MathsMaths QuestionsDifferential Equations Questions for CBSE Class 12th

Differential Equations Questions for CBSE Class 12th

The solution of differential equation 2 y sin ⁡ x d y d x = 2 sin ⁡ x cos ⁡ x − y 2 cos ⁡ x at x = π 2 , y = 1 is

The solution of differential equation x y 3 ( 1 + cos ⁡ x ) − y d x + x d y = 0 is

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    Solution of the equation x d y − y + x y 3 ( 1 + log ⁡ x ) d x = 0 is

    The solution of x + y d y / d x y − x d y / d x = x sin 2 ⁡ x 2 + y 2 y 3 is given by

    If y ( t ) is the solution of the differential equation ( 1 + t ) dy dt − yt = 1 and y ( 0 ) = − 1 then y ( 1 ) is

    If the solution of the differential equation d y d x = 1 x c o s y + s i n 2 y i s x = c e s i n y – k ( 1 + s i n y ) , t h e n t h e v a l u e o f k i s

    The solution of the differential equation x d y d x = y + x e y / x is

    Solution of the differential equation d y d x = 2 x − y + 1 x + y + 2 is

    The solution of differential equation e − x ( y + 1 ) d y + cos 2 ⁡ x − sin ⁡ 2 x y d x = 0 , if y ( 0 ) = 1 , is

    The solution of differential equation x d y y 2 e x y + e x / y = y d x e x / y − y 2 e x y , is

    The solution of differential equation 3 x 1 − x 2 y 2 d y d x + 2 x 2 − 1 y 3 = a x 3 is

    Solution of the differential equation x + 2 y 3 d y d x = y is

    If f ( 1 ) = 0 and d f d x > f ( x ) ∀ x ≥ 1 , then

    x cos ⁡ x d y d x + y ( x sin ⁡ x + cos ⁡ x ) = 1 then xy =

    If x + 2 y 3 d y d x = y then

    The orthogonal trajectories of the family of ellipses x 2 + 2 y 2 − y = c are family of parabola y = c x 2 + k , where k =

    If y = y ( x ) satisfies the differential equation 8 x ( 9 + x ) dy = ( 4 + 9 + x ) − 1 dx , x > 0 and y ( 0 ) = 7 then y ( 256 ) =

    The function y = f ( x ) is the solution of the differential equation dy dx + xy x 2 − 1 = x 4 + 2 x 1 − x 2 in x ∈ ( − 1 , 1 ) satisfying f ( 0 ) = 0 the ∫ − 3 2 3 / 2 f ( x ) d x is

    The solution of the differential equation y + sin x . cos 2 x y cos 2 x y d x + x cos 2 x y + sin y d y = 0 is

    The solution of the differential equation sin x + cos x d y + cos x − sin x d x = 0 is

    The solution of the differential equation d y d x = 2 x y − 3 y + 2 x − 3 is

    The solution of the differential equation 1 − x 2 sin − 1 x d y + y d x = 0 is

    If x 2 y d x − x 3 + y 3 d y = 0 , y 0 = 1 and y 3 log y = k   x 3 then k =

    If ϕ x = ∫ ϕ x − 2 d x and ϕ 1 = 0 then ϕ x =

    The length of latus rectum of the conic passing through the origin and having the property that normal at each point (x, y) intersects the x-axis at x + 1 , 0 is:

    If y 1 and y 2 are two solution to the differential equation d y d x + p x y = Q x and y = y 1 + c k then k =

    Solve the differential equation 1 − x y + x 2 y 2 d x = x 2 d y

    Solution of differential equation d y d x + x sin 2 ⁡ y = sin ⁡ y cos ⁡ y is

    General solution of x d y d x + y = y 2 x 3 cos ⁡ x , is

    The acute angle between the two line passing through the origin and satisfying the differential equation ( 4 x − 3 y ) d x + ( 2 y − 3 x ) d y = 0

