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Differential Equation

87489 The differential $y \frac{d y}{d x} + x =$ c represents

1 a family of parabolas

2 a family of circles whose centres are on the

x

axis

3 a family of hyperbolas

4 a family of circles whose centres are on the yaxis

Explanation:

(B) : Given,
$\begin{array}{l} y \frac{d y}{d x} + x = c \\ y \frac{d y}{d x} = c - x \\ y \frac{d y}{d y} = (c - x) d x \end{array}$
On integration both sides, we get
$\begin{array}{l} \int y d d = \int (c - x) dx \\ \frac{y^{2}}{2} = cx - \frac{x^{2}}{2} + k \\ \frac{x^{2}}{2} + \frac{y^{2}}{2} = cx + k \\ x^{2} + y^{2} - 2 cx - 2 k = 0 \end{array}$
This equation is represent the circle whose centre are on the x -axis.

Karnataka CET-2008

Differential Equation

87490 The differential equation of the family of circles passing through the origin and having their centres on the $x$ -axis is

1

x^{2} = y^{2} + x y \frac{d y}{d x}

2

x^{2} = y^{2} + 3 x y \frac{d y}{d x}

3

y^{2} = x^{2} + 2 x y \frac{d y}{d x}

4

y^{2} = x^{2} - 2 x y \frac{d y}{d x}

Explanation:

(C) : Given that family circle passing through the origin and centre on $x$ - axis.
Let centre be $(h, 0)$ and radius be $h$ .
Now equation of family of circle,
$(x - h)^{2} + (y - k)^{2} = h^{2}$
$(x - h)^{2} + (y - 0)^{2} = h^{2}$
$(x - h)^{2} + y^{2} = h^{2}$
$x^{2} + h^{2} - 2 xh + y^{2} = h^{2}$
$x^{2} + y^{2} - 2 xh = 0$
On differentiating both sides w.r.t.x, we get-
$2 x + 2 y \frac{d y}{d x} - 2 h = 0$
$x + y \frac{d y}{d x} = h$
On putting the value of $h$ in equation (i)
$x^{2} + y^{2} - 2 x (x + y \frac{d y}{d x}) = 0$
$x^{2} + y^{2} - 2 x^{2} - 2 xy \frac{dy}{dx} = 0$
$- x^{2} + y^{2} - 2 x y \frac{d y}{d x} = 0$
$x^{2} - y^{2} + 2 xy \frac{dy}{dx} = 0$
$y^{2} = x^{2} + 2 x y \frac{d y}{d x}$

Karnataka CET-2009

Differential Equation

87491 The integrating factor of the differential equation $\sin y (\frac{d y}{d x}) = \cos y (1 - x \cos y)$ is

1

e^{- x}

2

e^{- y}

3

e^{\sin y}

4

e^{\cos y}

Explanation:

(A) : Given,
$\sin y (\frac{d y}{d x}) = \cos y (1 - x \cos y)$
$\frac{\sin y}{\cos y} (\frac{d y}{d x}) = 1 - x \cos y$
$\tan y (\frac{d y}{d x}) = 1 - x \cos y$
$\tan y (\frac{d y}{d x}) = 1 - \frac{x}{\sec y}$
$\tan y \cdot \sec y (\frac{d y}{d x}) = \sec y - x$
Put, $\sec y = t$
$\sec y \cdot \tan y (\frac{d y}{d x}) = \frac{d t}{d x}$
Then equation becomes,
$\frac{dt}{dx} - t = - x$
Which is the form of $\frac{d y}{d x} + P y = Q$
$P = - 1$ and $Q = - X$
I.F. $= e^{\int - 1 dx} = e^{- x}$

MHT CET-2020

Differential Equation

87493 Solution of the differential equation $\frac{d y}{d x} + 2 y = e^{- x}$ is

1

{ye}^{x} = x + c

2

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

3

{ye}^{2 x} = x + c

4

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

Explanation:

(D) : Given,
$\frac{d y}{d x} + 2 y = e^{- x}$
Now, which is form of $\frac{d y}{d x} + P y = Q$
$\begin{aligned} I.F & = e^{\int 2 dx} \\ = e^{2 x} \end{aligned}$
$P = 2 and Q = e^{- x}$
For particular solution,
$\begin{array}{l} y \cdot IF = \int Q \cdot (IF) dx + c \\ ∴ y \cdot e^{2 x} = \int e^{2 x} \cdot e^{- x} dx + c \\ {ye}^{2 x} = \int e^{x} dx + c \\ {ye}^{2 x} = e^{x} + c \end{array}$

