MCQ Questions for Class 12 Commerce Maths Differential Equations Quiz 9

Find the solution of

$\displaystyle \left ( 3\tan x+4\cot y-7 \right )\sin ^{2}ydx$

$\displaystyle -\left ( 4\tan x+7\cot y-5 \right )\cos ^{2}xdy=0$ .

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$\displaystyle 3\frac{\tan ^{2}x}{2}-7\tan x+7\frac{\cot ^{2}y}{2}-5\cot y+4\left ( \cot y\tan x \right )=c$
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$\displaystyle 8\frac{\tan ^{2}x}{2}-7\tan x+7\frac{\cot ^{2}y}{2}-5\cot y+4\left ( \cot y\tan x \right )=c$
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$\displaystyle 3\frac{\tan ^{2}x}{2}+7\tan x+7\frac{\cot ^{2}y}{2}-5\cot y+4\left ( \cot y\tan x \right )=c$
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$\displaystyle 8\frac{\tan ^{2}x}{2}+7\tan x+7\frac{\cot ^{2}y}{2}-5\cot y+4\left ( \cot y\tan x \right )=c$

Find the curve for which the sum of the lengths of the tangent and subtangent at any of its point is proportional to the product of the coordinates of the point of tangency, the proportionality factor is equal to k.

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$y\, =\, \displaystyle \frac {1}{k}\, ln\, |\, c\, (k^2x^2-1)|$
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$y\, =\, \displaystyle \frac {1}{k}\, ln\, |\, c\, (kx^2-1)|$
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$y\, =\, \displaystyle \frac {1}{k}\, ln\, |\, c\, (kx^2+1)|$
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$y\, =\, \displaystyle \frac {1}{k}\, ln\, |\, c\, (k^2x^2+1)|$

Explanation

Length of tangent at any point of curve

$= y\sqrt {1+\, \left ( \displaystyle \frac {dx}{dy} \right )^2}\,$
Length of subtangent

$= y\, \displaystyle \frac {dx}{dy}\,$

$y\sqrt {1+\, \left ( \displaystyle \frac {dx}{dy} \right )^2}\, +\, y\, \displaystyle \frac {dx}{dy}\, =\, kxy$

$\Rightarrow\, \sqrt {1+\, \left ( \displaystyle \frac {dx}{dy} \right )^2}\, =\, kx\, -\, \displaystyle \frac {dx}{dy}$

$\Rightarrow\, 1\, +\, \left ( \displaystyle \frac {dx}{dy} \right )^2\, =\, k^2x^2\, +\, \left ( \displaystyle \frac {dx}{dy} \right )^2\, -2kx\, \displaystyle \frac {dx}{dy}$

$\Rightarrow\, 2kx\, \displaystyle \frac {dx}{dy}\, =\, k^2x^2\, -\, 1\quad \Rightarrow\, \displaystyle \int \frac {2kxdx}{k^2x^2-1}\, =\, \int dy$

$\Rightarrow\, =\, \displaystyle \frac {1}{k}\, ln\, |c(k^2x^2-1)|$

Solve:

$(1-x^2)\, (1-y)\, dx\, =\, xy\, (1+y)\, dy$

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$\ln(x) (1-y)^2\, =\, c\, +\, \displaystyle \frac {1}{2}\, y^2\, -\, 2y\, +\, \displaystyle \frac {1}{2}\, x^2$
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$\ln(x) (1-y)^2\, =\, c\, -\, \displaystyle \frac {1}{2}\, y^2\, -\, 2y\, +\, \displaystyle \frac {1}{2}\, x^2$
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$\ln(x) (1-y)^2\, =\, c\, -\, \displaystyle \frac {1}{2}\, y^2\, +\, 2y\, +\, \displaystyle \frac {1}{2}\, y^2$
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$\ln(x) (1+y)^2\, =\, c\, +\, \displaystyle \frac {1}{2}\, y^2\, -\, 2y\, +\, \displaystyle \frac {1}{2}\, x^2$

Explanation

$(1-x^2)\, (1-y)\, dx\, =\, xy\, (1+y)\, dy$

$\displaystyle \frac { 1-x^{ 2 } }{ x } dx=\frac { { y }^{ 2 }+y }{ 1-y } dy$
Integrating both sides, we get

$\displaystyle \log { x } -\frac { x^{ 2 } }{ 2 } =-\int { \left(y+2+ \frac { 2 }{ y-1 } \right)dy}$

$\displaystyle \Rightarrow \log { x } -\frac { x^{ 2 } }{ 2 } =-\frac { y^{ 2 } }{ 2 } -2y+2\log { (y-1) }$

$\log (x) (1-y)^2\, =\, c\, -\, \displaystyle \frac {1}{2}\, y^2\, -\, 2y\, +\, \displaystyle \frac {1}{2}\, x^2$

Solve:

$\displaystyle \frac {dy}{dx}\, +\, \displaystyle \frac {\sqrt {(x^2-1)\, (y^2-1)}}{xy}\, =0$

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$\sqrt {x^2+1}\, -\, \text{cosec}^{-1}x\, -\, \sqrt {y^2-1}\, =\,c$
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$\sqrt {x^2-1}\, -\, \sec^{-1}x\, +\, \sqrt {y^2-1}\, =\,c$
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$\sqrt {x^2-1}\, -\, \text{cosec}^{-1}x\, +\, \sqrt {y^2-1}\, =\,c$
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$\sqrt {x^2+1}\, -\, \sec^{-1}x\, -\, \sqrt {y^2-1}\, =\,c$

Explanation

Given differential eqn can be written as

$\displaystyle \frac { dy }{ dx } \, =-\, \frac { \sqrt { (x^{ 2 }-1)\, (y^{ 2 }-1) } }{ xy } \,$

$\Rightarrow \displaystyle \frac { y }{ \sqrt { y^{ 2 }-1 } } dy\, =-\, \frac { \sqrt { x^{ 2 }-1 } }{ x } dx\,$
Integrating both sides

$\displaystyle \int { \frac { y }{ \sqrt { y^{ 2 }-1 } } dy\, } =-\, \int { \frac { \sqrt { x^{ 2 }-1 } }{ x } dx\, }$

Put

$y^2-1=t$

$\Rightarrow 2ydy=dt$

Also, put

$\sqrt { x^{ 2 }-1 } =u$

$\Rightarrow x^2-1=u^2$

$\Rightarrow xdx=udu$

$\displaystyle \int { \frac { 1 }{ 2\sqrt { t } } dt\, } =-\, \int { \frac { u }{ 1+{ u }^{ 2 } } udu\, }$

$\displaystyle \Rightarrow \int { \frac { 1 }{ 2\sqrt { t } } dt\, } =-\, \int { \frac { { u }^{ 2 } }{ 1+{ u }^{ 2 } } du\, }$

$\displaystyle \Rightarrow \sqrt { t } =-\, \int { (1-\frac { 1 }{ 1+{ u }^{ 2 } } )du\, }$

$\displaystyle \Rightarrow \sqrt { { y }^{ 2 }-1 } =-u+\tan ^{ -1 }{ u } +C$

$\displaystyle \Rightarrow \sqrt { { y }^{ 2 }-1 } =-\sqrt { { x }^{ 2 }-1 } +\tan ^{ -1 }{ \sqrt { { x }^{ 2 }-1 } } +C$

