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

Find the solution of $$\displaystyle \left ( 3\tan x+4\cot y-7 \right )\sin ^{2}ydx$$ $$\displaystyle -\left ( 4\tan x+7\cot y-5 \right )\cos ^{2}xdy=0$$.
  • $$\displaystyle 3\frac{\tan ^{2}x}{2}-7\tan x+7\frac{\cot ^{2}y}{2}-5\cot y+4\left ( \cot y\tan x \right )=c$$
  • $$\displaystyle 8\frac{\tan ^{2}x}{2}-7\tan x+7\frac{\cot ^{2}y}{2}-5\cot y+4\left ( \cot y\tan x \right )=c$$
  • $$\displaystyle 3\frac{\tan ^{2}x}{2}+7\tan x+7\frac{\cot ^{2}y}{2}-5\cot y+4\left ( \cot y\tan x \right )=c$$
  • $$\displaystyle 8\frac{\tan ^{2}x}{2}+7\tan x+7\frac{\cot ^{2}y}{2}-5\cot y+4\left ( \cot y\tan x \right )=c$$
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.
  • $$y\, =\, \displaystyle \frac {1}{k}\, ln\, |\, c\, (k^2x^2-1)|$$
  • $$y\, =\, \displaystyle \frac {1}{k}\, ln\, |\, c\, (kx^2-1)|$$
  • $$y\, =\, \displaystyle \frac {1}{k}\, ln\, |\, c\, (kx^2+1)|$$
  • $$y\, =\, \displaystyle \frac {1}{k}\, ln\, |\, c\, (k^2x^2+1)|$$
Solve:
$$(1-x^2)\, (1-y)\, dx\, =\, xy\, (1+y)\, dy$$
  • $$\ln(x) (1-y)^2\, =\, c\, +\, \displaystyle \frac {1}{2}\, y^2\, -\, 2y\, +\, \displaystyle \frac {1}{2}\, x^2$$
  • $$\ln(x) (1-y)^2\, =\, c\, -\, \displaystyle \frac {1}{2}\, y^2\, -\, 2y\, +\, \displaystyle \frac {1}{2}\, x^2$$
  • $$\ln(x) (1-y)^2\, =\, c\, -\, \displaystyle \frac {1}{2}\, y^2\, +\, 2y\, +\, \displaystyle \frac {1}{2}\, y^2$$
  • $$\ln(x) (1+y)^2\, =\, c\, +\, \displaystyle \frac {1}{2}\, y^2\, -\, 2y\, +\, \displaystyle \frac {1}{2}\, x^2$$
Solve:
$$\displaystyle \frac {dy}{dx}\, +\, \displaystyle \frac {\sqrt {(x^2-1)\, (y^2-1)}}{xy}\, =0$$
  • $$\sqrt {x^2+1}\, -\, \text{cosec}^{-1}x\, -\, \sqrt {y^2-1}\, =\,c$$
  • $$\sqrt {x^2-1}\, -\, \sec^{-1}x\, +\, \sqrt {y^2-1}\, =\,c$$
  • $$\sqrt {x^2-1}\, -\, \text{cosec}^{-1}x\, +\, \sqrt {y^2-1}\, =\,c$$
  • $$\sqrt {x^2+1}\, -\, \sec^{-1}x\, -\, \sqrt {y^2-1}\, =\,c$$
Solve:
$$\displaystyle \frac {dy}{dx}\, +\, \sin\, \displaystyle \frac {x+y}{2}\, =\, \sin\, \displaystyle \frac {x-y}{2}$$
  • $$\ln \left |\tan \, \displaystyle \frac {y}{2} \right |\, =\, c_1\, -2\, \cos\, \displaystyle \frac {x}{4}$$
  • $$\ln \left |\tan \, \displaystyle \frac {y}{4} \right |\, =\, c_1\, -2\, \sin\, \displaystyle \frac {x}{2}$$
  • $$\ln \left |\cot \, \displaystyle \frac {y}{4} \right |\, =\, c_1\, -2\, \cos\, \displaystyle \frac {x}{2}$$
  • $$\ln \left |\cot \, \displaystyle \frac {y}{2} \right |\, =\, c_1\, -2\, \sin\, \displaystyle \frac {x}{4}$$
Solve:
$$\displaystyle \frac {dy}{dx}\, =\, \displaystyle \frac {x(2\ln\, x\, +\, 1)}{\sin\, y\, +\, y\cos\, y}$$
  • $$y\, \sin\, y\, =\, x\, \ln\, x\, +\, c$$
  • $$y\, \cos\, y\, =\, x^2\, \ln\, x\, +\, c$$
  • $$y\, \cos\, y\, =\, x\, \ln\, x\, +\, c$$
