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Differential Equation

87307 The solution of the differential equation $\frac{d y}{d x} - 2 y \tan 2 x = e^{x} \sec 2 x$ is

1

y \sin 2 x = e^{x} + C

2

y \cos 2 x = e^{x} + C

3

y = e^{x} \cos 2 x + C

4

y \cos 2 x + e^{x} = C

Explanation:

(B) : We have differential equation-
$\frac{d y}{d x} - 2 y \tan 2 x = e^{x} \sec 2 x$
Which is a linear differential equation,
$P = - 2 \tan 2 x and Q = e^{x} \sec 2 x$
I.F, $= e^{\int p pdx} = e^{- \int 2 \tan 2 x d x} = e^{- [\frac{\log \sec 2 x}{2}]}$
$= (\sec 2 x)^{- 1} = \cos 2 x$
Now for general solution
$y \cdot \cos 2 x = \int \cos 2 x \cdot e^{x} \sec 2 x d x$
$y \cos 2 x = \int e^{x} d x$
$y \cos 2 x = e^{x} + C$

AP EAMCET-2013

Differential Equation

87308 If $X$ is a poisson variation with $P (X = 0) = 0.8$ , then the variance of $X$ is

1

\log_{e} 20

2

\log_{10} 20

3

\log_{e} 5 / 4

4 0

Explanation:

(C) : We have-
$P (X = 0) = 0.8$
We have that,
$P (X = 0) = \frac{e^{- λ} λ^{x}}{x!} \Rightarrow x = 0$
Then,
$\frac{e^{- λ} (λ)^{0}}{0!} = 0.8 \Rightarrow e^{- λ} = 0.8$
$e^{- λ} = 4 / 5$
$e^{- λ} = 5 / 4$
$λ = \log_{e} 5 / 4$

AP EAMCET-2004

Differential Equation

87309 The differential equation of the family of parabolas with vertex at $(0, - 1)$ and having axis along the $Y$ -axis is

1

y y^{'} + 2 x y + 1 = 0

2

x y^{'} + y + 1 = 0

3

x y^{'} - 2 y - 2 = 0

4

x y^{'} - y - 1 = 0

Explanation:

(C) : We have the family of parabola having axis along the y axis
$(x - h)^{2} = 4 a (y - k)$
Where $(h, k)$ be the vertex of parabola,
Given vertex $(h, k) \to (0, - 1)$
The equation become $(x - 0)^{2} = 4 a (y + 1)$
$x^{2} = 4 a (y + 1)$
differentiating w.r.t $x$ we get-
$2 x = 4 a \frac{d y}{d x} \Rightarrow a = \frac{x}{2} \frac{d x}{d y}$
Substituting he value of 'a' in equation we get-
$x^{2} = 4 \cdot \frac{x}{2} \frac{d x}{d y} (y + 1) \Rightarrow x \frac{d y}{d x} = 2 (y + 1)$
$x \frac{d y}{d x} - 2 y - 2 = 0$
$x y^{'} - 2 y - 2 = 0$

AP EAMCET-2014

Differential Equation

87310 The solution of $\frac{d y}{d x} + \frac{1}{x} = \frac{e^{y}}{x^{2}}$ is

1

2 x = (1 + C x^{2}) e^{y}

2

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

3

2 x^{2} = (1 + C x^{2}) e^{- y}

4

x^{2} = (1 + C x^{2}) e^{- y}

Explanation:

We have given,
$\frac{d y}{d x} + \frac{1}{x} = \frac{e^{y}}{x^{2}}$
$\Rightarrow \frac{d y}{d x} e^{- y} + \frac{1}{x} e^{- y} = \frac{1}{x^{2}}$
Let,
$v = e^{- y}$
Now equation becomes:
$- \frac{d v}{d x} + \frac{v}{x} = \frac{1}{x^{2}}$
Which is a linear differential equation in $v$ .
$P = - \frac{1}{x} and Q = \frac{- 1}{x^{2}}$
$I . F = e^{\int P d x} = e^{- \int \frac{1}{x} d x} = e^{- \log x} = \frac{1}{x}$
For the general solution we get:
$v (I . F) = \int (I . F) Q d x + C$
$\frac{1}{x} v = \int (\frac{1}{x^{3}}) d x \Rightarrow \frac{1}{x} v = \frac{x^{- 2}}{- 2} + C_{1}$
$v = \frac{- x^{- 2}}{2} + C_{1}$
$e^{- y} = \frac{1}{2 x^{2}} + C_{1}$
$\Rightarrow \frac{1}{2 x^{2}} + C_{1} = \frac{1}{2 x^{2}} + C_{1}$
$e^{- y} = \frac{1}{2 x} + C_{1}$
$2 x = e^{y} + 2 x C_{1} e^{y}$
$2 x = e^{y} (1 + C x^{2}) ∴ C = 2 C_{1}$

AP EAMCET-2015

Differential Equation

87307 The solution of the differential equation $\frac{d y}{d x} - 2 y \tan 2 x = e^{x} \sec 2 x$ is

1

y \sin 2 x = e^{x} + C

2

y \cos 2 x = e^{x} + C

3

y = e^{x} \cos 2 x + C

4

y \cos 2 x + e^{x} = C

Explanation:

(B) : We have differential equation-
$\frac{d y}{d x} - 2 y \tan 2 x = e^{x} \sec 2 x$
Which is a linear differential equation,
$P = - 2 \tan 2 x and Q = e^{x} \sec 2 x$
I.F, $= e^{\int p pdx} = e^{- \int 2 \tan 2 x d x} = e^{- [\frac{\log \sec 2 x}{2}]}$
$= (\sec 2 x)^{- 1} = \cos 2 x$
Now for general solution
$y \cdot \cos 2 x = \int \cos 2 x \cdot e^{x} \sec 2 x d x$
$y \cos 2 x = \int e^{x} d x$
$y \cos 2 x = e^{x} + C$

AP EAMCET-2013

Differential Equation

87308 If $X$ is a poisson variation with $P (X = 0) = 0.8$ , then the variance of $X$ is

1

\log_{e} 20

2

\log_{10} 20

3

\log_{e} 5 / 4

4 0

Explanation:

(C) : We have-
$P (X = 0) = 0.8$
We have that,
$P (X = 0) = \frac{e^{- λ} λ^{x}}{x!} \Rightarrow x = 0$
Then,
$\frac{e^{- λ} (λ)^{0}}{0!} = 0.8 \Rightarrow e^{- λ} = 0.8$
$e^{- λ} = 4 / 5$
$e^{- λ} = 5 / 4$
$λ = \log_{e} 5 / 4$

AP EAMCET-2004

Differential Equation

87309 The differential equation of the family of parabolas with vertex at $(0, - 1)$ and having axis along the $Y$ -axis is

1

y y^{'} + 2 x y + 1 = 0

2

x y^{'} + y + 1 = 0

3

x y^{'} - 2 y - 2 = 0

4

x y^{'} - y - 1 = 0

Explanation:

(C) : We have the family of parabola having axis along the y axis
$(x - h)^{2} = 4 a (y - k)$
Where $(h, k)$ be the vertex of parabola,
Given vertex $(h, k) \to (0, - 1)$
The equation become $(x - 0)^{2} = 4 a (y + 1)$
$x^{2} = 4 a (y + 1)$
differentiating w.r.t $x$ we get-
$2 x = 4 a \frac{d y}{d x} \Rightarrow a = \frac{x}{2} \frac{d x}{d y}$
Substituting he value of 'a' in equation we get-
$x^{2} = 4 \cdot \frac{x}{2} \frac{d x}{d y} (y + 1) \Rightarrow x \frac{d y}{d x} = 2 (y + 1)$
$x \frac{d y}{d x} - 2 y - 2 = 0$
$x y^{'} - 2 y - 2 = 0$