MHT CET-2020

Differential Equation

87494 The integrating factor of the differential equation $\frac{d y}{d x} + \frac{1}{x} y = x^{3} - 3$ is

1 -y

2 x

3 -x

4 y

Explanation:

(B) : Given,
$\frac{d y}{d x} + \frac{1}{x} y = x^{3} - 3$
The above differential equation of the form $\frac{d y}{d x} + P y = Q$
$P = \frac{1}{x}$ and $Q = x^{3} - 3$
I. F. $= e^{\int P Pdx}$
$= e^{\int \frac{1}{x} dx}$
$= e^{\log x} = x$

MHT CET-2020

Differential Equation

87489 The differential $y \frac{d y}{d x} + x =$ c represents

1 a family of parabolas

2 a family of circles whose centres are on the

x

axis

3 a family of hyperbolas

4 a family of circles whose centres are on the yaxis

Explanation:

(B) : Given,
$\begin{array}{l} y \frac{d y}{d x} + x = c \\ y \frac{d y}{d x} = c - x \\ y \frac{d y}{d y} = (c - x) d x \end{array}$
On integration both sides, we get
$\begin{array}{l} \int y d d = \int (c - x) dx \\ \frac{y^{2}}{2} = cx - \frac{x^{2}}{2} + k \\ \frac{x^{2}}{2} + \frac{y^{2}}{2} = cx + k \\ x^{2} + y^{2} - 2 cx - 2 k = 0 \end{array}$
This equation is represent the circle whose centre are on the x -axis.

Karnataka CET-2008

Differential Equation

87490 The differential equation of the family of circles passing through the origin and having their centres on the $x$ -axis is

1

x^{2} = y^{2} + x y \frac{d y}{d x}

2

x^{2} = y^{2} + 3 x y \frac{d y}{d x}

3

y^{2} = x^{2} + 2 x y \frac{d y}{d x}

4

y^{2} = x^{2} - 2 x y \frac{d y}{d x}

Explanation:

(C) : Given that family circle passing through the origin and centre on $x$ - axis.
Let centre be $(h, 0)$ and radius be $h$ .
Now equation of family of circle,
$(x - h)^{2} + (y - k)^{2} = h^{2}$
$(x - h)^{2} + (y - 0)^{2} = h^{2}$
$(x - h)^{2} + y^{2} = h^{2}$
$x^{2} + h^{2} - 2 xh + y^{2} = h^{2}$
$x^{2} + y^{2} - 2 xh = 0$
On differentiating both sides w.r.t.x, we get-
$2 x + 2 y \frac{d y}{d x} - 2 h = 0$
$x + y \frac{d y}{d x} = h$
On putting the value of $h$ in equation (i)
$x^{2} + y^{2} - 2 x (x + y \frac{d y}{d x}) = 0$
$x^{2} + y^{2} - 2 x^{2} - 2 xy \frac{dy}{dx} = 0$
$- x^{2} + y^{2} - 2 x y \frac{d y}{d x} = 0$
$x^{2} - y^{2} + 2 xy \frac{dy}{dx} = 0$
$y^{2} = x^{2} + 2 x y \frac{d y}{d x}$

Karnataka CET-2009

Differential Equation

87491 The integrating factor of the differential equation $\sin y (\frac{d y}{d x}) = \cos y (1 - x \cos y)$ is

1

e^{- x}

2

e^{- y}

3

e^{\sin y}

4

e^{\cos y}

Explanation:

(A) : Given,
$\sin y (\frac{d y}{d x}) = \cos y (1 - x \cos y)$
$\frac{\sin y}{\cos y} (\frac{d y}{d x}) = 1 - x \cos y$
$\tan y (\frac{d y}{d x}) = 1 - x \cos y$
$\tan y (\frac{d y}{d x}) = 1 - \frac{x}{\sec y}$
$\tan y \cdot \sec y (\frac{d y}{d x}) = \sec y - x$
Put, $\sec y = t$
$\sec y \cdot \tan y (\frac{d y}{d x}) = \frac{d t}{d x}$
Then equation becomes,
$\frac{dt}{dx} - t = - x$
Which is the form of $\frac{d y}{d x} + P y = Q$
$P = - 1$ and $Q = - X$
I.F. $= e^{\int - 1 dx} = e^{- x}$