  • $$y\, \sin\, y\, =\, x^2\, \ln\, x\, +\, c$$
A curve, whose concavity is directly proportional to the logarithm of its x-coordinate at any point of curve, is given by:
  • $${ c }_{ 1 }.{ x }^{ 2 }\left( 2\log { x } -3 \right) +{ c }_{ 2 }x+{ c }_{ 3 }$$
  • $${ c }_{ 1 }.{ x }^{ 2 }\left( 2\log { x } +3 \right) +{ c }_{ 2 }x+{ c }_{ 3 }$$
  • $${ c }_{ 1 }.{ x }^{ 2 }\left( 2\log { x } \right) +{ c }_{ 2 }$$
  • None of these
Solve the following differential equation:

$$\dfrac{dy}{dx}=(1+x^2)(1+y^2)$$
  • $$\tan^{-1} y=x+\dfrac{x^3}{3}+C$$
  • $$\tan^{-1}y=\dfrac{x^2}{2}+x+C$$
  • $$\sec^{-1}y=x+C$$
  • None of these
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:
  • a parabola
  • an ellipse
  • a hyperbola
  • a circle
If $$y=\sin ^{ -1 }{ \left[ \sqrt { x-ax } -\sqrt { a-ax }  \right]  } $$, then $$\cfrac { dy }{ dx } $$ is equal to
  • $$\cfrac { 1 }{ \sin { \sqrt { a-ax } } } $$
  • $$\sin { \sqrt { x } } .\sin { \sqrt { a } } $$
  • $$\cfrac { 1 }{ 2\sqrt { x } \sqrt { 1-x } } $$
  • $$0$$
The solution of $$\cfrac { dy }{ dx } ={ 2 }^{ y-x }$$ is
  • $${ 2 }^{ x }+{ 2 }^{ y }=c$$
  • $${ 2 }^{ x }-{ 2 }^{ y }=c$$
  • $$\cfrac { 1 }{ { 2 }^{ x } } -\cfrac { 1 }{ { 2 }^{ y } } =c$$
  • $$\cfrac { 1 }{ { 2 }^{ x } } +\cfrac { 1 }{ { 2 }^{ y } } =c$$
The order of the differential equation whose solution is $$y=a\cos { x } +b\sin { x } +c{ e }^{ -x }$$, is
  • $$3$$
  • $$1$$
  • $$2$$
  • $$4$$
$$\cfrac { d }{ dx } \left[ \tan ^{ -1 }{ \cfrac { \sqrt { 1+{ x }^{ 2 } } -1 }{ x }  }  \right] $$ is equal to
  • $$\cfrac { 1 }{ 1+{ x }^{ 2 } } $$
  • $$\cfrac { 2 }{ 1+{ x }^{ 2 } } $$
  • $$\cfrac { { x }^{ 2 } }{ 2\sqrt { 1+{ x }^{ 2 } } \sqrt { 1+{ x }^{ 2 }-1 } } $$
  • $$\cfrac { 1 }{ 2\left( 1+{ x }^{ 2 } \right) } $$
Solution of differential equation $$sec\, x\, dy - cosec\, y \,dx = 0$$ is
  • $$cos\, x+ sin\, y=c$$
  • $$sin\, x+ cos\, y=c$$
  • $$sin\, y-cos\, x=c$$
  • $$cos\, y- sin\, x=c$$
The solution of the differential equation $$\dfrac { dy }{ dx } =2{ e }^{ x-y }+{ x }^{ 2 }{ e }^{ -y }$$ is
  • $${ e }^{ y }=2{ e }^{ x }+\dfrac { { x }^{ 3 } }{ 3 } +C$$
  • $${ e }^{ -y }=2{ e }^{ x }+\dfrac { { x }^{ -3 } }{ 3 } +C$$
  • $${ e }^{ -y }=2{ e }^{ x }+\dfrac { { x }^{ 3 } }{ 3 } +C$$
  • $${ e }^{ y }=2{ e }^{ -x }+\dfrac { { x }^{ 3 } }{ 3 } +C$$
General solution of $$y\dfrac{dy}{dx}+by^2 = a \cos x, 0 < x < 1$$ is:
(here $$c$$ is an arbitrary constant)
  • $$y^2 = 2a(2b\sin x + \cos x) + ce^{-2bx}$$
  • $$(4b^2+1)y^2 = 2a(\sin x + 2b \cos x)+ ce^{-2bx}$$
  • $$(4b^2+1)y^2 = 2a(\sin x + 2b \cos x)+ ce^{2bx}$$
  • $$y^2 = 2a(2b\sin x + \cos x) + ce^{2bx}$$
The solution of the differential equation $$\displaystyle\frac{d^{2}y}{dx^2}+3y=-2x$$ is.