    The solution of 2 y cos ⁡ y 2 d y d x − 2 x + 1 sin ⁡ y 2 = ( x + 1 ) 3 is

    The solution of d.E y 2 x 2 y + e x d x − e x + y 3 d y = 0 is

    The solution of differential equation 2 y + x y 3 d x + x + x 2 y 2 d y = 0 is

    Solution of differential equation d z d x + z x log ⁡ z = z x 2 ( log ⁡ z ) 2 is

    The solution of sin ⁡ x y ( y d x − x d y ) = x y 3 ( x d y + y d x ) is

    The solution of sec 2 ⁡ θ d θ + tan ⁡ θ ( 1 − r tan ⁡ θ ) d r = 0 is

    Solution of differential equation d t d x = t d d x ( g ( x ) ) − t 2 g ( x ) is

    The solution of differential equation x + y y 1 y − x y 1 = x 2 + 2 y 2 + y 4 x 2 is

    If d y d x = x y + 2 x + 3 y + 6 then y ( − 1 ) − e 2 y ( − 3 ) =

    Solution of differential equation x 2 1 x 2 + y 2 ( x d y + y d x ) + y 2 ( x d y − y d x ) = 0 is

    The substation required to change ( 3 y − 7 x + 7 ) d x + ( 7 y − 3 x + 3 ) d y = 0 to a homogeneous differential equation is

    The solution of the differential equation ( y + x x y ( x + y ) ) d x + ( y x y ( x + y ) − x ) d y = 0 is

    Solve d y d x = x 3 y 3 − x y

    The solution of the DE 2 x 3 y d y + 1 − y 2 x 2 y 2 + y 2 − 1 d x = 0 is

    If x ⋅ ln ⁡ x ⋅ d y d x + y = 2 ln x , y ( e ) = 2 then y e 2 =

    If y − cos ⁡ x ⋅ d y d x = y 2 ( 1 − sin ⁡ x ) cos ⁡ x , y ( 0 ) = 1 then y π 3

    A curve passes through the point ( 1 , π / 6 ) . Let the slope of the curve at each point ( x , y ) be y x + sec ⁡ ( y / x ) , x > 0 then the equation of the curve is

    If d y d x = x y + y x y + x , then the solution of the differential equation is

    The solution of the differential equation x d y − y d x = x 2 + y 2 d x is

    The real value of m for which the substitution y = u m will transform the differential equation 2 x 4 y d y d x + y 4 = 4 x 6 into a homogeneous equation is

    The solution of the differential equation d y d x = x + y 2 x + 2 y − 2 is

    The solution of y 5 x + y − x d y d x = 0 is

    x + y d y d x y − x d y d x = x 2 + 2 y 2 + y 4 x 2 , t h e s o l u t i o n o f t h e d i f f e r e n t i a l e q u a t i o n i s k y x − 1 x 2 + y 2 = C , t h e n k =

    The solution of d y d x = x + y − 1 + x + y log x + y is given by

    The solution of the differential equation x d x y = f x y f 1 x y d x is

    A solution of the differential equation y x y − 1 d x + x y ln x d y = 0 , when y 2 = 1 is

    x 2 y 3 + x y d y d x = 1 , y ( 1 ) = 0 , y = 2 when x =

    If y = y(x) is the solution of the differential equation, e y d y d x − 1 = e x , such that y(0) = 0, then y(1) is equal to:

    Let y=y(x) be the solution curve of the differential equation y 2 − x d y d x = 1 , satisfying y(0) = 1. This curve intersects the x-axis at a point whose abscissa is:

    Let y=y(x) be a function of x satisfying y 1 − x 2 = k − x 1 − y 2 where k is a constant y 1 2 = − 1 4 and . Then d y d x at x = 1 2 is equal to:

    Let y = y ( x ) be a solution of the differential equation, 1 – x 2 dy dx + 1 – y 2 = 0 , | x | < 1 . If y 1 2 = 3 2 , then y – 1 2 is equal to:

    If for x ≥ 0 , y = y ( x ) is the solution of the differential equation, ( x + 1 ) d y = ( x + 1 ) 2 + y – 3 d x , y ( 2 ) = 0 , then y ( 3 ) is equal to