$\displaystyle \Rightarrow \sqrt { { y }^{ 2 }-1 } =-\sqrt { { x }^{ 2 }-1 } +\sec ^{ -1 }{ x} +C$ (

$\tan ^{ -1 }{ \sqrt { { x }^{ 2 }-1 } }=z \Rightarrow \sqrt { { x }^{ 2 }-1 } =\tan z \Rightarrow \sec z=x$ )

Solve:

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

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$\ln \left |\tan \, \displaystyle \frac {y}{2} \right |\, =\, c_1\, -2\, \cos\, \displaystyle \frac {x}{4}$
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$\ln \left |\tan \, \displaystyle \frac {y}{4} \right |\, =\, c_1\, -2\, \sin\, \displaystyle \frac {x}{2}$
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$\ln \left |\cot \, \displaystyle \frac {y}{4} \right |\, =\, c_1\, -2\, \cos\, \displaystyle \frac {x}{2}$
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$\ln \left |\cot \, \displaystyle \frac {y}{2} \right |\, =\, c_1\, -2\, \sin\, \displaystyle \frac {x}{4}$

Explanation

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

we know

$\sin (A-B)-\sin (A+B)=-\cos A\sin B$

$\Rightarrow\, \displaystyle \frac {dy}{dx}\, =\, -\, 2\, \cos \, \displaystyle \frac {x}{2}\, \sin \displaystyle \frac {y}{2}$

$\Rightarrow\, \text{cosec}\, \displaystyle \frac {y}{2}\, dy\, =\, -2\, \cos \, \displaystyle \frac {x}{2}\, dx$

On integration we get

$\Rightarrow\, 2\, \left (\ln \left | \tan \, \displaystyle \frac {y}{4} \right | \right )\, =\, 2\, \left (-2\, \sin \, \displaystyle \frac {x}{2}\right )\, +\, c$

$\Rightarrow\, \ln \left |\tan \, \displaystyle \frac {y}{4} \right |\, =\, c_1\, -2\, \sin \, \displaystyle \frac {x}{2}$

Solve:

$\displaystyle \frac {dy}{dx}\, =\, \displaystyle \frac {x(2\ln\, x\, +\, 1)}{\sin\, y\, +\, y\cos\, y}$

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$y\, \sin\, y\, =\, x\, \ln\, x\, +\, c$
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$y\, \cos\, y\, =\, x^2\, \ln\, x\, +\, c$
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$y\, \cos\, y\, =\, x\, \ln\, x\, +\, c$
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$y\, \sin\, y\, =\, x^2\, \ln\, x\, +\, c$

Explanation

Given

$\displaystyle \frac {dy}{dx}\, =\, \displaystyle \frac {x(2\ln\, x\, +\, 1)}{\sin\, y\, +\, \cos\, y}$

$\dfrac { dy }{ dx } (\sin\, y\, +\, y\cos\, y)\, =\, 2x\ln\, x\, +\, x$

$\dfrac { dy }{ dx } \times \sin\, y\, +\, y\times \dfrac { d(\sin y) }{ dx } \, =\dfrac { \, { d(x }^{ 2 }) }{ dx } \times \ln\, x\, +\, { x }^{ 2 }\times \dfrac { d(\ln x) }{ dx }$

On integration

$y\sin y={ x }^{ 2 }\ln x+c$

A curve, whose concavity is directly proportional to the logarithm of its x-coordinate at any point of curve, is given by:

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${ c }_{ 1 }.{ x }^{ 2 }\left( 2\log { x } -3 \right) +{ c }_{ 2 }x+{ c }_{ 3 }$
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${ c }_{ 1 }.{ x }^{ 2 }\left( 2\log { x } +3 \right) +{ c }_{ 2 }x+{ c }_{ 3 }$
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${ c }_{ 1 }.{ x }^{ 2 }\left( 2\log { x } \right) +{ c }_{ 2 }$
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None of these

Explanation

Let

$P(x,y)$ be any point on the curve
According to given conditon,

$\displaystyle \frac{d^2 y}{dx^2}=k\log x$ ....(1)
where

$k$ is constant of proportionality.
Integrating both sides of eqn (1),

$\displaystyle \frac{dy}{dx}=k\int \log x.1 dx$

$\displaystyle \frac{dy}{dx}=k(x\log x-x)+c_2$
Again integrating, we get

$y=k \int {x\log x} -k\int xdx +\int c_2 dx$

$\Rightarrow \displaystyle y=k [\log x \frac{x^2}{2}-\int \frac{x^2}{2} \frac{1}{x}dx]-k\frac{x^2}{2}+c_2 x+c_3$

$\Rightarrow \displaystyle y=\frac { k }{ 4 } [2{ x }^{ 2 }\log x-x^{ 2 }-2x^{ 2 }]+c_{ 2 }x+c_{ 3 }$

$\Rightarrow \displaystyle y={ c }_{ 1 }[2{ x }^{ 2 }\log x-3x^{ 2 }]+c_{ 2 }x+c_{ 3 }$

Solve the following differential equation:

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

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$\tan^{-1} y=x+\dfrac{x^3}{3}+C$
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$\tan^{-1}y=\dfrac{x^2}{2}+x+C$
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$\sec^{-1}y=x+C$
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None of these

Explanation

The given differential eqaution :

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

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

Integrating both sides,

$\int \dfrac{dy}{1+y^2}=\int (1+x^2) \ dx$

$\tan^{-1}y = x+\dfrac{x^3}{3}+C$

The equation of the curve in which the portion of the tangent included between the coordinate axes is bisected at the point of contact is:

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a parabola
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an ellipse
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a hyperbola
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a circle

Explanation

Let the equation of curve be

$y=f(x)$

Let

$P(x,y)$ be any point on the curve.

Equation of tangent at

$P(x,y)$ is

$Y-y=y'(X-x)$

$\Rightarrow y'X-Y=xy'-y$

$\Rightarrow \displaystyle \dfrac{X}{\dfrac{xy'-y}{y'}}+\dfrac{Y}{y-xy'}=1$

So, the intercepts of the tangent on the axis are

$\dfrac{xy'-y}{y'}$ and

$y-xy'$

So, the point

$A$ on

$x$ -axis is

$(\dfrac{xy'-y}{y'} ,0)$ and

$B$ on

$y$ -axis is

$(0,y-xy')$ .

