MCQ Questions for Class 12 Commerce Maths Differential Equations Quiz 1

For

$x\epsilon R, x\neq 0$ , if

$y(x)$ is a differentiable function such that

$x\int_{1}^{x}y (t) dt = (x + 1) \int_{1}^{x} t y (t) dt$ , then

$y (x)$ equals:
(Where C is a constant)

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$Cx^{3} e^{\dfrac {1}{x}}$
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$\dfrac {C}{x^{2}} e^{-\dfrac {1}{x}}$
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$\dfrac {C}{x} e^{-\dfrac {1}{x}}$
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$\dfrac {C}{x^{3}} e^{-\dfrac {1}{x}}$

Explanation

$x\int_{1}^{x}y (t) dt = x \int_{1}^{x} ty (t) dt + \int_{1}^{x} ty (t) dt$
differentiate w.r. to x.

$\int_{1}^{x} y (t) dt + x [y (x) - y(1)] = \int_{1}^{x} ty (t) dt + x [xy (x) - y(1)] + xy (x) - y (1)$

$\int_{1}^{x} y (t) dt = \int_{1}^{x}t y (t) dt + x^{2} y (x) - y (1)$
diff. again w.r to x

$y(x) - y(1) = xy (x) - y(1) + 2xy (x) + x^{2}y^{'} (x)$

$(1 - 3x) y (x) = x^{2}y^{'} (x)$

$\dfrac {y^{'}(x)}{y(x)} = \dfrac {1 - 3x}{x^{2}}$

$\dfrac {1}{y}\dfrac {dy}{dx} = \dfrac {1 - 3x}{x^{2}} \Rightarrow ln\ y = -\dfrac {1}{x} - 3 ln\ x$

$ln (y x^{3}) = -\dfrac {1}{x}+lnc$

$yx^{3} = ce^{-\dfrac {1}{x}}$

$y = c\dfrac {e^{-\dfrac {1}{x}}}{x^{3}}$ or

$y = \dfrac {ce^{-\dfrac {1}{x}}}{x^{3}}$

The differential equation

$\displaystyle \frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=\frac{\sqrt{1-\mathrm{y}^{2}}}{\mathrm{y}}$ determines a family of circles with:

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variable radii and a fixed centre at $(0,1)$
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variable radii and a fixed centre at $(0, -1)$
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fixed radius 1 and variable centres along the x-axis
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fixed radius 1 and variable centres along the y-axis

Explanation

$\displaystyle \frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=\frac{\sqrt{1-\mathrm{y}^{2}}}{\mathrm{y}}$

$\displaystyle \Rightarrow\int\frac{\mathrm{y}}{\sqrt{1-\mathrm{y}^{2}}}$ dy

$=\displaystyle \int \mathrm{d}\mathrm{x}$

$\Rightarrow-\sqrt{1-\mathrm{y}^{2}}=\mathrm{x}+\mathrm{c}$

$\Rightarrow(\mathrm{x}+\mathrm{c})^{2}+\mathrm{y}^{2}=1$ centre

$(-\mathrm{c}, 0)$ ; radius

$\sqrt{\mathrm{c}^{2}-\mathrm{c}^{2}+1}=1$ .

Check whether the function is homogenous or not. If yes then find the degree of the function

$g(x)=x^2-8x^3$ .

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Not homogenous
0%
Homogenous with degree 4
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Homogenous with degree 2
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None of these

Explanation

Given function is

$g\left( x \right) ={ x }^{ 2 }-8{ x }^{ 3 }$

To check its Homogenitiy, we find

$g(kx)$ which gives,

$g\left( kx \right) ={ { k }^{ 2 }x }^{ 2 }-8{ k }^{ 3 }{ x }^{ 3 }$

Now,

$\cfrac { g\left( kx \right) }{ g\left( x \right) } =\cfrac { { k }^{ 2 }{ x }^{ 2 }-8{ k }^{ 3 }{ x }^{ 3 } }{ { x }^{ 2 }-8{ x }^{ 3 } } \neq { k }^{ n }$ where

$k>0$ and

$n$ is an integer.

$\therefore$ this function is not homogenous.

Solution of differential equation

$\displaystyle \frac{dy}{dx} = sin x + 2x$ , is

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$y = x^2 - cos x + c$
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$y = cos x + x^2 + c$
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$y = cos x + 2$
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$y = cos x + 2 + C$

Explanation

$\displaystyle \frac{dy}{dx} = sin x + 2x$

$y = - cos x + \displaystyle \frac{2x^2}{2} = x^2 - cos x + c$

Which of the following is/are correct regarding homogenous functions?

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A function is defined as a homogeneous function if its argument or input is multiplied by some factor, then the result is multiplied by some exponent of this factor.
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Denoted by: $f(kx)= k^n f(v)$ , where $n$ is the degree
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Denoted by: $f(kx)= nk f(v)$ , where $n$ is the degree
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None of these

Explanation

a function

$f:R\rightarrow R$ is called a homogenous function when

$f\left( kx \right) ={ k }^{ n }f\left( x \right)$ where

$k>0$ and

$n$ is an integer and the constant

$k$ is called the degree of the homogenous function.