  • $$c_1\cos\sqrt{3}x+c_2\sin\sqrt{3}x-\dfrac{2}{3}x$$
  • $$c_1\cos\sqrt{3}x+c_2\sin\sqrt{3}x-\dfrac{4}{5}$$
  • $$c_1\cos\sqrt{3}x+c_2\sin\sqrt{3}x-2x^2+\dfrac{4}{9}$$
  • $$c_1\cos\sqrt{3}x+c_2\sin\sqrt{3}x-\dfrac{2}{3}x^2+\dfrac{4}{9}$$
The general solution of the differential equation $$\displaystyle \frac{dy}{dx}+\frac{y}{x}=3x$$ is.
  • $$y=x+\frac{c}{x}$$
  • $$y=x^2+\frac{c}{x}$$
  • $$y=x-\frac{c}{x}$$
  • $$y=x^2-\frac{c}{x^2}$$
The differential equation $$y\dfrac{dy}{dx}=a-x$$ represents
  • a family of circles with centre on y-axis
  • a family of circles with centre at origin
  • a family of circles with given radius
  • a 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
  • $$3\ln 5$$
  • $$\ln 5$$
  • $$2\ln 5$$
  • None of these

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
  • $$y=2x^3+6x+1$$
  • $$y=2x^3+4x+3$$
  • $$y=3x^2+6x+1$$
  • None of these
The solution of $$e^{2x-3y}dx+e^{2y-3x}dy=0$$ is
  • $$e^{5x}+e^{5y}=C$$
  • $$e^{5x}-e^{5y}=C$$
  • $$e^{5x+5y}=C$$
  • None of these
Solution of the equation $$\dfrac {dy}{dx} = \dfrac {y\dfrac {d(\phi (x))}{dx} - y^{2}}{\phi (x)}$$ is
  • $$y = \dfrac {\phi(x) + c}{x}$$
  • $$y = \dfrac {\phi(x)}{x} + c$$
  • $$y = \dfrac {\phi(x)}{x + c}$$
  • $$y = \phi(x) + x + c$$
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
  • $${ x }^{ 2 }\cos { y } +{ x }^{ 3 }y-{ y }^{ 2 }/2=c$$
  • $${ x }^{ 2 }\sin { y } +{ x }^{ 3 }y-{ y }^{ 2 }/2=c\quad $$
  • $${ x }^{ 2 }\sin { y } -{ x }^{ 3 }y-{ y }^{ 2 }/2=c$$
  • $${ x }^{ 2 }\cos { y } -{ x }^{ 3 }y+{ y }^{ 2 }/2=c$$
The solution of the differential equation $$\dfrac{dy}{dx}=\dfrac{y}{x}+\dfrac{\theta (y/x)}{\theta ' (y/x)}$$ is 
  • $$X\theta (y/x)=k$$
  • $$\theta (y/x)=kx$$
  • $$y\theta (y/x)=k$$
  • $$\theta (y/x)=ky$$
The solution of differential equation $$x\cos^{2}y dx = y\cos^{2} x  dy$$ is
  • $$x\tan x - y\tan y - \log\left(\dfrac{\sec x}{\sec y}\right) = c$$
  • $$y\tan x - x\tan y - \log (\sec x \cdot \sec y) = c$$
  • $$x\tan x - y\tan y + \log (\sec x\cdot \sec y) = c$$
  • None of the above
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
  • $$y=\log _{ e }{ (x) } +C$$
  • $$y={ (\log _{ e }{ x } ) }^{ 2 }+C\quad $$
  • $$y=\pm \sqrt { { (\log _{ e }{ x } ) }^{ 2 }+2C } $$
  • $$xy={ x }^{ y }+K$$
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 
  • $$\dfrac{1}{3x}+\dfrac{2x^2}{3}$$
  • $$\dfrac{-1}{3x}+\dfrac{4x^2}{3}$$
  • $$\dfrac{-1}{x}+\dfrac{2}{x^2}$$
  • $$\dfrac{1}{x}$$
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 
  • $$y=log_{ e }{ \left| \sec { \dfrac { 1 }{ x }  } +tan{ \dfrac { 1 }{ x }  } \right|  }-1$$.