Now,

$P$ is mid-point of

$AB$

So coordinates of

$P$ are

$(\dfrac{xy'-y}{2y'}, \dfrac{y-xy'}{2})$

$\Rightarrow x=\dfrac{xy'-y}{2y'}$ and

$y=\dfrac{y-xy'}{2}$

$\Rightarrow \displaystyle xy'-y=2y'x$ and

$xy'-y=-2y$

$\Rightarrow 2y'x=-2y$

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

Integrating both sides, we get

$\log y=-\log x +\log C$

$\Rightarrow \log xy=\log C$

$\Rightarrow xy=C$

$y=\sin ^{ -1 }{ \left[ \sqrt { x-ax } -\sqrt { a-ax } \right] }$ , then

$\cfrac { dy }{ dx }$ is equal to

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$\cfrac { 1 }{ \sin { \sqrt { a-ax } } }$
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$\sin { \sqrt { x } } .\sin { \sqrt { a } }$
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$\cfrac { 1 }{ 2\sqrt { x } \sqrt { 1-x } }$
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$0$

Explanation

Given

$y=\sin ^{ -1 }{ \left[ \sqrt { x-ax } -\sqrt { a-ax } \right] }$

Put

$x=\sin ^{ 2 }{ \theta } ,a=\sin ^{ 2 }{ \phi }$ then

$y=\sin ^{ -1 }{ \left[ \sqrt { \sin ^{ 2 }{ \theta } -\sin ^{ 2 }{ \phi } \sin ^{ 2 }{ \theta } } -\sqrt { \sin ^{ 2 }{ \phi } -\sin ^{ 2 }{ \phi } \sin ^{ 2 }{ \theta } } \right] }$

$=\sin ^{ -1 }{ \left[ \sqrt { \sin ^{ 2 }{ \theta } (1-\sin ^{ 2 }{ \phi } ) } -\sqrt { \sin ^{ 2 }{ \phi } (1-\sin ^{ 2 }{ \theta } ) } \right] }$

$=\sin ^{ -1 }{ \left[ \sin { \theta } \cos { \phi } -\sin { \phi } \cos { \theta } \right] }$

$=\sin ^{ -1 }{ \left[ \sin { \left( \theta -\phi \right) } \right] } =\theta -\phi$

$=\sin ^{ -1 }{ \sqrt { x } } -\sin ^{ -1 }{ \sqrt { a } }$

$\therefore \cfrac { dy }{ dx } =\cfrac { 1 }{ \sqrt { 1-{ \left( \sqrt { x } \right) }^{ 2 } } } .\cfrac { 1 }{ 2\sqrt { x } } =\cfrac { 1 }{ \sqrt { 1-x } } .\cfrac { 1 }{ 2\sqrt { x } }$

The solution of

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

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${ 2 }^{ x }+{ 2 }^{ y }=c$
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${ 2 }^{ x }-{ 2 }^{ y }=c$
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$\cfrac { 1 }{ { 2 }^{ x } } -\cfrac { 1 }{ { 2 }^{ y } } =c$
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$\cfrac { 1 }{ { 2 }^{ x } } +\cfrac { 1 }{ { 2 }^{ y } } =c$

Explanation

Given

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

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

On integrating both sides, we get

$\cfrac { { 2 }^{ -y } }{ \log { 2 } } (-1)=\cfrac { { 2 }^{ -x } }{ \log { 2 } } (-1)+{ c }_{ 1 }$

$\Rightarrow -\cfrac { { 2 }^{ -y } }{ \log { 2 } } =\cfrac { -{ 2 }^{ -x } }{ \log { 2 } } +{ c }_{ 1 }$

$\Rightarrow -{ 2 }^{ -y }=-{ 2 }^{ -x }+{ c }_{ 1 }\log { 2 }$

$\therefore \quad \cfrac { 1 }{ { 2 }^{ x } } -\cfrac { 1 }{ { 2 }^{ y } } =c\quad \left( c={ c }_{ 1 }\log { 2 } \right)$

The order of the differential equation whose solution is

$y=a\cos { x } +b\sin { x } +c{ e }^{ -x }$ , is

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$3$
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$1$
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$2$
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$4$

Explanation

Given DE has 3 different independent constants

$a,b,c$

Therefore order of given differential equation is

$3$

$\cfrac { d }{ dx } \left[ \tan ^{ -1 }{ \cfrac { \sqrt { 1+{ x }^{ 2 } } -1 }{ x } } \right]$ is equal to

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$\cfrac { 1 }{ 1+{ x }^{ 2 } }$
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$\cfrac { 2 }{ 1+{ x }^{ 2 } }$
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$\cfrac { { x }^{ 2 } }{ 2\sqrt { 1+{ x }^{ 2 } } \sqrt { 1+{ x }^{ 2 }-1 } }$
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$\cfrac { 1 }{ 2\left( 1+{ x }^{ 2 } \right) }$

Explanation

Let

$y=\cfrac { d }{ dx } \left[ \tan ^{ -1 }{ \cfrac { \sqrt { 1+{ x }^{ 2 } } -1 }{ x } } \right]$

Put

$x=\tan { \theta }$

$\Rightarrow \theta =\tan ^{ -1 }{ x }$

Then,

$y=\cfrac { d }{ dx } \left[ \tan ^{ -1 }{ \cfrac { \sqrt { 1+\tan ^{ 2 }{ \theta } } -1 }{ \tan { \theta } } } \right]$

$=\cfrac { d }{ dx } \left[ \tan ^{ -1 }{ \cfrac { \sqrt { \sec ^{ 2 }{ \theta } } -1 }{ \tan { \theta } } } \right]$

$=\cfrac { d }{ dx } \left[ \tan ^{ -1 }{ \cfrac { \cfrac { 1 }{ \cos { \theta } } -1 }{ \cfrac { \sin { \theta } }{ \cos { \theta } } } } \right]$

$=\cfrac { d }{ dx } \left[ \tan ^{ -1 }{ \left( \cfrac { 1-\cos { \theta } }{ sin\theta } \right) } \right]$

$=\cfrac { d }{ dx } \tan ^{ -1 }{ \left( \tan { \cfrac { \theta }{ 2 } } \right) }$

$=\cfrac { d }{ dx } \left( \cfrac { \theta }{ 2 } \right) =\cfrac { 1 }{ 2 } \cfrac { d }{ dx } \left( \tan ^{ -1 }{ x } \right)$

$=\cfrac { 1 }{ 2 } \cfrac { 1 }{ \left( 1+{ x }^{ 2 } \right) }$

Solution of differential equation

$sec\, x\, dy - cosec\, y \,dx = 0$ is

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$cos\, x+ sin\, y=c$
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$sin\, x+ cos\, y=c$
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$sin\, y-cos\, x=c$
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$cos\, y- sin\, x=c$

Explanation

$sec\,x\,dy-cosec\,y-dx = 0$

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

By variable separable from

$\displaystyle sin\,y\,dy = cos\,x\,dx$

integrating,

$\displaystyle \int siny\,dy = \int cosx\,dx$

$-cos\,y = sin\,x+c$

$\boxed { sinx+cosy+c =0}$

$\boxed {sinx+cosy = c}$

1117080_460573_ans_ddc5ab14a5b8441a878145a0d1cb7ed7.JPG

The solution of the differential equation

$\dfrac { dy }{ dx } =2{ e }^{ x-y }+{ x }^{ 2 }{ e }^{ -y }$ is

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${ e }^{ y }=2{ e }^{ x }+\dfrac { { x }^{ 3 } }{ 3 } +C$
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${ e }^{ -y }=2{ e }^{ x }+\dfrac { { x }^{ -3 } }{ 3 } +C$
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${ e }^{ -y }=2{ e }^{ x }+\dfrac { { x }^{ 3 } }{ 3 } +C$
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${ e }^{ y }=2{ e }^{ -x }+\dfrac { { x }^{ 3 } }{ 3 } +C$

Explanation

Given,

$\dfrac { dy }{ dx } =2{ e }^{ x-y }+{ x }^{ 2 }{ e }^{ -y }$

$\Rightarrow \dfrac { dy }{ dx } =2{ e }^{ x }\cdot \dfrac { 1 }{ { e }^{ y } } +{ x }^{ 2 }\cdot \dfrac { 1 }{ { e }^{ y } }$

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

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

On integrating both sides, we get

$\displaystyle\int { { e }^{ y }dy } =\displaystyle\int { \left( 2{ e }^{ x }+{ x }^{ 2 } \right) dx }$

$\Rightarrow { e }^{ y }=2{ e }^{ x }+\dfrac { { x }^{ 3 } }{ 3 } +C$
which is the required solution.