The solution of

$\dfrac{dy}{dx}=e^{logx}$ is:

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$2y=x^{2}+c$
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$y=x^{2}+c$
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$y^{2}=x+c$
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$xy=x^{2}+c$

Explanation

$\dfrac{dy}{dx}=e^{log\ x}=x$

$\Rightarrow \int dy=\int xdx+c$

$\Rightarrow y=\frac{x^{2}}{2}+c$

$\Rightarrow 2y=x^{2}+c$

The solution of

$\dfrac{dy}{dx}=\dfrac{2x}{3y^{2}}$ is:

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

Explanation

$\dfrac{dy}{dx}=\dfrac { 2x }{ { 3y }^{ 2 } }$

$\Rightarrow \int 3y^{2}dy=\int2xdx+c$

$\rightarrow \dfrac{3y^{3}}{3}=\dfrac{2x^{2}}{2}+c$

$\rightarrow y^{3}-x^{2}=c$

The solution of

$\dfrac{dy}{dx}-\dfrac{2xy}{1+x^{2}}=0$ is

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$y=c\left ( 1+x^{2} \right )$
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$y=c\sqrt{1+x^{2}}$
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$y=\dfrac{c}{1+x^{2}}$
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$y=\dfrac{c}{\sqrt{1+x^{2}}}$

Explanation

$\dfrac{dy}{dx}=\dfrac{2xy}{1+x^{2}}$

$\Rightarrow\displaystyle \int \dfrac{dy}{y}=\int \dfrac{2x}{1+x^{2}}dx$

$\Rightarrow \log\ y=\log(1+x^{2})c$

$y=c(1+x^{2})$

The solution of

$x^{2} \cfrac{dy}{dx}=2$ is

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$y=-\cfrac{2}{x}+c$
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$y=\cfrac{x}{2}+c$
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$y=\cfrac{2}{x}+c$
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$y = 2x + c$

Explanation

$x^{2}\cfrac{dy}{dx}=2$

$\int dy=\int \cfrac{2}{x^{2}}dx+c$

$\Rightarrow y=\cfrac{-2}{x}+c$

$\Rightarrow y+\cfrac{2}{x}=c$

Check whether the function is homogenous or not. If yes then find the degree of the function

$g(x)=8x^4$ .

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Homogenous with degree 4
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Homogenous with degree 1
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Not homogenous
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None of these

Explanation

Given function is

$g\left( x \right) =8{ x }^{ 4 }$

To check the homogenity of the function we find

$g\left( kx \right)$ which is,

$g\left( kx \right) =8{ k }^{ 4 }{ x }^{ 4 }$

$\cfrac { g\left( kx \right) }{ g\left( x \right) } =\cfrac { 8{ k }^{ 4 }{ x }^{ 4 } }{ 8{ x }^{ 4 } } ={ k }^{ 4 }$

$\therefore$ The function is homogenous with

$k>0$ and is of degree

$4$ .

$y = e^{-x}\cos 2x$ then which of the following differential equations is satisfied?

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$\dfrac {d^{2}y}{dx^{2}} + 2 \dfrac {dy}{dx} + 5y = 0$
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$\dfrac {d^{2}y}{dx^{2}} + 5 \dfrac {dy}{dx} + 2y = 0$
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$\dfrac {d^{2}y}{dx^{2}} - 5 \dfrac {dy}{dx} - 2y = 0$
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$\dfrac {d^{2}y}{dx^{2}} + 2 \dfrac {dy}{dx} - 5y = 0$

Explanation

$\therefore y_{1} = - y - 2e^{-x} \sin 2x ..... (1)$

$\therefore y_{2} = -y_{1} + 2e^{-x}\sin 2x - 4e^{-x} \cos 2x$

$= -y_{1} - y - y_{1} - 4y$

$\Rightarrow y_{2} + 2y_{1} + 5y = 0$

What is the solution of the differential equation

$\dfrac {ydx - xdy}{y^{2}} = 0$ ?
where

$c$ is an arbitrary constant.

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$xy = c$
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$y = cx$
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$x + y = c$
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$x - y = c$

Explanation

$\dfrac { ydx-xdy }{ { y }^{ 2 } } =0\\ \Rightarrow ydx=xdy$

$\Rightarrow \dfrac { dx }{x } =\dfrac { dy }{ y }$

Integrating both the sides, we get

$\displaystyle \int \dfrac { dx }{x } =\int \dfrac { dy }{ y }$

$\Rightarrow \ln(cx)=\ln(y)\\ \Rightarrow y=cx$

where

$c$ is an arbitrary constant.

What is the solution of the differential equation

$\sin \left (\dfrac {dy}{dx}\right ) - a = 0$ ?
where

$c$ is an arbitrary constant.

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$y = x \sin^{-1} a + c$
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$x = y \sin^{-1} a + c$
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$y = x + x \sin^{-1} a + c$
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$y = \sin^{-1} a + c$

Explanation

$\sin\left( \dfrac { dy }{ dx } \right) =a$

$\Rightarrow \dfrac { dy }{ dx } ={\sin }^{ -1 }a$

$\Rightarrow dy={\sin }^{ -1 }a\ dx$

Integrating both the sides, we get

$\displaystyle \int dy=\int {\sin }^{ -1 }a\ dx$

$\Rightarrow y=x\ {\sin }^{ -1 }a+c$

where

$c$ is a constant.

Check whether the function is homogenous or not. If yes then find the degree of the function

$g(x)=4-x^2$ .

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Homogenous with degree 2
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Not homogenous
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Homogenous with degree 4
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None of these

Explanation

Given function is

$g\left( x \right) =4-{ x }^{ 2 }$

To check the Homogenity of the function we find

$g\left( kx \right)$

$g\left( kx \right) =4-{ k }^{ 2 }{ x }^{ 2 }$

$\cfrac { g\left( kx \right) }{ g\left( x \right) } =\cfrac { 4-{ k }^{ 2 }{ x }^{ 2 } }{ 4-{ x }^{ 2 } } \neq { k }^{ n }$

Where

$k>0$ and

$n$ is an integer.