MHT CET-2020

Differential Equation

87493 Solution of the differential equation $\frac{d y}{d x} + 2 y = e^{- x}$ is

1

{ye}^{x} = x + c

2

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

3

{ye}^{2 x} = x + c

4

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

Explanation:

(D) : Given,
$\frac{d y}{d x} + 2 y = e^{- x}$
Now, which is form of $\frac{d y}{d x} + P y = Q$
$\begin{aligned} I.F & = e^{\int 2 dx} \\ = e^{2 x} \end{aligned}$
$P = 2 and Q = e^{- x}$
For particular solution,
$\begin{array}{l} y \cdot IF = \int Q \cdot (IF) dx + c \\ ∴ y \cdot e^{2 x} = \int e^{2 x} \cdot e^{- x} dx + c \\ {ye}^{2 x} = \int e^{x} dx + c \\ {ye}^{2 x} = e^{x} + c \end{array}$

MHT CET-2020

Differential Equation

87494 The integrating factor of the differential equation $\frac{d y}{d x} + \frac{1}{x} y = x^{3} - 3$ is

1 -y

2 x

3 -x

4 y

Explanation:

(B) : Given,
$\frac{d y}{d x} + \frac{1}{x} y = x^{3} - 3$
The above differential equation of the form $\frac{d y}{d x} + P y = Q$
$P = \frac{1}{x}$ and $Q = x^{3} - 3$
I. F. $= e^{\int P Pdx}$
$= e^{\int \frac{1}{x} dx}$
$= e^{\log x} = x$

MHT CET-2020

Differential Equation

87489 The differential $y \frac{d y}{d x} + x =$ c represents

1 a family of parabolas

2 a family of circles whose centres are on the

x

axis

3 a family of hyperbolas

4 a family of circles whose centres are on the yaxis

Explanation:

(B) : Given,
$\begin{array}{l} y \frac{d y}{d x} + x = c \\ y \frac{d y}{d x} = c - x \\ y \frac{d y}{d y} = (c - x) d x \end{array}$
On integration both sides, we get
$\begin{array}{l} \int y d d = \int (c - x) dx \\ \frac{y^{2}}{2} = cx - \frac{x^{2}}{2} + k \\ \frac{x^{2}}{2} + \frac{y^{2}}{2} = cx + k \\ x^{2} + y^{2} - 2 cx - 2 k = 0 \end{array}$
This equation is represent the circle whose centre are on the x -axis.

Karnataka CET-2008

Differential Equation

87490 The differential equation of the family of circles passing through the origin and having their centres on the $x$ -axis is

1

x^{2} = y^{2} + x y \frac{d y}{d x}

2

x^{2} = y^{2} + 3 x y \frac{d y}{d x}

3

y^{2} = x^{2} + 2 x y \frac{d y}{d x}

4

y^{2} = x^{2} - 2 x y \frac{d y}{d x}

Explanation:

(C) : Given that family circle passing through the origin and centre on $x$ - axis.
Let centre be $(h, 0)$ and radius be $h$ .
Now equation of family of circle,
$(x - h)^{2} + (y - k)^{2} = h^{2}$
$(x - h)^{2} + (y - 0)^{2} = h^{2}$
$(x - h)^{2} + y^{2} = h^{2}$
$x^{2} + h^{2} - 2 xh + y^{2} = h^{2}$
$x^{2} + y^{2} - 2 xh = 0$
On differentiating both sides w.r.t.x, we get-
$2 x + 2 y \frac{d y}{d x} - 2 h = 0$
$x + y \frac{d y}{d x} = h$
On putting the value of $h$ in equation (i)
$x^{2} + y^{2} - 2 x (x + y \frac{d y}{d x}) = 0$
$x^{2} + y^{2} - 2 x^{2} - 2 xy \frac{dy}{dx} = 0$
$- x^{2} + y^{2} - 2 x y \frac{d y}{d x} = 0$
$x^{2} - y^{2} + 2 xy \frac{dy}{dx} = 0$
$y^{2} = x^{2} + 2 x y \frac{d y}{d x}$