    A function y = f ( x ) satisfies the differential equation f ( x ) · sin 2 x – cos x + 1 + sin 2 x f ‘ ( x ) = 0 with initial condition y ( 0 ) = 0 . The value of f ( π / 6 ) =

    An equation of the curve satisfying x d y – y d x = x 2 – y 2 d x and y ( 1 ) = 0 is

    The solution of the differential equation d y d x = x y + x + y + 1 is

    Let y = y ( x ) be the solution of the differential equation sin x d y d x + y cos x = 4 x , ∈ ( 0 , π ) . If y π 2 = 0 , then y π 6 is equal to

    Solution of the differential equation x d x + y d y x d x − y d y = y 3 x 3 is given by (Base of all log = e )

    If a function y = f ( x ) satisfies the differential equation f ( x ) ⋅ sin ⁡ 2 x − cos ⁡ x + 1 + sin 2 ⁡ x f ′ ( x ) = 0 with initial condition y ( 0 ) = 0 , then the value of f π 6 is equal to

    The slope of the tangent at ( x , y ) to a curve passing through 1 , π 4 is given by y x − cos 2 ⁡ y x then the equation of the curve is

    The solution of differential equation d y d x = 1 − y 2 y determines family of circles with

    The general solution of the differential equation d y d x + sin x + y 2 = sin x − y 2 is

    The solution of the differential equation x 4 + y 4 d x − x y 3 d y = 0 is

    Solution of the differential equation x sin y x d y = y sin y x − x d x is

    The solution of the differential equation y d y d x = x   e x 2 + y 2 is

    The solution of the differential equation, y   d x + x + x 2 y d y = 0 is

    The solution of the differential equation d y d x = 2 x y − 3 y + 2 x − 3 is

    The solution of the differential equation x − y    a r c     tan y x d x + x    a r c T a n y x d y = 0 is of the form log x 4 x 2 + y 2 + k   tan − 1 y x = c then the value of k is

    The solution of y 1 + x − 1 + sin y d x + x + log e x + x cos y d y = 0 is

    The equation of the curve satisfying the differential equation y 2 x 2 + 1 = 2 x d y d x passing through the point (1,1) is

    Solution of the differential equation y − x d y d x = a y 2 + d y d x

    The solution of the differential equation x d y d x + y = x 3 y 6 is

    If the slope of the tangent to the curve is maximum at x = 1 and curve has minim value 1 at x = 0, then the curve which also satisfies the equation d 3 y d x 3 = 4 x − 3 is

    The solution of the differential equation d y = tan 2 x + y d x

    Solution to the differential equation x + x 3 3 ! + x 5 5 ! + …. 1 + x 2 2 ! + x 4 4 ! …. = d x − d y d x + d y

    The solution of the differential equation y d x + x d y x cos y x = x d y − y d x y sin y x is

    The solution of y 2 x 2 y + e x d x − e x + y 3 d y = 0 , if y 0 = 1 is

    The slope of the tangent at (x, y) to a curve passing through 1 , π 4 is given by y x − cos 2 y x , then the equation of the curve is

    The solution of the differential equation y 2 d y = x x d y − y d x e x / y is

    The solution of differential equation 1 − x y + x 2 y 2 d x = x 2 d y is

    The general solution of the differential equation y 2 + e 2 x d y − y 3 d x = 0 , (c being the constant of integration), is

    If for x ≥ 0 , y = y ( x ) is the solution of the differential equation, ( x + 1 ) d y = ( x + 1 ) 2 + y – 3 d x , y ( 2 ) = 0 , then y ( 3 ) is equal to

    The solution of the differential equation y 1 + 3 x 2 y e x 3 d x = x d y is

    If x d y − y d x + x cos ⁡ ( ln ⁡ x ) d x = 0 , y ( 1 ) = 1 , then y ( e )

    The solution of differential equation y d x − 1 + e − x y 2 d y = 0 is

    Solution of x y 4 + y d x − x d y = 0 is

    The solution of differential equation x d x − y d y x d y − y d x = 1 + x 2 − y 2 x 2 − y 2 is

    The solution of the differential equation d y d x = ( 1 + y ) 2 x ( 1 + y ) − x 2 is given as ln ⁡ ( 1 + y ) =