$\therefore$ The function is not homogenous.

Which of the following is true regarding the function

$f(x, y)= x^4 \sin \dfrac{x}{y}$ ?

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Not homogenous
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Homogenous with degree 2
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Homogenous with degree 3
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None of the above

Explanation

given function is

$f\left( x,y \right) ={ x }^{ 4 }\sin { \cfrac { x }{ y } }$

Now,

$f\left( kx,ky \right) ={ k }^{ 4 }{ x }^{ 4 }\sin { \cfrac { x }{ y } }$ ,

$\cfrac { f\left( kx,ky \right) }{ f\left( x,y \right) } ={ k }^{ 4 }$

$\therefore$ the function is homogenous with Degree

$4$ .

Find the value of

$k$ for the function:

$2x^2y+3xyz+z^k$ to be homogenous.

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$6$
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$3$
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$2$
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None of these

Explanation

Let the given function be

$f\left( x,y,z \right) =2{ x }^{ 2 }y+3xyz+{ z }^{ k }$

Given that the function is homogenous,

$\Longrightarrow \cfrac { f\left( \alpha x,\alpha y,\alpha z \right) }{ f\left( x,y,z \right) } =\cfrac { 2{ x }^{ 2 }y\alpha ^{ 3 }+3xyz{ \alpha }^{ 3 }+{ z }^{ k }{ \alpha }^{ k } }{ 2{ x }^{ 2 }y+3xyz+{ z }^{ k } } ={ \alpha }^{ n }$

where

$\alpha$ is the degree of he function and

$n$ is a positive integer.

$\therefore$ by comparing the terms in the numerator we get

$k=3$

so that we can take common in the numerator such that it cancels with the denominator.

$\dfrac{dy}{dx}=x^{-3}$ then

$y$

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$-\dfrac{1}{2x^2}+c$
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$\dfrac{1}{2x^2}+c$
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$-\dfrac{1}{3x^3}+c$
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$-\dfrac{1}{4x^4}+c$

Explanation

$\displaystyle{\frac { dy }{ dx } ={ x }^{ -3 }\\ dy={ x }^{ -3 }dx\\ \int { dy } =\int { { x }^{ -3 }dx } \\ y=\frac { { x }^{ -2 } }{ -2 } +c\\ y=-\frac { 1 }{ 2{ x }^{ 2 } } +c}$

The number of arbitrary constants in the particular solution of the differential equation of order

$3$ is ______.

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$0$
0%
$1$
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$2$
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$3$

Explanation

The number of arbitary constants in the particular solution of the differential equation of order

$3$ is

$0$

The number of arbitrary constant in the general solution of differential equation of order

$3$ is _________.

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$0$
0%
$2$
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$3$
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$4$

Explanation

As per the note given the number of arbitrary constants in third order differential equation are

$3$ (Three).

What is the solution of the differential equation

$\dfrac {dx}{dy} + \dfrac {x}{y} - y^{2} = 0$ ?
where

$c$ is an arbitrary constant.

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$xy = x^{4} + c$
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$xy = y^{4} + c$
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$4xy = y^{4} + c$
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$3xy = y^{3} + c$

Explanation

$\dfrac { dx }{ dy } +\dfrac { x }{ y } -{ y }^{ 2 }=0\\ \Rightarrow \dfrac { dx }{ dy } =\dfrac { { y }^{ 3 }-x }{ y } \\ \Rightarrow ydx={ y }^{ 3 }dy-xdy\\ \Rightarrow xdy+ydx={ y }^{ 3 }dy\\ \Rightarrow d(xy)=d\left(\dfrac { { y }^{ 4 } }{ 4 }\right)$

Integrating both the sides, we get

${ y }^{ 4 }+c=4xy$

What is the general solution of the differential equation

$x\, dy - y\, dx \,y^2$ ?

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$x = cy$
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$y^2=cx$
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$x + xy - cy = 0$
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None of the above

Explanation

$xdy-{ y }^{ 3 }dx=0\\ rearranging\quad \\ \cfrac { dx }{ x } =\cfrac { dy }{ { y }^{ 3 } } \\ integrating\quad both\quad sides\\ \ln { x } =\frac{{ y }^{ -2 }}{-2}+c\\$

The solution of the differential equation

$x^4\dfrac{dy}{dx}+x^3y+cosec(xy)=0$ is equal to

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$2 \cos (xy)+x^{-2}=C$
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$2\cos(xy)+y^{-2}=C$
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$2\sin(xy)+x^{-2}=C$
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$2\sin (xy)+y^{-2}+C$

Explanation

Given,

$x^4\dfrac{dy}{dx}+x^3y+cosec(xy)=0$

Put

$xy=t$

$y+x\dfrac{dy}{dx}=\dfrac{dt}{dx}$

Substituting it we get;

$x^3\left(\dfrac{dt}{dx}-\dfrac{t}{x}\right)+x^2t+cosec(t)=0$

$\Rightarrow x^3\dfrac{dt}{dx}+cosec(t)=0$

$\Rightarrow \sin(t)dt+x^{-3}dx=0$

After Integrating we get;

$2\cos(xy)+x^{-2}=C$

The differential equation

$y\cfrac { dy }{ dx } +x=a$ where '

$a$ ' is any constant represents:

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A set of straight lines
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A set of ellipses
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A set of circles
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None of the above

Explanation

$y\frac { dy }{ dx } +x=a$

$\Rightarrow ydy+xdx=adx$

Integrating both sides,

$\Rightarrow \dfrac { { y }^{ 2 } }{ 2 } +\dfrac { { x }^{ 2 } }{ 2 } =ax+c\\ \Rightarrow { x }^{ 2 }+{ y }^{ 2 }-2ax-2c=0\\ \Rightarrow { (x-a) }^{ 2 }+{ y }^{ 2 }={ a }^{ 2 }+2c$

Clearly, this equation represents circle with centre

$(a,0)$ and radius

$\sqrt { { a }^{ 2 }+2c }.$
Option C is correct.