General solution of

$y\dfrac{dy}{dx}+by^2 = a \cos x, 0 < x < 1$ is:

(here

$c$ is an arbitrary constant)

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$y^2 = 2a(2b\sin x + \cos x) + ce^{-2bx}$
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$(4b^2+1)y^2 = 2a(\sin x + 2b \cos x)+ ce^{-2bx}$
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$(4b^2+1)y^2 = 2a(\sin x + 2b \cos x)+ ce^{2bx}$
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$y^2 = 2a(2b\sin x + \cos x) + ce^{2bx}$

Explanation

Let us take

$y^{2}=t$ , which means

$dt=2ydy$

Given equation will transform into

$\cfrac{dt}{dx}+2bt=2a\cos x$

The integration factor is

$e^{2bx}$

The general solution is

$te^{2bx} = \displaystyle \int { 2a\cos x{ e }^{ 2bx }dx } +c$

$\Rightarrow te^{2bx}=\cfrac{4ab\cos x+2a\sin x}{4b^{2}+1}e^{2bx}+c$

$\Rightarrow (4b^{2}+1)y^{2}=2a(\sin x+2b\cos x)+Ce^{-2bx}$

Therefore the correct option is

$B$

The solution of the differential equation

$\displaystyle\frac{d^{2}y}{dx^2}+3y=-2x$ is.

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$c_1\cos\sqrt{3}x+c_2\sin\sqrt{3}x-\dfrac{2}{3}x$
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$c_1\cos\sqrt{3}x+c_2\sin\sqrt{3}x-\dfrac{4}{5}$
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$c_1\cos\sqrt{3}x+c_2\sin\sqrt{3}x-2x^2+\dfrac{4}{9}$
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$c_1\cos\sqrt{3}x+c_2\sin\sqrt{3}x-\dfrac{2}{3}x^2+\dfrac{4}{9}$

Explanation

$\dfrac { { d }^{ 2 }y }{ d{ x }^{ 2 } } +3y=-2x.......(i)$

This is second order non homogeneous differential equation.

Its solution is given as $CF+PI$

For $CF$ part

$\dfrac { { d }^{ 2 }y }{ d{ x }^{ 2 } } +3y=0....(ii)\\ y={ { { c }_{ 1 } }e }^{ mx }\\ \dfrac { dy }{ dx } ={ c }_{ 1 }m{ e }^{ mx }\\ \dfrac { { d }^{ 2 }y }{ d{ x }^{ 2 } } ={ c }_{ 1 }{ m }^{ 2 }{ e }^{ mx }$

Substituting in $(ii)$ , we get
${ m }^{ 2 }{ e }^{ mx }+3{ e }^{ mx }=0\\ { e }^{ mx }({ m }^{ 2 }+3)=0\\ { m }^{ 2 }+3=0\\ \Rightarrow m=\sqrt { -3 } \\ m=i\sqrt { 3 } \\ y=c_{1}{ e }^{ i\sqrt { 3 } x }={ c }_{ 1 }(\cos { \sqrt { 3 } x } +i\sin { \sqrt { 3 } x } )\\ y={ c }_{ 1 }\cos { \sqrt { 3 } x } +{ c }_{ 2 }\sin { \sqrt { 3 } x } \quad \quad$

For $PI$ part

Let $y=cx+d$

$\dfrac { dy }{ dx } =c\\ \dfrac { { d }^{ 2 }y }{ d{ x }^{ 2 } } =0$

Substituting in $(i)$

$0+3(cx+d)=-2x\\ 3cx+3d=-2x$

Comparing both sides

$c=-\dfrac { 2 }{ 3 } ,d=0$

So solution for $PI$ part is

$y=-\dfrac { 2 }{ 3 } x$

General solution is $CF+PI$

${ c }_{ 1 }\cos { \sqrt { 3 } x } +{ c }_{ 2 }\sin { \sqrt { 3 } x } -\dfrac { 2 }{ 3 } x\quad$

The general solution of the differential equation

$\displaystyle \frac{dy}{dx}+\frac{y}{x}=3x$ is.

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

Explanation

Comparing above equation with first order linear differential equation

$\dfrac{dy}{dx}+Py=Q$

We get

$P=\dfrac{1}{x}$ and

$Q=3x$

So integration factor

$I.F. =e^{\int Pdx}=e^{\int (1/x)dx}=e^{\ln x}=x$

Hence solution is given by,

$y(I.F.)=\displaystyle \int Q(I.F.)d x$

$\Rightarrow xy=\displaystyle \int 3x^2dx$

$\Rightarrow xy=x^3+c$

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

The differential equation

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

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a family of circles with centre on y-axis
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a family of circles with centre at origin
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a family of circles with given radius
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a family of circles with centre on x-axis

Explanation

Given: $y\dfrac { dy }{ dx } =a-x$

Integrating on both sides

$\int { ydx } =\int { adx } -\int { xdx }$

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

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

${ (x-a) }^{ 2 }+{ y }^{ 2 }=r^{ 2 }$

Hence family of circles with centre on x axis

A particle, initially at origin moves along $x$ axis according to the rule $\dfrac{dx}{dt}=x+4$ . The time taken by the particle to traverse a distance of $96$ units is

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$3\ln 5$
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$\ln 5$
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$2\ln 5$
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None of these

Explanation

Given

$\frac{dx}{dt}=x+4$

$\Rightarrow \frac{dx}{x+4}=dt$

By applying integration on both sides , we get

$\ln{(x+4)}=t+k$

$\Rightarrow x=e^{t}-4+C$

$t=0$ ,

$x=0$ , which implies

$C=3$

Therefore the equation becomes

$x=e^{t}-1$

Given that

$x=96$ , which implies

$t=\ln{97}$

The equation of the curves, satisfying the differential equation $\dfrac{d^2y}{dx^2}(x^2+1)=2x\dfrac{dy}{dx}$ passing through the point $(0,1)$ and having the slope of tangent at $x=0$ as $6$ is

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$y=2x^3+6x+1$
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$y=2x^3+4x+3$
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$y=3x^2+6x+1$
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None of these

Explanation

Given

$\ddot { y } -\dfrac { 2x }{ { x }^{ 2 }+1 } \dot { y } =0$

It is in the form of linear equation

The integration factor is

${ e }^{ \int { -\frac { 2x }{ { x }^{ 2 }+1 } dx } }=\dfrac { 1 }{ { x }^{ 2 }+1 }$