Karnataka CET-2009

Differential Equation

87491 The integrating factor of the differential equation $\sin y (\frac{d y}{d x}) = \cos y (1 - x \cos y)$ is

1

e^{- x}

2

e^{- y}

3

e^{\sin y}

4

e^{\cos y}

Explanation:

(A) : Given,
$\sin y (\frac{d y}{d x}) = \cos y (1 - x \cos y)$
$\frac{\sin y}{\cos y} (\frac{d y}{d x}) = 1 - x \cos y$
$\tan y (\frac{d y}{d x}) = 1 - x \cos y$
$\tan y (\frac{d y}{d x}) = 1 - \frac{x}{\sec y}$
$\tan y \cdot \sec y (\frac{d y}{d x}) = \sec y - x$
Put, $\sec y = t$
$\sec y \cdot \tan y (\frac{d y}{d x}) = \frac{d t}{d x}$
Then equation becomes,
$\frac{dt}{dx} - t = - x$
Which is the form of $\frac{d y}{d x} + P y = Q$
$P = - 1$ and $Q = - X$
I.F. $= e^{\int - 1 dx} = e^{- x}$

MHT CET-2020

Differential Equation

87493 Solution of the differential equation $\frac{d y}{d x} + 2 y = e^{- x}$ is

1

{ye}^{x} = x + c

2

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

3

{ye}^{2 x} = x + c

4

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

Explanation:

(D) : Given,
$\frac{d y}{d x} + 2 y = e^{- x}$
Now, which is form of $\frac{d y}{d x} + P y = Q$
$\begin{aligned} I.F & = e^{\int 2 dx} \\ = e^{2 x} \end{aligned}$
$P = 2 and Q = e^{- x}$
For particular solution,
$\begin{array}{l} y \cdot IF = \int Q \cdot (IF) dx + c \\ ∴ y \cdot e^{2 x} = \int e^{2 x} \cdot e^{- x} dx + c \\ {ye}^{2 x} = \int e^{x} dx + c \\ {ye}^{2 x} = e^{x} + c \end{array}$

MHT CET-2020

Differential Equation

87494 The integrating factor of the differential equation $\frac{d y}{d x} + \frac{1}{x} y = x^{3} - 3$ is

1 -y

2 x

3 -x

4 y

Explanation:

(B) : Given,
$\frac{d y}{d x} + \frac{1}{x} y = x^{3} - 3$
The above differential equation of the form $\frac{d y}{d x} + P y = Q$
$P = \frac{1}{x}$ and $Q = x^{3} - 3$
I. F. $= e^{\int P Pdx}$
$= e^{\int \frac{1}{x} dx}$
$= e^{\log x} = x$

MHT CET-2020

Differential Equation

87489 The differential $y \frac{d y}{d x} + x =$ c represents

1 a family of parabolas

2 a family of circles whose centres are on the

x

axis

3 a family of hyperbolas

4 a family of circles whose centres are on the yaxis

Explanation:

(B) : Given,
$\begin{array}{l} y \frac{d y}{d x} + x = c \\ y \frac{d y}{d x} = c - x \\ y \frac{d y}{d y} = (c - x) d x \end{array}$
On integration both sides, we get
$\begin{array}{l} \int y d d = \int (c - x) dx \\ \frac{y^{2}}{2} = cx - \frac{x^{2}}{2} + k \\ \frac{x^{2}}{2} + \frac{y^{2}}{2} = cx + k \\ x^{2} + y^{2} - 2 cx - 2 k = 0 \end{array}$
This equation is represent the circle whose centre are on the x -axis.

Karnataka CET-2008

Differential Equation

87490 The differential equation of the family of circles passing through the origin and having their centres on the $x$ -axis is

1

x^{2} = y^{2} + x y \frac{d y}{d x}

2

x^{2} = y^{2} + 3 x y \frac{d y}{d x}

3

y^{2} = x^{2} + 2 x y \frac{d y}{d x}

4

y^{2} = x^{2} - 2 x y \frac{d y}{d x}

Explanation:

(C) : Given that family circle passing through the origin and centre on $x$ - axis.
Let centre be $(h, 0)$ and radius be $h$ .
Now equation of family of circle,
$(x - h)^{2} + (y - k)^{2} = h^{2}$
$(x - h)^{2} + (y - 0)^{2} = h^{2}$
$(x - h)^{2} + y^{2} = h^{2}$
$x^{2} + h^{2} - 2 xh + y^{2} = h^{2}$
$x^{2} + y^{2} - 2 xh = 0$
On differentiating both sides w.r.t.x, we get-
$2 x + 2 y \frac{d y}{d x} - 2 h = 0$
$x + y \frac{d y}{d x} = h$
On putting the value of $h$ in equation (i)
$x^{2} + y^{2} - 2 x (x + y \frac{d y}{d x}) = 0$
$x^{2} + y^{2} - 2 x^{2} - 2 xy \frac{dy}{dx} = 0$
$- x^{2} + y^{2} - 2 x y \frac{d y}{d x} = 0$
$x^{2} - y^{2} + 2 xy \frac{dy}{dx} = 0$
$y^{2} = x^{2} + 2 x y \frac{d y}{d x}$

Karnataka CET-2009

Differential Equation

87491 The integrating factor of the differential equation $\sin y (\frac{d y}{d x}) = \cos y (1 - x \cos y)$ is

1

e^{- x}

2

e^{- y}

3

e^{\sin y}

4

e^{\cos y}

Explanation:

(A) : Given,
$\sin y (\frac{d y}{d x}) = \cos y (1 - x \cos y)$
$\frac{\sin y}{\cos y} (\frac{d y}{d x}) = 1 - x \cos y$
$\tan y (\frac{d y}{d x}) = 1 - x \cos y$
$\tan y (\frac{d y}{d x}) = 1 - \frac{x}{\sec y}$
$\tan y \cdot \sec y (\frac{d y}{d x}) = \sec y - x$
Put, $\sec y = t$
$\sec y \cdot \tan y (\frac{d y}{d x}) = \frac{d t}{d x}$
Then equation becomes,
$\frac{dt}{dx} - t = - x$
Which is the form of $\frac{d y}{d x} + P y = Q$
$P = - 1$ and $Q = - X$
I.F. $= e^{\int - 1 dx} = e^{- x}$

MHT CET-2020

Differential Equation

87493 Solution of the differential equation $\frac{d y}{d x} + 2 y = e^{- x}$ is

1

{ye}^{x} = x + c

2

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

3

{ye}^{2 x} = x + c

4

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

Explanation:

(D) : Given,
$\frac{d y}{d x} + 2 y = e^{- x}$
Now, which is form of $\frac{d y}{d x} + P y = Q$
$\begin{aligned} I.F & = e^{\int 2 dx} \\ = e^{2 x} \end{aligned}$
$P = 2 and Q = e^{- x}$
For particular solution,
$\begin{array}{l} y \cdot IF = \int Q \cdot (IF) dx + c \\ ∴ y \cdot e^{2 x} = \int e^{2 x} \cdot e^{- x} dx + c \\ {ye}^{2 x} = \int e^{x} dx + c \\ {ye}^{2 x} = e^{x} + c \end{array}$

MHT CET-2020

Differential Equation

87494 The integrating factor of the differential equation $\frac{d y}{d x} + \frac{1}{x} y = x^{3} - 3$ is

1 -y

2 x

3 -x

4 y

Explanation:

(B) : Given,
$\frac{d y}{d x} + \frac{1}{x} y = x^{3} - 3$
The above differential equation of the form $\frac{d y}{d x} + P y = Q$
$P = \frac{1}{x}$ and $Q = x^{3} - 3$
I. F. $= e^{\int P Pdx}$
$= e^{\int \frac{1}{x} dx}$
$= e^{\log x} = x$

MHT CET-2020

Differential Equation

87489 The differential $y \frac{d y}{d x} + x =$ c represents

1 a family of parabolas

2 a family of circles whose centres are on the

x

axis

3 a family of hyperbolas

4 a family of circles whose centres are on the yaxis

Explanation:

(B) : Given,
$\begin{array}{l} y \frac{d y}{d x} + x = c \\ y \frac{d y}{d x} = c - x \\ y \frac{d y}{d y} = (c - x) d x \end{array}$
On integration both sides, we get
$\begin{array}{l} \int y d d = \int (c - x) dx \\ \frac{y^{2}}{2} = cx - \frac{x^{2}}{2} + k \\ \frac{x^{2}}{2} + \frac{y^{2}}{2} = cx + k \\ x^{2} + y^{2} - 2 cx - 2 k = 0 \end{array}$
This equation is represent the circle whose centre are on the x -axis.