What is the solution of the differential equation

$\dfrac{dy}{dx} +\dfrac{y}{x} = 0$ ?
Where c is a constant.

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$xy = c$
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$x = cy$
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$y = cx$
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None of the above

Explanation

Given expression is

$\dfrac { dy }{ dx } +\dfrac { y }{ x } =0\\ \Rightarrow \dfrac { dy }{ dx } =-\dfrac { y }{ x } \\ \Rightarrow \dfrac { dy }{ y } +\dfrac { dx }{ x } =0$

Integrating we get

$\displaystyle \int { \dfrac { dy }{ y } } +\int { \frac { dx }{ x } } =0\\ \Rightarrow \ln { y } +\ln { x=c } \\ \Rightarrow xy=c$

The differential equation of

$y=c{ x }^{ 3 },c$ is a arbitrary constant is _______

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$x.\cfrac { dy }{ dx } =3y$
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${ x }^{ 3 }.\cfrac { dy }{ dx } =3y$
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$3x\cfrac { dy }{ dx } =y$
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$x.\cfrac { dy }{ dx } =3{ y }^{ 2 }$

Explanation

$$y=cx^{3}$$

Differentiating both sides

$\dfrac{dy}{dx}=\dfrac{d}{dx}(cx^{3})$

$\dfrac{dy}{dx}=c\dfrac{d}{dx}(x^{3})$

$\dfrac{dy}{dx}=3cx^{2}$

Multiplying x on both sides

$x\dfrac{dy}{dx}=x\times 3cx^{2}$

$x\dfrac{dy}{dx}=3cx^{3}$

Substituting

$y=cx^{3}$

$x\dfrac{dy}{dx}=3y$

The equation of the curve passing through

$(0,1)$ which is a solution of the differential equation

$\left( 1+{ y }^{ 2 } \right) dx+\left( 1+{ x }^{ 2 } \right) dy=0$ is given by

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$\tan ^{ -1 }{ \left( x \right) } +\tan ^{ -1 }{ \left( y \right) =0 }$
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$\tan ^{ -1 }{ \left( x \right) } +\tan ^{ -1 }{ \left( y \right) =\dfrac {\pi}{4}}$
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$\sin h ^{ -1 }{ \left( x \right) } +\sin h ^{ -1 }{ \left( y \right) =0 }$
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$\sinh ^{ -1 }{ \left( x \right) } +\sinh ^{ -1 }{ \left( y \right) } \log { \left( 1+\sqrt { 2 } \right) =0 }$

Explanation

$\begin{array}{l} \left( { 1+{ y^{ 2 } } } \right) dx+\left( { 1+{ x^{ 2 } } } \right) dy=0 \\ \Rightarrow \left( { 1+{ x^{ 2 } } } \right) dy=-\left( { 1+{ y^{ 2 } } } \right) dx \\ \Rightarrow \int { \dfrac { { dy } }{ { d+{ y^{ 2 } } } } =\int { -\dfrac { { dx } }{ { 1+{ x^{ 2 } } } } } } \\ \Rightarrow { \tan ^{ -1 } }y=-{ \tan ^{ -1 } }x+c \end{array}$

Since, this curve passes through

$(0,1)$

so,

$\begin{array}{l} { \tan ^{ -1 } }\left( 1 \right) =-{ \tan ^{ -1 } }\left( 0 \right) +c \\ \Rightarrow \dfrac { \pi }{ 4 } =c \end{array}$

Therefore required equation of the curve is

$\begin{array}{l} { \tan ^{ -1 } }y=-{ \tan ^{ -1 } }x+\dfrac { \pi }{ 4 } \\ \Rightarrow { \tan ^{ -1 } }x+{ \tan ^{ -1 } }y=\dfrac { \pi }{ 4 } \end{array}$

The solution of

$\dfrac{ydx-xdy}{y^{2}}=0$ represents a family of

0%
Straight lines passing through the origin
0%
Circles
0%
Parabola
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Hyperbola

Explanation

$\dfrac{{ydx - xdy}}{{{y^2}}} = 0$

$\Rightarrow d\left( {\dfrac{x}{y}} \right) = 0$

integrating both sides

$\Rightarrow \dfrac{x}{y} = c$

$\Rightarrow x = cy$

This is a straight passing through origin

The

$D.E$

$y\dfrac { dy }{ dx } +x=a$ represents

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a circle whose centra is on $X-$ axis
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a circle whose centre is on $Y-$ axis
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a circle whose centre is origin
0%
a parabola