Therefore the general solution is

$\dot { y } \dfrac { 1 }{ { x }^{ 2 }+1 } =\int { 0dx } =c$

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

Given that at

$x=0$ , the value of

$\dot { y }$ is

$6$

$\Rightarrow c=6$

Therefore, the solution is

$\dfrac{dy}{dx}=6(x^{2}+1)$

By integrating on both sides, we get

$y=2x^{3}+6x+C$

It passes through point

$(0,1)$ , so we get

$C=1$

Therefore, the general solution is

$y=2x^{3}+6x+1$

The solution of

$e^{2x-3y}dx+e^{2y-3x}dy=0$ is

0%
$e^{5x}+e^{5y}=C$
0%
$e^{5x}-e^{5y}=C$
0%
$e^{5x+5y}=C$
0%
None of these

Explanation

${ e }^{ 2x-3y }dx + { e }^{ 2x-3y }dy=0\\ \Rightarrow \cfrac { { e }^{ 2x } }{ { e }^{ 3y } } dx+\cfrac { { e }^{ 2y } }{ { e }^{ 3x } } dy=0 \\ \Rightarrow \cfrac { { e }^{ 5x }dx + { e }^{ 5y }dy }{ { e }^{ 3y }{ e }^{ 3x } } =0\\ \Rightarrow { e }^{ 5x }dx + { e }^{ 5y }dy=0$

Integrating n both sides.

$\cfrac { { e }^{ 5x } }{ 5 } +\cfrac { { e }^{ 5y } }{ 5 } =k\\ \Rightarrow { e }^{ 5x }+{ e }^{ 5y }=k$

Solution of the equation

$\dfrac {dy}{dx} = \dfrac {y\dfrac {d(\phi (x))}{dx} - y^{2}}{\phi (x)}$ is

0%
$y = \dfrac {\phi(x) + c}{x}$
0%
$y = \dfrac {\phi(x)}{x} + c$
0%
$y = \dfrac {\phi(x)}{x + c}$
0%
$y = \phi(x) + x + c$

The solution of the differential equation

$\left( 2x\cos { y } +3{ x }^{ 2 }y \right) dx+\left( { x }^{ 3 }-{ x }^{ 2 }\sin { y } -y \right) dy=0$ , is given by

0%
${ x }^{ 2 }\cos { y } +{ x }^{ 3 }y-{ y }^{ 2 }/2=c$
0%
${ x }^{ 2 }\sin { y } +{ x }^{ 3 }y-{ y }^{ 2 }/2=c\quad$
0%
${ x }^{ 2 }\sin { y } -{ x }^{ 3 }y-{ y }^{ 2 }/2=c$
0%
${ x }^{ 2 }\cos { y } -{ x }^{ 3 }y+{ y }^{ 2 }/2=c$

The solution of the differential equation

$\dfrac{dy}{dx}=\dfrac{y}{x}+\dfrac{\theta (y/x)}{\theta ' (y/x)}$ is

0%
$X\theta (y/x)=k$
0%
$\theta (y/x)=kx$
0%
$y\theta (y/x)=k$
0%
$\theta (y/x)=ky$

The solution of differential equation

$x\cos^{2}y dx = y\cos^{2} x dy$ is

0%
$x\tan x - y\tan y - \log\left(\dfrac{\sec x}{\sec y}\right) = c$
0%
$y\tan x - x\tan y - \log (\sec x \cdot \sec y) = c$
0%
$x\tan x - y\tan y + \log (\sec x\cdot \sec y) = c$
0%
None of the above

Explanation

$x\cos^{2}y dx = y\cos^{2} x dy$

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

$\Rightarrow x\sec^{2}x dx = y\sec^{2}y dy$
On integration, we get

$\Rightarrow x\tan x - \int 1\cdot \tan x dx$

$= y \tan y - \int 1 \cdot \tan y dy$

$\Rightarrow x\tan x - \log \sec x = y\tan y - \log \sec y + c$

$\Rightarrow x\tan x - y\tan y - (\log \sec x- \log \sec y) = c$

$\Rightarrow x\tan x - y\tan y - \log\left(\dfrac{\sec x}{\sec y}\right) = c$ .

The solution of the differential equation

$x=1+xy\cfrac { dy }{ dx } +\cfrac { { \left( xy \right) }^{ 2 } }{ 2! } { \left( \cfrac { dy }{ dx } \right) }^{ 2 }+\cfrac { { \left( xy \right) }^{ 3 } }{ 3! } { \left( \cfrac { dy }{ dx } \right) }^{ 3 }+...\quad \quad$
is

0%
$y=\log _{ e }{ (x) } +C$
0%
$y={ (\log _{ e }{ x } ) }^{ 2 }+C\quad$
0%
$y=\pm \sqrt { { (\log _{ e }{ x } ) }^{ 2 }+2C }$
0%
$xy={ x }^{ y }+K$

Explanation

Given Differential equation is the expansion of

$e^{xy\frac{dy}{dx}}$

So,

$x=e^{xy\frac{dy}{dx}}$

Taking log on both sides, We get

$log x=xy\dfrac{dy}{dx}$

$\dfrac{log x}{x}dx=ydy$

Integrating both sides

$\Rightarrow \dfrac{(log x)^2}{2}+C=\dfrac{y^2}{2}$

$\Rightarrow y=\pm\sqrt{(log_ex)^2+2C}$

Let f(x) be differentiable on the interval

$(0,\infty)$ such that

$f(1)=1$ and

$\displaystyle \lim_{t \rightarrow x }\dfrac{t^2f(x)-x^2f(t)}{t-x}=1$ for each

$x>0$ . Then f(x) is

0%
$\dfrac{1}{3x}+\dfrac{2x^2}{3}$
0%
$\dfrac{-1}{3x}+\dfrac{4x^2}{3}$
0%
$\dfrac{-1}{x}+\dfrac{2}{x^2}$
0%
$\dfrac{1}{x}$

The solution of the differential equation,

$x^2\dfrac{dy}{dx}.cos\dfrac{1}{x}=-1$ , where

$y\rightarrow -1$ as

$x\rightarrow \infty$ is

0%
$y=log_{ e }{ \left| \sec { \dfrac { 1 }{ x } } +tan{ \dfrac { 1 }{ x } } \right| }-1$ .
0%
$y=\dfrac{x+1}{x\, sin\dfrac{1}{x}}$
0%
$y=cos\dfrac{1}{x}+sin\dfrac{1}{x}$
0%
$y=\dfrac{x+1}{x\, cos\dfrac{1}{x}}$

Explanation

Given,

${ x }^{ 2 }\dfrac { dy }{ dx } .cos{ \dfrac { 1 }{ x } }=-1$

$\Rightarrow dy=\dfrac { 1 }{ cos{ \dfrac { 1 }{ x } } } .\dfrac { -1 }{ { x }^{ 2 } } dx$

Let us assume $\dfrac { 1 }{ x } =t$ .

Therefore, $\dfrac { -1 }{ { x }^{ 2 } } dx=dt$ .