    The solution of differential equation x d y − y d x x 2 + y 2 + tan − 1 ⁡ x y 3 ( x d y + y d x ) = 0 is

    The solution of d.E x 2 + y 2 + x d x − 2 x 2 + 2 y 2 − y d y = 0 is

    The solution of e x y 2 − 1 y x y 2 d y + y 3 d x + { y d x − x d y } = 0 is

    Solve x y 2 − e 1 x 3 d x − x 2 y d y = 0

    The curve satisfying the equation d y d x = y x + y 3 x y 3 − x and passing through the point ( 4 , − 2 ) is

    The solution of differential equation d y d x = 1 − x ( y − x ) − x 3 ( y − x ) 3 is

    The solution of the differential equation e x 2 + e y 2 y d y d x + e x 2 x y 2 − x = 0 is

    If x 2 x 2 − 1 d y d x + x x 2 + 1 y = x 2 − 1 then y x − 1 x − ln ⁡ x =

    Integrating factor of x d y d x = 2 y + 2 x 4 + x 2

    Solution of the differential equation 1 + y + x 2 y d x + x + x 3 d y = 0 is

    The solution of d y d x + a y = e m x is (where a + m = 0 )

    The solution of ( 6 x + 7 y − 4 ) d x + ( 7 x − 4 y + 3 ) d y = 0 is

    The solution of the differential equation d y d x = x − 2 y + 1 2 x − 4 y is

    If d y d x + y x = x 2 then 2 y ( 2 ) − y ( 1 ) =

    If d y d x + y sec ⁡ x = tan ⁡ x then ( 2 + 1 ) y π 4 − y ( o ) =

    If x d y = 2 y + 2 x 4 + x 2 d x , y ( 1 ) = 0 then y ( e ) =

    The curve such that at an point of it the square of the ordinate is twice the product of the x-intercept of the normal and the obscissa, is

    The curve satisfy the differential equation 1 − x 2 y 1 + x y = a x are

    If d y d x + x sin ⁡ 2 y = x 3 cos 2 ⁡ y , y ( 0 ) = 0 then tan ⁡ y ( 1 ) =

    The solution of primitive integral equation x 2 + y 2 dy = xydx is y = y ( x ) . If y ( 1 ) = 1 and y x 0 = e then x 0 is

    For the primitive integral equation ydx + y 2 dy = xdy , x ∈ R , y > 0 y = y ( x ) , y ( 1 ) = 1 then y – 3 is

    If y = y ( x ) and 2 + sin ⁡ x y + 1 dy dx = − cos ⁡ x , y ( 0 ) = 1 then y ( π / 2 ) is

    If dy dx = y + 3 > 0 and y ( 0 ) = 2 then y ( log ⁡ 2 ) is

    The order of the differential equation whose general solution is given by y = c 1 + c 2 cos ⁡ x + c 3 − c 4 e x + c 5 where c 1 , c 2 , c 3 , c 4 , c 5 are arbitrary constants is

    The solution of the DE cos ⁡ x d y = y ( sin ⁡ x − y ) d x , 0 < x < π 2 is

    The population p(t) at time ‘t’ of a certain mouse species satisfies differential equation d dt p ( t ) = 0.5 p ( t ) − 450 . If P ( 0 ) = 850 , then the time at which the population becomes zero is

    Let y ( x ) be the solution of the differential equation ( xlogx ) dy dx + y = 2 xlogx , ( x ≥ 1 ) then y (e) is

    If a curve y=f(x) passes through the point (1,-1) and satisfies the differential equation y(1+xy)dx=xdy then f − 1 2 is

    Let y = y ( x ) be the solution of the D.E sin ⁡ x d y d x + y cos ⁡ x = 4 x , x ∈ ( 0 , π ) . If ⇒ y ( π / 2 ) = 0 then y ( π / 6 ) is

    The x-intercept of the tangent to a curve is equal to the coordinate of the point of contact. The equation of the curve through the point (1,1) is

    The normal to a curve at p(x,y) meets the x-axis at G. If the distance of G from the origin is twice the abscisa of P, then the curve is a