Explanation

$y\frac{{dy}}{{dx}} + x = a$

$\Rightarrow ydy + xdx = adx$

Integrating both sides

$\Rightarrow \dfrac{{{y^2}}}{2} + \dfrac{{{x^2}}}{2} = ax + c$

$\Rightarrow {x^2} + {y^2} = 2ax + c$

$\Rightarrow {x^2} + {y^2} - 2ax + c = 0$

on comparing it with

${x^2} + {y^2} + 2gx + + 2fy + c = 0$

here ,

$g = - a,f = 0$

centre =

$\left( { - g, - f} \right) = \left( {a,0} \right)$

therefore , it represent a circle with centre at X- axis

The solution of

$\left( { x }^{ 2 }{ y }^{ 3 }+{ x }^{ 2 } \right) dx+\left( { y }^{ 2 }{ x }^{ 3 }+{ y }^{ 2 } \right) dy=0$ is

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$\left( { x }^{ 3 }+1 \right) \left( { y }^{ 3 }+1 \right) =c$
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$\left( { x }^{ 3 }-1 \right) \left( { y }^{ 3 }-1 \right) =c$
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$\left( { x }^{ 3 }-1 \right) \left( { y }^{ 3 }+1 \right) =c$
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$\left( { x }^{ 3 }+1 \right) \left( { y }^{ 3 }-1 \right) =c$

Explanation

$\left( {{x^2}{y^3} + {x^2}} \right)dx + \left( {{y^2}{x^3} + {y^2}} \right)dy = 0$

$\Rightarrow {x^2}\left( {1 + {y^3}} \right)dx + {y^2}\left( {1 + {x^3}} \right)dy = 0$

$\Rightarrow {y^2}\left( {1 + {x^3}} \right)dy = - {x^2}\left( {1 + {y^3}} \right)dx$

$\Rightarrow \dfrac{1}{3}\int {\dfrac{{3{y^2}}}{{{y^3} + 1}}} dy = \dfrac{1}{3}\int { - \dfrac{{3{x^2}}}{{{x^3} + 1}}} dx$

$\Rightarrow \dfrac{1}{3}\log \left( {{y^3} + 1} \right) = - \dfrac{1}{3}\log \left( {{x^3} + 1} \right) + \log c$

$\Rightarrow \log \left( {{y^3} + 1} \right) + \log \left( {{x^3} + 1} \right) = \log c$

$\Rightarrow \left( {{x^3} + 1} \right)\left( {{y^3} + 1} \right) = c$

Which of the following transformation reduce the differential equation

$\dfrac{{dz}}{{dx}} + \dfrac{z}{x}\log z = \dfrac{z}{{{x^2}}}{\left( {\log z} \right)^2}$ into the form

$\dfrac{{dv}}{{dx}} + P\left( x \right)v = Q\left( x \right)$

0%
$v=logz$
0%
$v=e^2$
0%
$v=\dfrac{1}{logz}$
0%
$v=(logz)^2$

Explanation

Given,

$\dfrac {dz}{dx}+\dfrac {z}{x} \log {z}=\dfrac {z}{x^2}(logz)^2$

divide both side by

$z(logz)^2$

$\dfrac {dz}{z(logz)^2dx}+\dfrac {z}{z(logz)^2x} \log {z}=\dfrac {z}{z(logz)^2x^2}(logz)^2$

$\dfrac {dz}{z(logz)^2dx}+\dfrac {1}{(logz)x} =\dfrac {1}{x^2}$

Let

$v=\dfrac {1}{logz}$ ..............1

and

$\dfrac {dv}{dx}=\dfrac {-1}{z(logz)^2}\dfrac {dz}{dx}$ ................2

So , by 1 & 2

$-\dfrac {dv}{dx}+\dfrac {v}{x}=\dfrac {1}{x^2}$

$\dfrac {dv}{dx}-\dfrac {v}{x}=-\dfrac {1}{x^2}$

so, here

$v=\dfrac {1}{logz}$

The differential equation for all the straight lines which are at a unit distance from the origin is

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${\left( {y - x\frac{{dy}}{{dx}}} \right)^2} = 1 - {\left( {\dfrac{{dy}}{{dx}}} \right)^2}$
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${\left( {y + x\frac{{dy}}{{dx}}} \right)^2} = 1 + {\left( {\dfrac{{dy}}{{dx}}} \right)^2}$
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${\left( {y - x\frac{{dy}}{{dx}}} \right)^2} = 1+{\left( {\dfrac{{dy}}{{dx}}} \right)^2}$
0%
${\left( {y + x\frac{{dy}}{{dx}}} \right)^2} = 1 - {\left( {\dfrac{{dy}}{{dx}}} \right)^2}$

Explanation

Let the line be :

$y=mx+c$

$\Rightarrow 1=\dfrac {|c|} {\sqrt {m^{2}+1}}$

$\Rightarrow y=mx\pm \sqrt {m^{2}+1}$

$\Rightarrow \dfrac {dy} {dx} =m$

$\Rightarrow y=x\dfrac {dy} {dx} \pm \sqrt {\left (\dfrac {dy} {dx} \right) ^{2}+1}$

$\Rightarrow \left (y-x\dfrac {dy} {dx} \right) ^{2}=1+\left (\dfrac {dy} {dx} \right) ^{2}$

Which of the following differential equation is linear ?

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$\dfrac{d^2y}{dx^2}+x\dfrac{dy}{dx}+2y=0$
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$\dfrac{d^2y}{dx^2}+y\dfrac{dy}{dx}+x=0$
0%
$\dfrac{d^{2}y}{dx^2}+\dfrac{y}{x}+\sin y=x^2$
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$(1+x)\dfrac{dy}{dx}-xy=1$

Explanation

Only in the equation

$(1+x)\dfrac{dy}{dx}-xy=1$ , both the terms

$\dfrac{dy}{dx}$ and

$y$ are linear. Hence this equation is a linear differential equation.