  • $$y=\dfrac{x+1}{x\, sin\dfrac{1}{x}}$$
  • $$y=cos\dfrac{1}{x}+sin\dfrac{1}{x}$$
  • $$y=\dfrac{x+1}{x\, cos\dfrac{1}{x}}$$
Consider the differential equation $${ y }^{ 2 }dx+\left( x-\cfrac { 1 }{ y }  \right) dy=0$$. If $$y(A)=1$$, then $$x$$ is given by
  • $$1-\cfrac { 1 }{ y } +\cfrac { { e }^{ \cfrac { 1 }{ y } } }{ e } $$
  • $$4-\cfrac { 2 }{ y } -\cfrac { { e }^{ \cfrac { 1 }{ y } } }{ e } $$
  • $$3-\cfrac { 1 }{ y } +\cfrac { { e }^{ \cfrac { 1 }{ y } } }{ e } $$
  • $$1+\cfrac { 1 }{ y } -\cfrac { { e }^{ \cfrac { 1 }{ y } } }{ e } $$
Solution of the differential equation 
$$\frac{dy}{dx}$$ = $$\frac{3 x^2 y^4 + 2 xy}{x^2 - 2 x^3 y^3}$$ is
  • $$x^2 y^3 + {x}{y^2}$$ = c
  • $$x^3 y^2 + {x^2}{y}$$ = c
  • $${x^2}{y^3} + xy^2$$ = c
  • $$x^3 y^2 - {x^2}{y}$$ = 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
  • $$y = x\cot (c - x)$$
  • $$\cos - t\left (\dfrac {y}{x}\right ) = (-x + c)$$
  • $$y = x\tan (c - x)$$
  • 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.
  • $$2\sqrt{e}$$
  • $$-2\sqrt{e}$$
  • $$\dfrac{-2}{\sqrt{e}}$$
  • $$\dfrac{2}{\sqrt{e}}$$
If $$y_1(x)$$ 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
  • $$\frac{1}{y(x)} \int y_1(x) dx$$
  • $$y_1(x) \int \frac{r(x)}{y_1(x)} dx$$
  • $$ \int r(x)y_1(x) dx$$
  • none of these
The solution of the differential equation $$x^{2} \dfrac{dy}{dx} = x^{2} + xy + y^{2}$$ is-
  • $$tan^{-1}\dfrac{y}{x} = log x + c$$
  • $$tan^{-1}\dfrac{y}{x} = -log x + c$$
  • $$sin^{-1}\dfrac{y}{x} = log x + c$$
  • $$tan^{-1}\dfrac{x}{y} = log x + c$$
Solution of differential equation $$\frac{dy}{dx} = sin(x+y) + cos(x+y)$$ is
  • $$log|1+tan \frac{x+y}{2}| = y+c$$
  • $$log|2+sec \frac{x+y}{2}| = x+c$$
  • $$log|1+tan (x+y)| = y+c$$
  • none of these
Solution of differential equation $$\dfrac{dy}{dx} = \sin (x+y) + \cos(x+y)$$ is
  • $$log |1 + \tan\frac{x+y}{2} | = y+c$$
  • $$log |1 + \sec\frac{x+y}{2} | = x+c$$
  • $$log |1 + \tan(x+y) | = y+c$$
  • none of these
If $$y={e}^{4x}+2{e}^{-x}$$ satisfies the relation $$\dfrac {{d}^{3}y}{{dx}^{3}}+A\dfrac {dy}{dx}+By=0$$,then value of $$A$$ and $$B$$ respectively are 
  • $$-13, 14$$
  • $$-13, -12$$
  • $$-13, 12$$
  • $$12, -13$$
The general solution of $$\dfrac {dy}{dx} = \dfrac {ax + h}{by + k}$$ represents a circle only when
  • $$a = b = 0$$
  • $$a = -b\neq 0$$
  • $$a = b \neq 0, h = k$$
  • $$a = b \neq 0$$
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?
  • $$f(2) < 1 -log_e 2$$
  • $$f(2) > 1-log_e 2$$
  • $$g(1) > 1 -log_e 2$$
  • $$g(1) < 1 -log_e 2$$
$$Solution\quad of\quad the\quad equation:\quad (1-{ x }^{ 2 })dy\quad +\quad xydx=\quad x{ y }^{ 2 }dx$$
  • $$({ y-1 }^{ )2 }{ (1- }{ x }^{ 2 })=0$$
  • $$({ y-1 }^{ )2 }{ (1- }{ x }^{ 2 })={ c }^{ 2 }{ y }^{ 2 }$$
  • $$({ y-1 }^{ )2 }{ (1+ }{ x }^{ 2 })={ c }^{ 2 }{ y }^{ 2 }$$
  • None of these
What is the solution of $$(1+2x)dy-(1-2y)dx=0$$?