$\Rightarrow dy=\dfrac { 1 }{ cos{ t } } dt$

$\Rightarrow dy=\sec { t } dt$

Integrating both sides, we get,

$\Rightarrow \int { dy } =\int { \sec { t } } dt$

$\Rightarrow y=log_{ e }{ \left| \sec { t } +tan{ t } \right| }+C$

$\Rightarrow y=log_{ e }{ \left| \sec { \dfrac { 1 }{ x } } +tan{ \dfrac { 1 }{ x } } \right| }+C$

Now, given that $y\rightarrow -1\quad as\quad x\rightarrow \infty$

${ \Rightarrow lim }_{ x\rightarrow \infty }y=-1$

${ \Rightarrow lim }_{ x\rightarrow \infty }\left( log_{ e }{ \left| \sec { \dfrac { 1 }{ x } } +tan{ \dfrac { 1 }{ x } } \right| }+C \right) =-1$

${ \Rightarrow }\left( log_{ e }{ \left| \sec { 0 } +tan{ 0 } \right| }+C \right) =-1$

$\Rightarrow C=-1$ .

Therefore, the solution of given differential equation is:

$y=log_{ e }{ \left| \sec { \dfrac { 1 }{ x } } +tan{ \dfrac { 1 }{ x } } \right| }-1$ .

Consider the differential equation

${ y }^{ 2 }dx+\left( x-\cfrac { 1 }{ y } \right) dy=0$ . If

$y(A)=1$ , then

$x$ is given by

0%
$1-\cfrac { 1 }{ y } +\cfrac { { e }^{ \cfrac { 1 }{ y } } }{ e }$
0%
$4-\cfrac { 2 }{ y } -\cfrac { { e }^{ \cfrac { 1 }{ y } } }{ e }$
0%
$3-\cfrac { 1 }{ y } +\cfrac { { e }^{ \cfrac { 1 }{ y } } }{ e }$
0%
$1+\cfrac { 1 }{ y } -\cfrac { { e }^{ \cfrac { 1 }{ y } } }{ e }$

Solution of the differential equation

$\frac{dy}{dx}$ =

$\frac{3 x^2 y^4 + 2 xy}{x^2 - 2 x^3 y^3}$ is

0%
$x^2 y^3 + {x}{y^2}$ = c
0%
$x^3 y^2 + {x^2}{y}$ = c
0%
${x^2}{y^3} + xy^2$ = c
0%
$x^3 y^2 - {x^2}{y}$ = c

The solution of the differential equation

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

0%
$y = x\cot (c - x)$
0%
$\cos - t\left (\dfrac {y}{x}\right ) = (-x + c)$
0%
$y = x\tan (c - x)$
0%
none of these

If x and y are independent variables and

$f(x)=(\int^x_0e^{x-y}f'(y)dy)-(x^2-x+1)e^x$ where f(x) is a differentiable function, then

$f(\dfrac{-1}{2})$ equals.

0%
$2\sqrt{e}$
0%
$-2\sqrt{e}$
0%
$\dfrac{-2}{\sqrt{e}}$
0%
$\dfrac{2}{\sqrt{e}}$

$y_1(x)$ is a solution of the differential equation

$\frac{dy}{dx} + f(x) + y = 0$ , then a solution of differential equation

$\frac{dy}{dx} + f(x) + y = r (x)$ is

0%
$\frac{1}{y(x)} \int y_1(x) dx$
0%
$y_1(x) \int \frac{r(x)}{y_1(x)} dx$
0%
$\int r(x)y_1(x) dx$
0%
none of these

The solution of the differential equation

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

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

Explanation

$x^{2}\cfrac{dy}{dx}=x^{2}+xy+y^{2}$

$\cfrac{dy}{dx}=1+\cfrac{y}{x}+(\cfrac{y}{x})^{2}$

Let

$\cfrac{y}{x}=v$

$y=vx$

$\cfrac{dy}{dx}=v+x\cfrac{dv}{dx}$

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

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

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

$\tan^{-1}\,v=\log x+C$

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

Solution of differential equation

$\frac{dy}{dx} = sin(x+y) + cos(x+y)$ is

0%
$log|1+tan \frac{x+y}{2}| = y+c$
0%
$log|2+sec \frac{x+y}{2}| = x+c$
0%
$log|1+tan (x+y)| = y+c$
0%
none of these

Solution of differential equation

$\dfrac{dy}{dx} = \sin (x+y) + \cos(x+y)$ is

0%
$log |1 + \tan\frac{x+y}{2} | = y+c$
0%
$log |1 + \sec\frac{x+y}{2} | = x+c$
0%
$log |1 + \tan(x+y) | = y+c$
0%
none of these

$y={e}^{4x}+2{e}^{-x}$ satisfies the relation

$\dfrac {{d}^{3}y}{{dx}^{3}}+A\dfrac {dy}{dx}+By=0$ ,then value of

$A$ and

$B$ respectively are

0%
$-13, 14$
0%
$-13, -12$
0%
$-13, 12$
0%
$12, -13$

The general solution of

$\dfrac {dy}{dx} = \dfrac {ax + h}{by + k}$ represents a circle only when

0%
$a = b = 0$
0%
$a = -b\neq 0$
0%
$a = b \neq 0, h = k$
0%
$a = b \neq 0$

Explanation

Given that

$\dfrac{dy}{dx} = \dfrac{ax + h}{by + k}$

$\Rightarrow dy \; (by + k) = dx \; (ax + h)$

Integrating LHS and RHS,

$b\dfrac{y^2}{2} + ky + c = a\dfrac{x^2}{2} + h x + c$

$\Rightarrow ax^2 - b y^2 + 2hx - 2ky + c = 0$

The general equation of a cirlce is given as:

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

Upon comparing the obtained equation with the general equation of a circle,

$ax^2 - b y^2 + 2hx - 2ky + c = 0$

represents a cirlce if

$a = -b \neq 0$ .

Let f

$:\mathbb{R}\rightarrow \mathbb{R}$ and

$g:\mathbb{R}\rightarrow \mathbb{R}$ be two non-constant differentiable functions. If

$f'(x)=(e^{(f(x)-g(x))})g'(x)$ for all

$x\epsilon \mathbb{R}$ , and

$f(1)=g(2)=1$ , then which of the following statement(s) is (are) TRUE?

0%
$f(2) < 1 -log_e 2$
0%
$f(2) > 1-log_e 2$
0%
$g(1) > 1 -log_e 2$
0%
$g(1) < 1 -log_e 2$

Explanation

$f'(x)=e^{(f(x)-g(x))}g'(x)\forall x\epsilon \mathbb{R}$

$\Rightarrow e^{-f(x)}\cdot f'(x)-e^{-g(x)}g'(x)=0$

$\Rightarrow \displaystyle\int(e^{-f(x)}f'(x)-e^{-g(x)}\cdot g'(x))dx=C$

$\Rightarrow -e^{-f(x)}+e^{-g(x)}=C$

$\Rightarrow -e^{-f(1)}+e^{-g(1)}=-e^{-f(2)}+e^{-g(2)}$

$\Rightarrow -\displaystyle\dfrac{1}{e}+e^{-g(1)}=-e^{-f(2)}+\dfrac{1}{e}$

$\Rightarrow e^{-f(2)}+e^{-g(1)}=\dfrac{2}{e}$

$\therefore$

$e^{-f(2)} < \dfrac{2}{e}$ and

$e^{-g(1)} < \dfrac{2}{e}$

$\Rightarrow -f(2) < log_{e}2 -1$ and

$-g(1) < log_{e}2-1$

$\Rightarrow f(2) > 1 -log_{e}2$ and

$g(1) > 1-log_{e}2$ .