Solution of the different equation,

$ydx - xdy + x{y^2}dx = 0$ can be.

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$2x + {x^2}y = \lambda y$
0%
$2y + {y^2}x = \lambda y$
0%
$2y - {y^2}x = \lambda y$
0%
none of these

Explanation

$\cfrac{ydx-xdy}{y^2} = d(\cfrac{x}{y})$

Hence,

$\cfrac{ydx-xdy}{y^2} + xdx = 0$

$d(\cfrac{x}{y}) + xdx = 0$

Integrating both sides we get,

$\cfrac{x}{y} + \cfrac{x^2}{2} = C$

$2x+x^2 y = \lambda y$

Solve

$(1+\cos x)dy=(1-\cos x)dx$ .

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$y = \displaystyle \cot \cfrac{x}{2} - x +C$
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$y = \displaystyle \tan \cfrac{x}{2} - x +C$
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$y = \displaystyle \sin \cfrac{x}{2} - x +C$
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None of these

Explanation

$dy = \cfrac{1-\cos x}{1+\cos x} dx$

Integrating both sides,we get

$y = \displaystyle \int -\cfrac{\cos x+1-2}{\cos x+1}dx$

$y = \displaystyle \int -dx+ \int 2\cfrac{1}{\cos x+1}dx$

$y = \displaystyle -x+\int \cfrac{\sec^2 \cfrac{x}{2}}{2} dx$

$y = \displaystyle \tan \cfrac{x}{2} - x +C$

The solution of the differential equation

$\dfrac{dy}{dx} = \dfrac{3y - 7x - 3}{3x - 7y + 7}$ is

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$(y - x - 2)^5 (y + x - 5)^7 = c$
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$(y - x - 5)^2 (y + x - 1)^7 = c$
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$(y - x - 7)^2 (y + x - 5) = c$
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$(y - x - 1)^2 (y + x - 1)^5 = c$

Explanation

$\dfrac{dy}{dx}=\dfrac{3y-7x-3}{3x-7y+7}$ put

$y=Y+1$ and

$x=X$

$\dfrac{dy}{dx}=\dfrac{dY}{dX}$

$\dfrac{dY}{dX}=\dfrac{3Y-7X}{3X-7Y}$ put

$Y=vX$

$\Rightarrow \dfrac{dY}{dX}=v+X\dfrac{dv}{dX}$

$\Rightarrow V+X\dfrac{dv}{dX}=\dfrac{3v-7}{3-7v}\Rightarrow \dfrac{3v-7v^2-3v+7}{(3-7v)}=-X\dfrac{dv}{dX}$

$X\dfrac{dv}{dX}=\dfrac{7(v+1)(v-1)}{(3-7v)}$

$\Rightarrow \boxed{\frac{dv(3-7v)}{(v+1)(v-1)}=7\frac{dx}{X}}$

$\dfrac{3-7v}{(v+1)(v-1)}=\dfrac{7-4-7v}{(v+1)(v-1)}=\dfrac{7(1-v)}{(v+1)(v-1)}-\dfrac{4}{(v+1)(v-1)}$

$=\dfrac{-7}{(v+1)}-\dfrac{4}{(v+1)(v-1)}$

$\Rightarrow \dfrac{3-7v}{(v+1)(v-1)}=\dfrac{-7}{(v+1)}-\dfrac{4}{(v+1)(v-1)}=\dfrac{-7}{(v+1}-\dfrac{4}{2}\left[\dfrac{1}{(v-1)}-\dfrac{1}{(v+1)}\right]=\dfrac{-7+2}{(v+1)}\left[\dfrac{1}{(v+1)}-\dfrac{1}{(v-1)}\right]$

$\boxed{\dfrac{3-7v}{(v+1)(v-1)}=\left[\dfrac{-5}{(v+1)}-\dfrac{2}{(v-1)}\right]}$

$\Rightarrow dv\left[\dfrac{-5}{(v+1)}-\dfrac{2}{(v-1)}\right]$

$7\dfrac{dx}{x}$

$\Rightarrow |(v+1)^5(v-1)^2|=7|n||x|$

$\Rightarrow X^7(v+1)^5(v-1)^2 = c$

$\Rightarrow x^7\dfrac{(Y+x)^5(Y-x)^2}{x^7}=c$

$\Rightarrow (y+x)^5(y-x)^2=c$

$\Rightarrow (y-1+x)^5(y-1-x)^2=c$

$\Rightarrow \boxed{(y+x-1)^5(y-x-1)^2=c}$

$\dfrac{dy}{dx}+y\tan x=\sin 2x$ and

$y(0)=1$ , then

$y(\pi)=?$

0%
$1$
0%
$-1$
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$-5$
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$5$

Explanation

$I.F. = e^{\int tanxdx}$

$\implies I.F. = e^{-lncosx}$

$I.F. = \cfrac{1}{cosx}$

Multiplying with

$I.F.$ in the given equation,we get-

$\cfrac{dy}{cosx dx} + \cfrac{ysinx}{cos^2x} = 2sinx$

$d(\cfrac{y}{cosx}) = 2sinxdx$

Integrating both sides, we get-

$\cfrac{y}{cosx} = -2cosx + C$

$y(0) = 1$

$1 = -2+C$

$C = 3$

For

$x=\pi$ we get

$\cfrac{y}{-1} = 2+3$

$y = -5$

The solution of the differential equaton

$3{e^x}\tan ydx + \left( {1 - {e^x}} \right){\sec ^2}ydy = 0$ is

0%
$\tan y = c{\left( {1 - {e^x}} \right)^3}$
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${\left( {1 - {e^x}} \right)^3}\tan y = c$
0%
$\tan y = c\left( {1 - {e^x}} \right)$
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$\left( {1 - {e^x}} \right)\tan y = c$