  • $$x-y-2xy=c$$
  • $$y-x-2xy=c$$
  • $$y+x-2xy=c$$
  • $$x+y+2xy=c$$
Let $$f(x)$$ be a differential function and satisfy $$f(0) =2$$, and $$f'(x) = f(x)$$ . Find 
  • The range of the function $$f(x)\ is\ (0,\infty)$$
  • The value of the function when $$x= ln2\ is\ 3$$
  • The area enclosed by $$y =f(x)$$ in the  $$2nd$$ quadrant is bounded.
  • None of these
The solution of $$\left( {\cos ecx\log y} \right)dy + \left( {{x^2}y} \right)dx = 0$$ is
  • $${{\log y} \over 2} + \left( {2 - {x^2}} \right)\cos x + 2\sin x = c$$
  • $${\left( {{{\log y} \over 2}} \right)^2} + \left( {2 - {x^2}} \right)\cos x + 2x\sin x = c$$
  • $${{{{\left( {\log y} \right)}^2}} \over 2} + \left( {2 - {x^2}} \right)\cos x + 2x\sin x = c$$
  • none of these
What is the solution of the differential equation $$xdy+ydx=0$$
  • $$xy=c$$
  • $$y=cx$$
  • $$x+y=c$$
  • $$x-y=c$$
Solve: $$\dfrac{{dy}}{{dx}} = 1 + x + y + xy$$
  • $${1+y=k{e}^\dfrac{x^2+2x}{4}}$$
  • $${1+y=k{e}^\dfrac{x^2+2x}{2}}$$
  • $${1+y=k{e}^\dfrac{x^2+4x}{2}}$$
  • $${1+y=k{e}^\dfrac{x^2+4x}{4}}$$
$$\frac{{dy}}{{dx}} + \frac{{{y^2} + y + 1}}{{{x^2} + x + 1}} = 0$$
  • $${\tan ^{ - 1}}\left( {2x + 1} \right) + {\tan ^{ - 1}}\left( {2y + 1} \right) = c$$
  • $${\tan ^{ - 1}}\frac{{2x + 1}}{{\sqrt 3 }} + {\tan ^{ - 1}}\frac{{2y + 1}}{{\sqrt 3 }} = c$$
  • $${\tan ^{ - 1}}\frac{{2x}}{{\sqrt 5 }} + {\tan ^{ - 1}}\frac{{2y}}{{\sqrt 3 }} = c$$
  • $${\tan ^{ - 1}}\frac{{2x - 1}}{{\sqrt 3 }} + {\tan ^{ - 1}}\frac{{2y - 1}}{{\sqrt 3 }} = c$$
The solution of the differential equation $$(1+y^{2})+(x-e^{\tan -1}y)\dfrac {dy}{dx}=0$$, is-
  • $$xe^{2\tan^{-1}y}= e^{\tan^{-1}y}+k$$
  • $$(x-2)=ke^{-\tan^{-1}y}$$
  • $$2xe^{\tan^{-1}y}= e^{2\tan^{-1}y}+k$$
  • $$xe^{\tan^{-1}y}=\tan^{-1}y+k$$
The solution of $$\dfrac {dy}{dx}+\dfrac {y}{x}=x^{2}y^{6}$$ is
  • $$\dfrac{1}{x^{2}y^{5}}=\dfrac{5}{2x^{2}}+c$$
  • $$\dfrac{1}{x^{5}y^{5}}=\dfrac{5}{x^{2}}+c$$
  • $$\dfrac{1}{x^{5}y^{5}}=\dfrac{5}{2x^{2}}+c$$
  • $$\dfrac {1}{x^{2}y^{2}}=\dfrac {5}{2x^{2}}+c$$
Solution of $$\left(\dfrac{x+y-a}{x+y-b}\right)\left(\dfrac{dy}{dx}\right)=\left(\dfrac{x+y+a}{x+y+b}\right)$$.
  • $$log\left[(x+y)^2-ab\right]=\dfrac{2}{b-a}[x-y]+k$$
  • $$log\left[(x+y)^2+ab\right]=\dfrac{1}{b-a}[x+y]+k$$
  • $$\left(\dfrac{b-a}{2}\right)\left[log \left((x+y)^2-ab\right)\right]=x+c$$
  • $$2log(x+y)=\dfrac{x+y}{b-a}+k$$
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