$Solution\quad of\quad the\quad equation:\quad (1-{ x }^{ 2 })dy\quad +\quad xydx=\quad x{ y }^{ 2 }dx$

0%
$({ y-1 }^{ )2 }{ (1- }{ x }^{ 2 })=0$
0%
$({ y-1 }^{ )2 }{ (1- }{ x }^{ 2 })={ c }^{ 2 }{ y }^{ 2 }$
0%
$({ y-1 }^{ )2 }{ (1+ }{ x }^{ 2 })={ c }^{ 2 }{ y }^{ 2 }$
0%
None of these

Explanation

$(1-x^{2})dy+xydx=xy^{2}dx$

Dividing by

$y^{2}dx$ , we get(and also divisible by

$(1-x^{2})$

$\cfrac{1}{y^{2}}\cfrac{dy}{dx}+\cfrac{x}{1-x^{2}}\cfrac{1}{y}=\cfrac{x}{1-x^{2}}$

Now by putting

$\cfrac{-1}{y}=z$

$\cfrac{1}{y^{2}}dy=dz$

Therefore, equation becomes

$\cfrac{dz}{dx}-\cfrac{x}{1-x^{2}}dz=\cfrac{x}{(1-x^{2})}$

Now ,

$IF=e^{\int\cfrac{-x}{1-x^{2}}}=e^{+\cfrac{1}{2}\log(1-x^{2})}=\sqrt{1-x^{2}}$

Multiplying by IF , we get

$\cfrac{d}{dx}((\sqrt{1-x^{2}})z)$ =

$\cfrac{x}{\sqrt{1-x^{2}}}$

Now integrating on both the sides and taking

$\sqrt{1-x^{2}}=P$ on RHS, we get

$\cfrac{1}{2\sqrt{1-x^{2}}}\times(-2x)dx=dP$

$\cfrac{-x}{\sqrt{1-x^{2}}}dx=dP$

$\cfrac{x}{\sqrt{1-x^{2}}}dx=-dP$

$\int d((1-\sqrt{1-x^{2}})dz)=-\int-dP$

$(1-\sqrt{1-x^{2}})z=-P+C=-\sqrt{1-x^{2}}+C$

$(\sqrt{1-x^{2})}(z+1)=C$

$\int(\sqrt{1-x^{2})}(\cfrac{-1}{y}+1)=C$

Now squaring on both the sides, we get

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

What is the solution of

$(1+2x)dy-(1-2y)dx=0$ ?

0%
$x-y-2xy=c$
0%
$y-x-2xy=c$
0%
$y+x-2xy=c$
0%
$x+y+2xy=c$

Explanation

Consider

$(1+2x)dy-(1-2y)dx=0$

$\Rightarrow (1+2x)dy=(1-2y)dx$

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

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

$\Rightarrow \dfrac{1}{2}log(1+2x)=-\dfrac{1}{2}log(1-2y)+\dfrac{1}{2}log\:c$

$\Rightarrow log(1+2x)+log(1-2y)=log\:c$

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

$\Rightarrow x-y-2xy=c$

Let

$f(x)$ be a differential function and satisfy

$f(0) =2$ , and

$f'(x) = f(x)$ . Find

0%
The range of the function $f(x)\ is\ (0,\infty)$
0%
The value of the function when $x= ln2\ is\ 3$
0%
The area enclosed by $y =f(x)$ in the $2nd$ quadrant is bounded.
0%
None of these

Explanation

$f^{'}(x)=f(x)$

$\Rightarrow \cfrac{dy}{dx}=y$

$\Rightarrow\cfrac{dy}{y}=dx$

Now on integrating, we get

$\log y=x+C$

$f(0)=2\Rightarrow C=\log 2$

$\log y=x+C=x+\log2$

$y=2e^{x}$

Therefore, we can say that range of f(x) is

$(0,\infty )$ .

The solution of

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

0%
${{\log y} \over 2} + \left( {2 - {x^2}} \right)\cos x + 2\sin x = c$
0%
${\left( {{{\log y} \over 2}} \right)^2} + \left( {2 - {x^2}} \right)\cos x + 2x\sin x = c$
0%
${{{{\left( {\log y} \right)}^2}} \over 2} + \left( {2 - {x^2}} \right)\cos x + 2x\sin x = c$
0%
none of these

Explanation

$\left( {\mathrm cos ec \ x\log y} \right)dy + \left( {{x^2}y} \right)dx = 0$

$\cfrac{-\log y}{y}\,dy=\cfrac{x^{2}dx}{\csc x}=\sin x\, x^{2}dx$

Integrating on both the sides and using Product Formula in RHS; for LHS,

Let

$\log y=z$

$\cfrac{1}{y}dy=dz$

$\int -z\, dz=x^{2}\int\sin x-\int(\cfrac{d}{dx}x^{2}\int \sin x)dx$

$\cfrac{-z^{2}}{2}=-x^{2}\,\cos x+\int\,2x\,\cos x$

$\cfrac{-z^{2}}{2}=-x^{2}\,\cos x+2[x\int \cos x-\int(\cfrac{dx}{dx}\int\cos x) dx]$

$\cfrac{-z^{2}}{2}=-x^{2}\,\cos x+2x\sin x-2\int \sin x$

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

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

What is the solution of the differential equation

$xdy+ydx=0$

0%
$xy=c$
0%
$y=cx$
0%
$x+y=c$
0%
$x-y=c$

Explanation

Consider

$xdy+ydx=0$

$\Rightarrow xdy=-ydx$

$\Rightarrow \dfrac{1}{y}dy=-\dfrac{1}{x}dx$

Integrating both sides we get

$\Rightarrow ln\:|y|=-ln\:|x|+C$

$\Rightarrow ln\:|y|+ln\:|x|=C$

$\Rightarrow ln(|x||y|)=C$

$\Rightarrow |x||y|=e^C$

$\Rightarrow xy=e^C$

Let

$e^C=c$ we get

$\Rightarrow xy=c$

Solve:

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

0%
${1+y=k{e}^\dfrac{x^2+2x}{4}}$
0%
${1+y=k{e}^\dfrac{x^2+2x}{2}}$
0%
${1+y=k{e}^\dfrac{x^2+4x}{2}}$
0%
${1+y=k{e}^\dfrac{x^2+4x}{4}}$