AP EAMCET-2014

Differential Equation

87310 The solution of $\frac{d y}{d x} + \frac{1}{x} = \frac{e^{y}}{x^{2}}$ is

1

2 x = (1 + C x^{2}) e^{y}

2

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

3

2 x^{2} = (1 + C x^{2}) e^{- y}

4

x^{2} = (1 + C x^{2}) e^{- y}

Explanation:

We have given,
$\frac{d y}{d x} + \frac{1}{x} = \frac{e^{y}}{x^{2}}$
$\Rightarrow \frac{d y}{d x} e^{- y} + \frac{1}{x} e^{- y} = \frac{1}{x^{2}}$
Let,
$v = e^{- y}$
Now equation becomes:
$- \frac{d v}{d x} + \frac{v}{x} = \frac{1}{x^{2}}$
Which is a linear differential equation in $v$ .
$P = - \frac{1}{x} and Q = \frac{- 1}{x^{2}}$
$I . F = e^{\int P d x} = e^{- \int \frac{1}{x} d x} = e^{- \log x} = \frac{1}{x}$
For the general solution we get:
$v (I . F) = \int (I . F) Q d x + C$
$\frac{1}{x} v = \int (\frac{1}{x^{3}}) d x \Rightarrow \frac{1}{x} v = \frac{x^{- 2}}{- 2} + C_{1}$
$v = \frac{- x^{- 2}}{2} + C_{1}$
$e^{- y} = \frac{1}{2 x^{2}} + C_{1}$
$\Rightarrow \frac{1}{2 x^{2}} + C_{1} = \frac{1}{2 x^{2}} + C_{1}$
$e^{- y} = \frac{1}{2 x} + C_{1}$
$2 x = e^{y} + 2 x C_{1} e^{y}$
$2 x = e^{y} (1 + C x^{2}) ∴ C = 2 C_{1}$

AP EAMCET-2015

Differential Equation

87307 The solution of the differential equation $\frac{d y}{d x} - 2 y \tan 2 x = e^{x} \sec 2 x$ is

1

y \sin 2 x = e^{x} + C

2

y \cos 2 x = e^{x} + C

3

y = e^{x} \cos 2 x + C

4

y \cos 2 x + e^{x} = C

Explanation:

(B) : We have differential equation-
$\frac{d y}{d x} - 2 y \tan 2 x = e^{x} \sec 2 x$
Which is a linear differential equation,
$P = - 2 \tan 2 x and Q = e^{x} \sec 2 x$
I.F, $= e^{\int p pdx} = e^{- \int 2 \tan 2 x d x} = e^{- [\frac{\log \sec 2 x}{2}]}$
$= (\sec 2 x)^{- 1} = \cos 2 x$
Now for general solution
$y \cdot \cos 2 x = \int \cos 2 x \cdot e^{x} \sec 2 x d x$
$y \cos 2 x = \int e^{x} d x$
$y \cos 2 x = e^{x} + C$

AP EAMCET-2013

Differential Equation

87308 If $X$ is a poisson variation with $P (X = 0) = 0.8$ , then the variance of $X$ is

1

\log_{e} 20

2

\log_{10} 20

3

\log_{e} 5 / 4

4 0

Explanation:

(C) : We have-
$P (X = 0) = 0.8$
We have that,
$P (X = 0) = \frac{e^{- λ} λ^{x}}{x!} \Rightarrow x = 0$
Then,
$\frac{e^{- λ} (λ)^{0}}{0!} = 0.8 \Rightarrow e^{- λ} = 0.8$
$e^{- λ} = 4 / 5$
$e^{- λ} = 5 / 4$
$λ = \log_{e} 5 / 4$

AP EAMCET-2004

Differential Equation

87309 The differential equation of the family of parabolas with vertex at $(0, - 1)$ and having axis along the $Y$ -axis is