    Orthogonal trajectories of family of the curve x 2 3 + y 2 3 = a 2 3 , whose ‘ a ‘ is an arbitrary constant is

    A normal at p ( x , y ) on a curve meets the x -axis at Q and N is the foot of the ordinate at P . If NQ = x 1 + y 2 1 + x 2 , then the equation of the curve given that it passes through the point (3,1) is

    The solution of d y d x = x 2 + y 2 + 1 2 x y satisfying y ( 1 ) = 1 is given by

    Let y = y ( t ) be a solution of the differential equation dy dx = 1 xcosy + sin ⁡ 2 y is x = ce siny − k ( 1 + siny ) then the value of k is

    The solution of the differential equation d y d x − y = 1 − e − x and y ( 0 ) = y , has a finite value when x ∞ then the value of 2 y 0 is

    If x d y d x = x 2 + y − 2 , y ( 1 ) = 1 then y ( 2 ) is

    The solution of the differential equation x 2 dy dx cos ⁡ 1 x − ysin 1 x = − 1 When y − 1 as x ∞ is

    Let yey(t) be a solution to the differential equation y 1 + 2 ty = t 2 then 16 lim t ∞ y is

    If y = y ( x ) and it follows the relation 4 xe xy = y + 5 sin 2 ⁡ x , then y 1 ( 0 ) is

    The solution of the equation d y d x = x ( 2 log ⁡ x + 1 ) siny+ycosy is

    The general solution of the differential equation d y d x + sin ⁡ ( x + y ) 2 = sin ⁡ ( x − y ) 2 is

    A curve passes through ( 2 , 3 ) and satisfying the D.E ∫ 0 x ty ⁡ ( t ) d t = x 2 y ( x ) , x > 0 is

    If y + x dy dx = x φ ( xy ) φ 1 ( xy ) then φ ( xy ) is

    The solution of the D.E yy 1 = x y 2 x 2 + f y 2 x 2 f 1 y 2 x 2 is

    The solution of D.E is x + y d y d x y − x d y d x = x cos 2 ⁡ x 2 + y 2 y 3 is

    The solution of the D . E x x 2 + 1 d y d x = y 1 − x 2 + x 3 log ⁡ x is

    The solution of the D.E [ x coty + log ⁡ cos ⁡ x ] d y + ( logsiny-ytanx ) d x = 0 is

    The general solution of the D.E y 1 + y φ 1 ( x ) − φ ( x ) φ 1 ( x ) = 0 where φ ( x ) is a known function in

    The solution of the equation cos 2 ⁡ x d y d x − ( tan ⁡ 2 x ) y = cos 4 ⁡ x , | x | < π 4 when y ( π / 6 ) = 3 3 8 is

    The solution of the D.E 2 y + x y 3 d x + x + x 2 y 2 d y = 0 is

    The solution of the D.E x + 2 y 3 d y d x = y is

    The solution of the D.E 2 x 2 y dy dx = tan ⁡ x 2 y 2 − 2 xy 2 , given that y ( 1 ) = π 2 is

    The general solution of the differential equation y ( 1 − xy ) dx = x ( 1 + xy ) dy is

    If x 3 d y d x = y 3 + y 2 y 2 − x 2 and y = 1 when x = 1 then y = 2 when x is

    A curve y=f(x) has the property that if the tangent drawn at any point p(x,y) on the curve meets the co-ordinate axes at A and B, then P is the midpoint of AB. If y(1)=1 then y(2)

    The general solution of the differential equation y 5 x 4 y + 1 dx + x + 2 y + 2 x 5 y dy = 0 i s

    A curve passes through (2,0) and slope of the tangent at the point p ( x , y ) i s ( x + 1 ) 2 + y − 3 x + 1 , the area bounds by the curve and the x -axis is

    The general solution of d y d x = 2 y tan ⁡ x + tan 2 ⁡ x is

    Let y = y ( x ) be the solution of the differential equation x 2 + 1 2 d y d x + 2 x x 2 + 1 y = 1 such that y ( 0 ) = 0 . If 8 y ( 1 ) = k π then k is