Explanation

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

$\dfrac{dy}{dx}=y(x+1)+(x+1)$

$\dfrac{dy}{dx}=(x+1)(y+1)$

$\dfrac{dy}{y+1}=(x+1)dx$

Now, integrating both sides,

$\int\dfrac{dy}{y+1}=\int(x+1)dx$

$\Rightarrow$

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

Now, taking exponential both sides,

$\Rightarrow$

$y+1=e^{\dfrac{x^2}{2}+x+c}$

$\therefore$

$y=e^{\dfrac{x^2}{2}+x+c}-1$

$\frac{{dy}}{{dx}} + \frac{{{y^2} + y + 1}}{{{x^2} + x + 1}} = 0$

0%
${\tan ^{ - 1}}\left( {2x + 1} \right) + {\tan ^{ - 1}}\left( {2y + 1} \right) = c$
0%
${\tan ^{ - 1}}\frac{{2x + 1}}{{\sqrt 3 }} + {\tan ^{ - 1}}\frac{{2y + 1}}{{\sqrt 3 }} = c$
0%
${\tan ^{ - 1}}\frac{{2x}}{{\sqrt 5 }} + {\tan ^{ - 1}}\frac{{2y}}{{\sqrt 3 }} = c$
0%
${\tan ^{ - 1}}\frac{{2x - 1}}{{\sqrt 3 }} + {\tan ^{ - 1}}\frac{{2y - 1}}{{\sqrt 3 }} = c$

Explanation

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

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

On integrating, we get

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

$\displaystyle\int{\dfrac{dy}{{y}^{2}+2\times\dfrac{1}{2}y+\dfrac{1}{4}-\dfrac{1}{4}+1}}=-\displaystyle\int{\dfrac{dx}{{x}^{2}+2\times\dfrac{1}{2}x+\dfrac{1}{4}-\dfrac{1}{4}+1}}$

$\displaystyle\int{\dfrac{dy}{{\left(y+\dfrac{1}{2}\right)}^{2}+\dfrac{3}{4}}}=-\displaystyle\int{\dfrac{dx}{{\left(x+\dfrac{1}{2}\right)}^{2}+\dfrac{3}{4}}}$

$\displaystyle\int{\dfrac{dy}{{\left(y+\dfrac{1}{2}\right)}^{2}+{\left(\dfrac{\sqrt{3}}{2}\right)}^{2}}}=-\displaystyle\int{\dfrac{dx}{{\left(x+\dfrac{1}{2}\right)}^{2}+{\left(\dfrac{\sqrt{3}}{2}\right)}^{2}}}$

$\dfrac{2}{\sqrt{3}}{\tan}^{-1}{\left(\dfrac{y+\dfrac{1}{2}}{\dfrac{\sqrt{3}}{2}}\right)}=-\dfrac{2}{\sqrt{3}}{\tan}^{-1}{\left(\dfrac{x+\dfrac{1}{2}}{\dfrac{\sqrt{3}}{2}}\right)}+k$

$\dfrac{2}{\sqrt{3}}{\tan}^{-1}{\left(\dfrac{y+\dfrac{1}{2}}{\dfrac{\sqrt{3}}{2}}\right)}+\dfrac{2}{\sqrt{3}}{\tan}^{-1}{\left(\dfrac{x+\dfrac{1}{2}}{\dfrac{\sqrt{3}}{2}}\right)}=k$

${\tan}^{-1}{\left(\dfrac{y+\dfrac{1}{2}}{\dfrac{\sqrt{3}}{2}}\right)}+{\tan}^{-1}{\left(\dfrac{x+\dfrac{1}{2}}{\dfrac{\sqrt{3}}{2}}\right)}=\dfrac{\sqrt{3}}{2}k$

${\tan}^{-1}{\left(\dfrac{y+\dfrac{1}{2}}{\dfrac{\sqrt{3}}{2}}\right)}+{\tan}^{-1}{\left(\dfrac{x+\dfrac{1}{2}}{\dfrac{\sqrt{3}}{2}}\right)}=c$ where

$c=\dfrac{\sqrt{3}}{2}k$

${\tan}^{-1}{\left(\dfrac{2y+1}{\sqrt{3}}\right)}+{\tan}^{-1}{\left(\dfrac{2x+1}{\sqrt{3}}\right)}=c$ where

$c=\dfrac{\sqrt{3}}{2}k$

The solution of the differential equation

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

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

The solution of

$\dfrac {dy}{dx}+\dfrac {y}{x}=x^{2}y^{6}$ is

0%
$\dfrac{1}{x^{2}y^{5}}=\dfrac{5}{2x^{2}}+c$
0%
$\dfrac{1}{x^{5}y^{5}}=\dfrac{5}{x^{2}}+c$
0%
$\dfrac{1}{x^{5}y^{5}}=\dfrac{5}{2x^{2}}+c$
0%
$\dfrac {1}{x^{2}y^{2}}=\dfrac {5}{2x^{2}}+c$

Solution of

$\left(\dfrac{x+y-a}{x+y-b}\right)\left(\dfrac{dy}{dx}\right)=\left(\dfrac{x+y+a}{x+y+b}\right)$ .

0%
$log\left[(x+y)^2-ab\right]=\dfrac{2}{b-a}[x-y]+k$
0%
$log\left[(x+y)^2+ab\right]=\dfrac{1}{b-a}[x+y]+k$
0%
$\left(\dfrac{b-a}{2}\right)\left[log \left((x+y)^2-ab\right)\right]=x+c$
0%
$2log(x+y)=\dfrac{x+y}{b-a}+k$

Explanation

$\left( \cfrac { x+y-a }{ x+y-b } \right) \left( \cfrac { dy }{ dx } \right) =\left( \cfrac { x+y+a }{ x+y+b } \right)$

Let

$x+y=z$

$\Rightarrow 1+\cfrac { dy }{ dx } =\cfrac { dz }{ dx } \Rightarrow \cfrac { dy }{ dx } =\cfrac { dz }{ dx } -1\quad$

substituting

$\Rightarrow \left( \cfrac { z-a }{ z-b } \right) \left( \cfrac { dz }{ dx } -1 \right) =\left( \cfrac { z+a }{ z+b } \right)$

$\Rightarrow \cfrac { dz }{ dx } =1+\cfrac { \left( z+a \right) \left( z-b \right) }{ \left( z-a \right) \left( z+b \right) } =\cfrac { { z }^{ 2 }+bz-az+{ z }^{ 2 }-bz+az-ab }{ \left( z-a \right) \left( z+b \right) } =\cfrac { 2\left( { z }^{ 2 }-ab \right) }{ \left( z-a \right) \left( z+b \right) }$

$\Rightarrow \int { \cfrac { \left( { z }^{ 2 }-ab \right) +z\left( b-a \right) }{ { z }^{ 2 }-ab } } dz=\int { 2 } dx=2x+C$

$\Rightarrow \int { \left\{ 1+\cfrac { z\left( b-a \right) }{ { z }^{ 2 }-ab } \right\} } dz=2x+C\quad$

$\Rightarrow z+\cfrac { b-a }{ 2 } \ln { \left( { z }^{ 2 }-ab \right) } =2x+C$

$\Rightarrow \cfrac { b-a }{ 2 } \ln { \left( { z }^{ 2 }-ab \right) } =2x+C-(x+y)=x-y+C$

$\Rightarrow \log { \left[ { \left( x+y \right) }^{ 2 }-ab \right] } =\cfrac { 2 }{ b-a } \left( x-y \right) +C$

CBSE Questions for Class 12 Commerce Maths Differential Equations Quiz 9 - 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 9 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

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