Karnataka CET-2008

Differential Equation

87490 The differential equation of the family of circles passing through the origin and having their centres on the $x$ -axis is

1

x^{2} = y^{2} + x y \frac{d y}{d x}

2

x^{2} = y^{2} + 3 x y \frac{d y}{d x}

3

y^{2} = x^{2} + 2 x y \frac{d y}{d x}

4

y^{2} = x^{2} - 2 x y \frac{d y}{d x}

Explanation:

(C) : Given that family circle passing through the origin and centre on $x$ - axis.
Let centre be $(h, 0)$ and radius be $h$ .
Now equation of family of circle,
$(x - h)^{2} + (y - k)^{2} = h^{2}$
$(x - h)^{2} + (y - 0)^{2} = h^{2}$
$(x - h)^{2} + y^{2} = h^{2}$
$x^{2} + h^{2} - 2 xh + y^{2} = h^{2}$
$x^{2} + y^{2} - 2 xh = 0$
On differentiating both sides w.r.t.x, we get-
$2 x + 2 y \frac{d y}{d x} - 2 h = 0$
$x + y \frac{d y}{d x} = h$
On putting the value of $h$ in equation (i)
$x^{2} + y^{2} - 2 x (x + y \frac{d y}{d x}) = 0$
$x^{2} + y^{2} - 2 x^{2} - 2 xy \frac{dy}{dx} = 0$
$- x^{2} + y^{2} - 2 x y \frac{d y}{d x} = 0$
$x^{2} - y^{2} + 2 xy \frac{dy}{dx} = 0$
$y^{2} = x^{2} + 2 x y \frac{d y}{d x}$

Karnataka CET-2009

Differential Equation

87491 The integrating factor of the differential equation $\sin y (\frac{d y}{d x}) = \cos y (1 - x \cos y)$ is

1

e^{- x}

2

e^{- y}

3

e^{\sin y}

4

e^{\cos y}

Explanation:

(A) : Given,
$\sin y (\frac{d y}{d x}) = \cos y (1 - x \cos y)$
$\frac{\sin y}{\cos y} (\frac{d y}{d x}) = 1 - x \cos y$
$\tan y (\frac{d y}{d x}) = 1 - x \cos y$
$\tan y (\frac{d y}{d x}) = 1 - \frac{x}{\sec y}$
$\tan y \cdot \sec y (\frac{d y}{d x}) = \sec y - x$
Put, $\sec y = t$
$\sec y \cdot \tan y (\frac{d y}{d x}) = \frac{d t}{d x}$
Then equation becomes,
$\frac{dt}{dx} - t = - x$
Which is the form of $\frac{d y}{d x} + P y = Q$
$P = - 1$ and $Q = - X$
I.F. $= e^{\int - 1 dx} = e^{- x}$

MHT CET-2020

Differential Equation

87493 Solution of the differential equation $\frac{d y}{d x} + 2 y = e^{- x}$ is

1

{ye}^{x} = x + c

2

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

3

{ye}^{2 x} = x + c

4

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

Explanation:

(D) : Given,
$\frac{d y}{d x} + 2 y = e^{- x}$
Now, which is form of $\frac{d y}{d x} + P y = Q$
$\begin{aligned} I.F & = e^{\int 2 dx} \\ = e^{2 x} \end{aligned}$
$P = 2 and Q = e^{- x}$
For particular solution,
$\begin{array}{l} y \cdot IF = \int Q \cdot (IF) dx + c \\ ∴ y \cdot e^{2 x} = \int e^{2 x} \cdot e^{- x} dx + c \\ {ye}^{2 x} = \int e^{x} dx + c \\ {ye}^{2 x} = e^{x} + c \end{array}$

MHT CET-2020

Differential Equation

87494 The integrating factor of the differential equation $\frac{d y}{d x} + \frac{1}{x} y = x^{3} - 3$ is

1 -y

2 x

3 -x

4 y

Explanation:

(B) : Given,
$\frac{d y}{d x} + \frac{1}{x} y = x^{3} - 3$
The above differential equation of the form $\frac{d y}{d x} + P y = Q$
$P = \frac{1}{x}$ and $Q = x^{3} - 3$
I. F. $= e^{\int P Pdx}$
$= e^{\int \frac{1}{x} dx}$
$= e^{\log x} = x$

MHT CET-2020

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