Explanation

$3{e}^{x}\tan y dx+(1-{e}^{x})\sec ^{ 2 }{ y } dy=0$

$\cfrac { 3{ e }^{ x } }{ 1-{ e }^{ x } } dx+\cfrac { \sec ^{ 2 }{ y } dy\quad }{ \tan { y } } =0$

Integrating both sides

$-3\log{1-{ e }^{ x }}+\log{\tan y)}=\log{e}$

$\log{(\tan y)}=\log C+3\log{(1-{ e }^{ x })}$

$\log{(\tan y)}=\log { \left( C{ \left( 1-{ e }^{ x } \right) }^{ 3 } \right) }$

$\tan y=C{(1-{e}^{x})}^{3}$

Solve:

$\dfrac {dy}{dx} = y \cot x$

0%
$y=k\tan x$
0%
$y=k\csc x$
0%
$y=k\sin x$
0%
$y=k\cos x$

Explanation

$\dfrac{dy}{dx}=y \cot x$

$\Rightarrow \dfrac{dy}{y}= \cot x dx$

$\Rightarrow \dfrac{dy}{y} = \dfrac{\cos x}{\sin x} dx$

$\Rightarrow ln |y|+ln |c|= ln |\sin x|$

$\Rightarrow ln |Cy| = ln |\sin \, x|$

$\Rightarrow cy= \sin x$

$\Rightarrow y= k \sin x$ where k= constant

The D.E. obtained from

$(y - a)^2 = 4 (x - b)$ is given by

0%
$2y_2 + y_3 = 0$
0%
$2y_2 = y_3$
0%
$2y_1 + y^3_2 = 0$
0%
$2y_2 + y^3_1 = 0$

Explanation

$(y-a^2)=4(x-b)$

$y^2+a^2-2ay=4x-4b$

Differentiating w.r.t to x, we get

$2yy_1-2ay_1=4$

$yy_1-ay_1=2$

$yy_1-2=ay_1$

$y-\dfrac{2}{y_1}=a$

Differentiating both sides wrt

$x_1$ we get

$y_1=\dfrac{2}{y_1^2}y_2=0$

$y^3_1+2y_2=0$

$\therefore y^3_1+2y_2=0$

where

$y_1=\dfrac{dy}{dx}$ &

$y_2=\dfrac{d^2y}{dx^2}$ .

1147917_1209140_ans_6e271a6876ef495f8ada58ba2e2cfd47.jpg

The differential equation

$\dfrac{dy}{dx} = 2$ represents

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A straight line of slope $2$ units
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A circle with radius $2$ units
0%
A parabola with foci $2$ units
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None of these

Explanation

$\dfrac{dy}{dy}=2$

$\Rightarrow dy=2dx$

Integrating on both sides

$\int dy=\int 2dx$

$\Rightarrow y= 2x+c$

straight line of slope

$2$ .

1190563_1353869_ans_d3996c80be194d5aa2f81b0aec46ccb0.jpeg

The solution of the differential equation

$\left( 1+y^{ { 2 } } \right) +\left( x-e^{ { { \tan }^{ -1 }y } } \right) \dfrac { dy }{ dx } =0,$ is

0%
$( x - 2 ) = k e ^ { - \tan ^ { - 1 } y }$
0%
$2 x e ^ { 2 \tan ^ { - 1 } y } = e ^ { 2 \tan ^ { - 1 } y } = k$
0%
$x e ^ { \tan ^ { - 1 } y } = \tan ^ { - 1 } y + k$
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$xe^{ { 2\tan ^{ { -1 } } y } }=e^{ { \tan ^{ { -1 } } y } }+k$

Explanation

$\begin{array}{l} \left( { 1+{ y^{ 2 } } } \right) \frac { { dx } }{ { dy } } ={ e^{ { { \tan }^{ -1y } } } }-x \\ \frac { { dx } }{ { dy } } +\frac{x}{\left( { 1+{ y^{ 2 } } } \right)} =\frac { { { e^{ { { \tan }^{ -1y } } } } } }{ { \left( { 1+{ y^{ 2 } } } \right) } } \end{array}$

Its a linear differential equation with

$I.F. = {e^{{{\tan }^{ - 1}}y}}$

So,

We would be

$x.{e^{{{\tan }^{ - 1}}y}}$ =integral of

$\frac{{{e^{{{\tan }^{ - 1}}y}}.{e^{{{\tan }^{ - 1}}y}}}}{{\left( {1 + {y^2}} \right)}}$

Put

${\tan ^{ - 1y}} = t$

The order of the differential equation of the parabola whose axis is parallel to y-axis is:

0%
1
0%
2
0%
3
0%
4

Explanation

Equation of parabola :-

$(x-h)^{2}=4a(y-k)$
No of constants in the alove equation are 3

$\therefore$ order of diff eq

$=3$

The order of the differential equation of the curve

$y^{2}=4ax$ is:

0%
2
0%
1
0%
3
0%
Cant be determined

Explanation

$y^{2}=4ax$

$y^{2}=4ax$ has only one arbitrary constant i.e a.