1

y y^{'} + 2 x y + 1 = 0

2

x y^{'} + y + 1 = 0

3

x y^{'} - 2 y - 2 = 0

4

x y^{'} - y - 1 = 0

Explanation:

(C) : We have the family of parabola having axis along the y axis
$(x - h)^{2} = 4 a (y - k)$
Where $(h, k)$ be the vertex of parabola,
Given vertex $(h, k) \to (0, - 1)$
The equation become $(x - 0)^{2} = 4 a (y + 1)$
$x^{2} = 4 a (y + 1)$
differentiating w.r.t $x$ we get-
$2 x = 4 a \frac{d y}{d x} \Rightarrow a = \frac{x}{2} \frac{d x}{d y}$
Substituting he value of 'a' in equation we get-
$x^{2} = 4 \cdot \frac{x}{2} \frac{d x}{d y} (y + 1) \Rightarrow x \frac{d y}{d x} = 2 (y + 1)$
$x \frac{d y}{d x} - 2 y - 2 = 0$
$x y^{'} - 2 y - 2 = 0$

AP EAMCET-2014

Differential Equation

87310 The solution of $\frac{d y}{d x} + \frac{1}{x} = \frac{e^{y}}{x^{2}}$ is

1

2 x = (1 + C x^{2}) e^{y}

2

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

3

2 x^{2} = (1 + C x^{2}) e^{- y}

4

x^{2} = (1 + C x^{2}) e^{- y}

Explanation:

We have given,
$\frac{d y}{d x} + \frac{1}{x} = \frac{e^{y}}{x^{2}}$
$\Rightarrow \frac{d y}{d x} e^{- y} + \frac{1}{x} e^{- y} = \frac{1}{x^{2}}$
Let,
$v = e^{- y}$
Now equation becomes:
$- \frac{d v}{d x} + \frac{v}{x} = \frac{1}{x^{2}}$
Which is a linear differential equation in $v$ .
$P = - \frac{1}{x} and Q = \frac{- 1}{x^{2}}$
$I . F = e^{\int P d x} = e^{- \int \frac{1}{x} d x} = e^{- \log x} = \frac{1}{x}$
For the general solution we get:
$v (I . F) = \int (I . F) Q d x + C$
$\frac{1}{x} v = \int (\frac{1}{x^{3}}) d x \Rightarrow \frac{1}{x} v = \frac{x^{- 2}}{- 2} + C_{1}$
$v = \frac{- x^{- 2}}{2} + C_{1}$
$e^{- y} = \frac{1}{2 x^{2}} + C_{1}$
$\Rightarrow \frac{1}{2 x^{2}} + C_{1} = \frac{1}{2 x^{2}} + C_{1}$
$e^{- y} = \frac{1}{2 x} + C_{1}$
$2 x = e^{y} + 2 x C_{1} e^{y}$
$2 x = e^{y} (1 + C x^{2}) ∴ C = 2 C_{1}$

AP EAMCET-2015

Differential Equation

87307 The solution of the differential equation $\frac{d y}{d x} - 2 y \tan 2 x = e^{x} \sec 2 x$ is

1

y \sin 2 x = e^{x} + C

2

y \cos 2 x = e^{x} + C

3

y = e^{x} \cos 2 x + C

4

y \cos 2 x + e^{x} = C

Explanation:

(B) : We have differential equation-
$\frac{d y}{d x} - 2 y \tan 2 x = e^{x} \sec 2 x$
Which is a linear differential equation,
$P = - 2 \tan 2 x and Q = e^{x} \sec 2 x$
I.F, $= e^{\int p pdx} = e^{- \int 2 \tan 2 x d x} = e^{- [\frac{\log \sec 2 x}{2}]}$
$= (\sec 2 x)^{- 1} = \cos 2 x$
Now for general solution
$y \cdot \cos 2 x = \int \cos 2 x \cdot e^{x} \sec 2 x d x$
$y \cos 2 x = \int e^{x} d x$
$y \cos 2 x = e^{x} + C$