    Let f : [ 0 , 1 ] R be such that f ( xy ) = f ( x ) f ( y ) ∀ x , y ∈ [ 0 , 1 ] and f ( 0 ) ≠ 0 . If y = y ( x ) satisfies the differential equation dy dx = f ( x ) with y ( 0 ) = 1 then y ( 1 / 4 ) + y ( 3 / 4 ) is

    A function y = f ( x ) satisfies the differential equation f ( x ) sin ⁡ 2 x − cos ⁡ x + 1 + sin 2 ⁡ x f 1 ( x ) = 0 with initial condition y ( 0 ) = 0 then the value of 5 f ( π / 6 ) is

    Let x dy dx − y = x 2 xe x + e x − 1 ∀ x ∈ R − { 0 } such that y ( 1 ) = e − 1 . If y ( 2 ) = ky ( 1 ) ( y ( 1 ) + 2 ) the k is

    The solution of the differential equation x dy dx + 2 y = x 2 , ( x ≠ 0 ) with y ( 1 ) = 1 then y(2) is

    The degree of the differential equation d 2 y dx 2 2 + dy dx 2 = xsin ⁡ d 2 y dx 2 is

    The differential equation of all parabolas whose axes are parallel to the axis of ‘y’ is

    The differential equation all conics whose centre lie at the origin is of order

    The solution of the differential equation x 3 dy dx + 4 x 2 tany = e x secy satisfying y 1 =0 is

    The solution of the differential equation x x 2 + y 2 d y = y x 2 + y 2 − 1 d x is

    The general solution of the differential equation y x 2 y + e x dx − e x dy = 0 is

    The solution of the differential y xy + 2 x 2 y 2 dx + x xy − x 2 y 2 dy = 0 is given by

    The differential equation of all circles which passes through the origin and whose centre lies on y-axis is

    The equation of the curve whose slope is y − 1 x 2 + x and which passes through the point (1,0) is

    The solution of d y d x + 2 y t a n x = sin ⁡ x , given that y = 0 , x = π / 3 is

    The equation of the family of curves which intersect the hyperbola xy=2 orthogonally is

    The differential equation of the ellipse with centre at origin and ends of one axes of symmetry at ± 1 , 0 is

    A curve having the condition that the slope of the tangent at some point is two times the slope of the straight lines joining the same point to the origin of co-ordination is a/an

    The solution of x log ⁡ x d y d x + y = 2 log ⁡ x is

    The solution of d y d x = x − 2 y + 3 2 x − y + 5 is

    The solution of the differential equation 3 xy 1 − 3 y + x 2 − y 2 1 / 2 = 0 satisfying the condition y(1)=1 is

    The solution of the differential equation dy dx = sin ⁡ ( x + y ) tan ⁡ ( x + y ) − 1 is

    The solution of ydx − xdy + 3 x 2 y 2 e x 3 dx = 0 is

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

    The solution of y 3 cos ⁡ x dx − x e 1 / y 2 dy = 0 is

    The slope of a curve at any point on it is the reciprocal of twice the ordinate at that point. If the curve passes through (4,3) then its equation is lo g(y/3)= x 2 -k then k is

    The solution of d y d x + 3 x y = 1 x 2 , given that y = 2 , x = 1 is 2 x 3 y = x 3 + λ then λ is

    The solution of the differential equation ( 2 x − 4 y + 3 ) d y d x + ( x − 2 y + 1 ) = 0 is log ⁡ [ 4 ( x − 2 y ) + 5 ] = λ ( x − 2 y ) + c then λ is

    The solution of cos ⁡ y + ( x siny − 1 ) dy dx = 0 is xsecy = Ptany + c then ‘ P ′ is

    The equation of the curve passing through ( 0 , 1 ) which is a solution of differential equation 1 + y 2 dx + 1 + x 2 dy = 0 is λ Tan − 1 ⁡ x + Tan − 1 ⁡ y = π then λ is

    The general solution of y ‘ + x y = 4 x is given by

    The degree of the differential equation which has a solution y 2 = 4 a x + a 2 where a is arbitrary constant

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