So, differentiating once we get,

$\Rightarrow 2y\dfrac{dy}{dx}=4a$

$\Rightarrow 2y\dfrac{dy}{dx}=\dfrac{y^{2}}{x}$

$\Rightarrow \dfrac{dy}{dx}=\dfrac{y}{2x}$

The above equation implies,

Order

$=1$

The differential equation of all conics with the co-ordinate axes as axes is of the order

0%
1
0%
2
0%
3
0%
4

Explanation

The general equation of all conics with axes as coordinate axes is
$ax^{2}+by^{2}+c=0$

$\Rightarrow x^2+ \dfrac{b}{a}y^2+\dfrac{c}{a}=0$
Since, it has two arbitrary constants, the differential equation has the order 2.

The solution of

$\dfrac{d^{2}y}{dx^{2}}=0$ represents

0%
a straight line
0%
a circle
0%
a parabola
0%
a point

Explanation

$\dfrac{d}{dx}y_{1}=0$

$\Rightarrow \int dy_{1}=\int 0.dx+c_{1}$

$\Rightarrow \int dy_{1}=c_{1}$

$\Rightarrow y_{1}=c_{1}$

$\dfrac{dy}{dx}=c_{1}$

$\int dy=\int c_{1}\ dx+c_{2}$

$\Rightarrow y=c_{1}x+c_{2}\ straight\ line$

The order of the differenital equation of

$(x-a)^{2}+(y-b)^{2}=a^{2}$ is:

0%
4
0%
3
0%
2
0%
1

The differential equation $y\dfrac{dy}{dx}+x=A$ where A is a constant represents a set of:

0%
circles centred on y axis
0%
circles centred on x axis
0%
parabolas
0%
ellipses

Explanation

$y\ \dfrac{dy}{dx}= A-x$

$\Rightarrow y\ dy= (A-x)dx$

$\Rightarrow \dfrac{y^{2}}{2}= Ax-\dfrac{x^{2}}{2}+c$

$\dfrac{y^{2}}{2}= \dfrac{2Ax-x^{2}+c}{2}$

$y^{2}= -(x^{2}-2Ax+A^{2})+A^{2}+c$

$\Rightarrow (x-A)^{2}+y^{2}= A^{2}+c$

$\Rightarrow$ center of circle = (A,0)

The differential equation of all conics with centre at origin is of order

0%
2
0%
3
0%
4
0%
1

Explanation

The general equation of all conics with center at origin can be written as
$ax^{2}+2hxy+by^{2}+c=0$

Dividing by ' $a$ ', we get

$x^2+ \left(\dfrac{2h}{a} \right)xy+\left(\dfrac{b}{a} \right)y^2+\left(\dfrac{c}{a} \right)=0$
Since, it has three arbitrary constants.So, the differential equation is of order 3.

The differential equation of the family of circles whose center lies on

$x-$ axis and passing through origin is

0%
$x^{2}+y^{2}+\dfrac{dy}{dx}=0$
0%
$(y^{2}-x^2)dx-2xydy=0$
0%
$y^{2}dx+(x^{2}+2xy)dy=0$
0%
$xdy+ydx+x^{2}dx+y^{2}dy=0$

Explanation

General equation of a circle can be given as

$x^2+y^2+2gx+2fy+c=0$
Since it is given that this circle passes through origin and centre lies on

$x-$ axis

$\therefore$ Center of circle is

$(-g,0)$

So, the equation of a circle passing through origin and centre on

$x-$ axis is

$x^2+y^2+2gx=0$ ....(1)

$\Rightarrow g = -\dfrac{x^2+y^2}{2x}$

Differentiating (1) w.r.t

$x$

$2x+2y\dfrac{dy}{dx}+2g=0$ ....(2)
Substituting

$g$ in (2), we get

$\Rightarrow \displaystyle 2x+2y\frac{dy}{dx}-\frac{x^2+y^2}{x}=0$

$\Rightarrow \displaystyle 2x+2y\frac{dy}{dx}=\frac{x^2+y^2}{x}$

$\Rightarrow \displaystyle 2x^2+2xy\frac{dy}{dx}=x^2+y^2$
Re-arranging this, we get

$\Rightarrow (y^{2}-x^2)dx-2xydy=0$
which is the required differential equation.

Hence, option B.

The solution of

$\displaystyle \frac{dy}{dx}+\displaystyle \frac{\sqrt{1+y^{2}}}{\sqrt{1+x^{2}}}=0$ is:

0%
$\sin^{-1}x+\sin^{-1}y=c$
0%
$\tan^{-1}x+\tan^{-1}y=c$
0%
$\sinh^{-1}x+\sinh^{-1}y=c$
0%
$\cot^{-1}x+\cot^{-1}y=c$

Explanation

$\dfrac{dy}{\sqrt{1+y^{2}}}= -\dfrac{dx}{\sqrt{1+x^{2}}}$
Let

$y= \tan\theta$

$dy= \sec^{2}\theta \ d\ \theta$

$\Rightarrow \int \dfrac{dy}{\sqrt{1+y^{2}}}= \dfrac{\sec^{2}\theta\ d\ \theta }{sec\theta }$

$\Rightarrow \int \sec\theta\ d\ \theta = \log| \sec\theta +\tan\theta |$

$\Rightarrow \log |\sqrt{1+y^{2}}+y|= -\log |(\sqrt{1+x^{2}}+x)|+\log \ c$

$\Rightarrow (y+\sqrt{1+y^{2}})(x+\sqrt{1+x^{2}})= c$ , which can also be written as

$\displaystyle \sinh^{-1}x +\sinh^{-1}y=c$
Hence, option 'C' is correct.

CBSE Questions for Class 12 Commerce Maths Differential Equations Quiz 1 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Maths Differential Equations Quiz 1 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

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