / 3} = {(\log_{e} x)}^{1 / 3}$

WB JEE-2012

Differential Equation

87320 A solution of the differential equation
${(\frac{d y}{d x})}^{2} - x \frac{d y}{d x} + y = 0$ is :

1

y = 2 x

2

y = - 2 x

3

y = 2 x - 4

4

y = 2 x + 4

Explanation:

(C) : Given,
${(\frac{d y}{d x})}^{2} - x \frac{d y}{d x} + y = 0$
Let, $\frac{d y}{d x} = p$
$p^{2} - p x + y = 0$
$y = x p - p^{2}$
On Differentiating both sides w.r.t $x$ , we get-
$\frac{d y}{d x} = x \cdot \frac{d p}{d x} + p - 2 p \cdot \frac{d p}{d x}$
$\frac{d y}{d x} = (x - 2 p) \frac{d p}{d x} + p$
On putting the value $\frac{d y}{d x}$ in above equation,
$p = (x - 2 p) \frac{d p}{d x} + p \Rightarrow (x - 2 p) \frac{d p}{d x} = 0$
$\frac{d p}{d x} = 0$
On integrating w.r.t. $x$ , we get-
$p = c$
Then, $\frac{d y}{d x} = c \Rightarrow y = x . c - c^{2}$
$c = 2$
Then, $y = 2 x - 4$

AMU-2013

Differential Equation

87321 Solution of $(x + y)^{2} \frac{d y}{d x} = a^{2}$ (' $a$ ' being a constant) is

1

\frac{(x + y)}{a} = \tan \frac{y + C}{a}, C

is an arbitrary constant

2

x y = a \tan C x, C

is an arbitrary constant

3

\frac{x}{a} = \tan \frac{y}{C}, C

is an arbitrary constant

4

x y = \tan (x + C), C

is an arbitrary constant

Explanation:

(A) : Given,
Let, $x + y = v$
$(x + y)^{2} \frac{d y}{d x} = a^{2}$
$\Rightarrow 1 + \frac{d y}{d x} = \frac{d v}{d x} \Rightarrow \frac{d y}{d x} = \frac{d v}{d x} - 1$
$∴ v^{2} (\frac{dv}{dx} - 1) = a^{2}$
$\Rightarrow \frac{d v}{d x} = \frac{v^{2} + a^{2}}{v^{2}} \Rightarrow \frac{v^{2}}{v^{2} + a^{2}} d v = d x$
On integrating both sides, we get
$\int \frac{v^{2}}{v^{2} + a^{2}} d v = \int d x$
$\Rightarrow \int 1 - \frac{a^{2}}{v^{2} + a^{2}} dv = x + C^{'}$
$\Rightarrow v - \frac{a^{2}}{a} \tan^{- 1} \frac{v}{a} = x + C^{'}$
$\Rightarrow x + y - a \tan^{- 1} \frac{x + y}{a} = x + C^{'}$
$\Rightarrow y = a \tan^{- 1} \frac{x + y}{a} + C^{'}$
$\frac{y - C^{'}}{a} = \tan^{- 1} \frac{x + y}{a}$
$\Rightarrow (\frac{x + y}{a}) = \tan (\frac{y + C}{a})$ .
Hence, $\frac{(x + y)}{a} = \tan \frac{y + C}{a}, C$ is an arbitrary constant

WB JEE-2017

Differential Equation

87322 If $y = e^{{msin}^{- 1} x}$ then $(1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} - k y = 0$ , where $k$ is equal to

1

m^{2}

2 2

3 -1

4

- m^{2}

Explanation:

(A) : Given,
$(1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} - k y = 0$
$y = e^{{msin}^{- 1} x}$
$\Rightarrow \frac{d y}{d x} = e^{m \sin^{- 1} x} \cdot \frac{m}{\sqrt{1 - x^{2}}} \Rightarrow \sqrt{1 - x^{2}} \frac{d y}{d x} = m^{m \sin^{- 1} x}$
$\Rightarrow \sqrt{1 - x^{2}} \frac{d y}{d x} = m y$
$\Rightarrow \sqrt{1 - x^{2}} \frac{d^{2} y}{d x^{2}} + \frac{1}{2 \sqrt{1 - x^{2}}} (- 2 x) \frac{d y}{d x} = m \frac{d y}{d x}$
$\Rightarrow (1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} = m \sqrt{1 - x^{2}} \frac{d y}{d x}$
$\Rightarrow (1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} = m$ .(my)
$\Rightarrow (1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} - m^{2} y = 0$
On comparing equation (i) and (ii), we get-
$∴ k = m^{2}$

WB JEE-2017

Differential Equation

87318 The solution of $25 \frac{d^{2} y}{d x^{2}} - 10 \frac{d y}{d x} + y = 0$ , $y (0) = 1, y (1) = 2 e^{1 / 5}$ is

1

y = e^{5 x} + e^{- 5 x}

2

y = (1 + x) e^{5 x}

3

y = (1 + x) e^{x / 5}

4

y = (1 + x) e^{- x / 5}

Explanation:

(C) : We have given,
$25 \frac{d^{2} y}{d x^{2}} - 10 \frac{d y}{d x} + y = 0$
$y = e^{mx}$
$\frac{d y}{d x} = m e^{m x}$
And, $\frac{d^{2} y}{d x^{2}} = m^{2} e^{m x} \Rightarrow 25 \frac{d^{2} y}{d x^{2}} - 10 \frac{d y}{d x} + y = 0$
$25 m^{2} e^{m x} - 10 m e^{m x} + e^{m x} = 0$
$e^{m x} (25 m^{2} - 10 m + 1) = 0$
For auxiliary equation-
$25 m^{2} - 10 m + 1 = 0$
$(5 m - 1) (5 m - 1) = 0$
$m = \frac{1}{5}, \frac{1}{5}$
Since roots are real and equation for general solution.
Given,
$y = (c_{1} + c_{2} x) e^{x / 5}$
$y (0) = 1$
$c_{1} = 1$
$and, y (1) = 2 e^{1 / 5}$
$(c_{1} + c_{2}) e^{1 / 5} = 2 e^{1 / 5}$
$c_{1} + c_{2} = 2$
$1 + c_{2} = 2$
$c_{2} = 1$
Then general solution,
$y = (1 + 1 \cdot x) e^{x / 5}$
$y = (1 + x) e^{x /}$

WB JEE-2012

Differential Equation

87319 The integrating factor of the differential equation $3 x \log_{e} x \frac{d y}{d x} + y = 2 \log_{e} x$ is given by

1

{(\log_{e} x)}^{3}

2

\log_{e} (\log_{e} x)

3

\log_{e} x

4

{(\log_{e} x)}^{1 / 3}

Explanation:

(D) : We have differential equation,
$3 x \log_{e} x \frac{d y}{d x} + y = 2 \log_{e} x$
It can be written as-
$\frac{d y}{d x} + \frac{1}{3 x \log_{e} x} y = \frac{2}{3 x}$
Which is a linear form $\frac{d y}{d x} + P y = Q$ , where $P$ and $Q$ are function of $x$ and the integrating factor is given by the following formula $e^{\int Pdx}$
$∴ I F = e^{\int \frac{1}{3 x \log_{c} x}} dx$
Put, $\log_{e} x = t$
$\frac{1}{x} dx = dt$
$= e^{1 / 3 \frac{dt}{t}} = e^{1 / 3 \log t} = e^{\log t^{1 / 3}} = t^{1 / 3} = {(\log_{e} x)}^{1 / 3}$

WB JEE-2012

Differential Equation

87320 A solution of the differential equation
${(\frac{d y}{d x})}^{2} - x \frac{d y}{d x} + y = 0$ is :

1

y = 2 x

2

y = - 2 x

3

y = 2 x - 4

4

y = 2 x + 4

Explanation:

(C) : Given,
${(\frac{d y}{d x})}^{2} - x \frac{d y}{d x} + y = 0$
Let, $\frac{d y}{d x} = p$
$p^{2} - p x + y = 0$
$y = x p - p^{2}$
On Differentiating both sides w.r.t $x$ , we get-
$\frac{d y}{d x} = x \cdot \frac{d p}{d x} + p - 2 p \cdot \frac{d p}{d x}$
$\frac{d y}{d x} = (x - 2 p) \frac{d p}{d x} + p$
On putting the value $\frac{d y}{d x}$ in above equation,
$p = (x - 2 p) \frac{d p}{d x} + p \Rightarrow (x - 2 p) \frac{d p}{d x} = 0$
$\frac{d p}{d x} = 0$
On integrating w.r.t. $x$ , we get-
$p = c$
Then, $\frac{d y}{d x} = c \Rightarrow y = x . c - c^{2}$
$c = 2$
Then, $y = 2 x - 4$

AMU-2013

Differential Equation

87321 Solution of $(x + y)^{2} \frac{d y}{d x} = a^{2}$ (' $a$ ' being a constant) is

1

\frac{(x + y)}{a} = \tan \frac{y + C}{a}, C

is an arbitrary constant

2

x y = a \tan C x, C

is an arbitrary constant

3

\frac{x}{a} = \tan \frac{y}{C}, C

is an arbitrary constant

4

x y = \tan (x + C), C

is an arbitrary constant

Explanation:

(A) : Given,
Let, $x + y = v$
$(x + y)^{2} \frac{d y}{d x} = a^{2}$
$\Rightarrow 1 + \frac{d y}{d x} = \frac{d v}{d x} \Rightarrow \frac{d y}{d x} = \frac{d v}{d x} - 1$
$∴ v^{2} (\frac{dv}{dx} - 1) = a^{2}$
$\Rightarrow \frac{d v}{d x} = \frac{v^{2} + a^{2}}{v^{2}} \Rightarrow \frac{v^{2}}{v^{2} + a^{2}} d v = d x$
On integrating both sides, we get
$\int \frac{v^{2}}{v^{2} + a^{2}} d v = \int d x$
$\Rightarrow \int 1 - \frac{a^{2}}{v^{2} + a^{2}} dv = x + C^{'}$
$\Rightarrow v - \frac{a^{2}}{a} \tan^{- 1} \frac{v}{a} = x + C^{'}$
$\Rightarrow x + y - a \tan^{- 1} \frac{x + y}{a} = x + C^{'}$
$\Rightarrow y = a \tan^{- 1} \frac{x + y}{a} + C^{'}$
$\frac{y - C^{'}}{a} = \tan^{- 1} \frac{x + y}{a}$
$\Rightarrow (\frac{x + y}{a}) = \tan (\frac{y + C}{a})$ .
Hence, $\frac{(x + y)}{a} = \tan \frac{y + C}{a}, C$ is an arbitrary constant

WB JEE-2017

Differential Equation

87322 If $y = e^{{msin}^{- 1} x}$ then $(1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} - k y = 0$ , where $k$ is equal to

1

m^{2}

2 2

3 -1

4

- m^{2}

Explanation:

(A) : Given,
$(1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} - k y = 0$
$y = e^{{msin}^{- 1} x}$
$\Rightarrow \frac{d y}{d x} = e^{m \sin^{- 1} x} \cdot \frac{m}{\sqrt{1 - x^{2}}} \Rightarrow \sqrt{1 - x^{2}} \frac{d y}{d x} = m^{m \sin^{- 1} x}$
$\Rightarrow \sqrt{1 - x^{2}} \frac{d y}{d x} = m y$
$\Rightarrow \sqrt{1 - x^{2}} \frac{d^{2} y}{d x^{2}} + \frac{1}{2 \sqrt{1 - x^{2}}} (- 2 x) \frac{d y}{d x} = m \frac{d y}{d x}$
$\Rightarrow (1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} = m \sqrt{1 - x^{2}} \frac{d y}{d x}$
$\Rightarrow (1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} = m$ .(my)
$\Rightarrow (1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} - m^{2} y = 0$
On comparing equation (i) and (ii), we get-
$∴ k = m^{2}$

WB JEE-2017