AP EAMCET-2013

Differential Equation

87308 If $X$ is a poisson variation with $P (X = 0) = 0.8$ , then the variance of $X$ is

1

\log_{e} 20

2

\log_{10} 20

3

\log_{e} 5 / 4

4 0

Explanation:

(C) : We have-
$P (X = 0) = 0.8$
We have that,
$P (X = 0) = \frac{e^{- λ} λ^{x}}{x!} \Rightarrow x = 0$
Then,
$\frac{e^{- λ} (λ)^{0}}{0!} = 0.8 \Rightarrow e^{- λ} = 0.8$
$e^{- λ} = 4 / 5$
$e^{- λ} = 5 / 4$
$λ = \log_{e} 5 / 4$

AP EAMCET-2004

Differential Equation

87309 The differential equation of the family of parabolas with vertex at $(0, - 1)$ and having axis along the $Y$ -axis is

1

y y^{'} + 2 x y + 1 = 0

2

x y^{'} + y + 1 = 0

3

x y^{'} - 2 y - 2 = 0

4

x y^{'} - y - 1 = 0

Explanation:

(C) : We have the family of parabola having axis along the y axis
$(x - h)^{2} = 4 a (y - k)$
Where $(h, k)$ be the vertex of parabola,
Given vertex $(h, k) \to (0, - 1)$
The equation become $(x - 0)^{2} = 4 a (y + 1)$
$x^{2} = 4 a (y + 1)$
differentiating w.r.t $x$ we get-
$2 x = 4 a \frac{d y}{d x} \Rightarrow a = \frac{x}{2} \frac{d x}{d y}$
Substituting he value of 'a' in equation we get-
$x^{2} = 4 \cdot \frac{x}{2} \frac{d x}{d y} (y + 1) \Rightarrow x \frac{d y}{d x} = 2 (y + 1)$
$x \frac{d y}{d x} - 2 y - 2 = 0$
$x y^{'} - 2 y - 2 = 0$

AP EAMCET-2014

Differential Equation

87310 The solution of $\frac{d y}{d x} + \frac{1}{x} = \frac{e^{y}}{x^{2}}$ is

1

2 x = (1 + C x^{2}) e^{y}

2

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

3

2 x^{2} = (1 + C x^{2}) e^{- y}

4

x^{2} = (1 + C x^{2}) e^{- y}

Explanation:

We have given,
$\frac{d y}{d x} + \frac{1}{x} = \frac{e^{y}}{x^{2}}$
$\Rightarrow \frac{d y}{d x} e^{- y} + \frac{1}{x} e^{- y} = \frac{1}{x^{2}}$
Let,
$v = e^{- y}$
Now equation becomes:
$- \frac{d v}{d x} + \frac{v}{x} = \frac{1}{x^{2}}$
Which is a linear differential equation in $v$ .
$P = - \frac{1}{x} and Q = \frac{- 1}{x^{2}}$
$I . F = e^{\int P d x} = e^{- \int \frac{1}{x} d x} = e^{- \log x} = \frac{1}{x}$
For the general solution we get:
$v (I . F) = \int (I . F) Q d x + C$
$\frac{1}{x} v = \int (\frac{1}{x^{3}}) d x \Rightarrow \frac{1}{x} v = \frac{x^{- 2}}{- 2} + C_{1}$
$v = \frac{- x^{- 2}}{2} + C_{1}$
$e^{- y} = \frac{1}{2 x^{2}} + C_{1}$
$\Rightarrow \frac{1}{2 x^{2}} + C_{1} = \frac{1}{2 x^{2}} + C_{1}$
$e^{- y} = \frac{1}{2 x} + C_{1}$
$2 x = e^{y} + 2 x C_{1} e^{y}$
$2 x = e^{y} (1 + C x^{2}) ∴ C = 2 C_{1}$

AP EAMCET-2015