Lakshya Education MCQs

Question: The orthogonal trajectories of the family of curves an1y=xn are given by

A.xn+n2y =constant  
B.ny2+x2 =constant  
D.n2x−yn =constant 
Answer: Option B
: B

Differentiating, we have an1dydx=nxn1an1=nxn1dxdy
Putting this value in the given equation, we have nxn1dxdyy=xn
Replacing dydxby dydy, we have ny=xdxdy
nydy+xdx=0ny2+x2 =constant. Which is the required family of orthogonal trajectories.

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More Questions on This Topic :

Question 1. The solution of dydx+1=ex+y is
  1.    e−(x+y)+x+c=0
  2.    e−(x+y)−x+c=0
  3.    ex+y+x+c=0
  4.    ex+y−x+c=0
Answer: Option A
: A

Question 2. If integrating factor of x(1x2)dy+(2x2yyax3)dx=0 is ePdx, then P is equal to
  1.    2x2−ax3x(1−x2)  
  2.    (2x2−1)  
  3.    2x2−1ax3  
  4.    (2x2−1)x(1−x2)
Answer: Option D
: D

Question 3. The solution of the differential equation (1+y2)+(xetan1y)dydx=0, is
  1.    2x etan−1y,=e2tan−1y+k
  2.    x etan−1y,=etan−1y+k
  3.    x e2tan−1y,=e−tan−1y+k
  4.    (x−2)k etan−1y
Answer: Option A
: A

Solution is x.etan1y
Question 4. A curve is such that the mid point of the portion of the tangent intercepted between the point where the tangent is drawn and the point where the tangent meets y-axis, lies on the line y = x. If the curve passes through (1, 0), then the curve is
  1.    2y=x2−x
  2.    y=x2−x
  3.    y=x−x2
  4.    y=2(x−x2)  
Answer: Option C
: C

The point on y-axis is (0,yxdydx)
According to given condition,
Putting yx=v we get
xdvdx=v1lnyx1=ln|x|+c1yx=x (as f(1)=0).
Question 5. Solution to the differential equation x+x33!+x55!+.....1+x22!+x44!+.....=dxdydx+dy is 
  1.    2ye2x=C.e2x+1  
  2.    2ye2x=C.e2x−1  
  3.    ye2x=C.e2x+2  
  4.    2xe2y=C.ex−1
Answer: Option B
: B

Applying C and D, we get
or 2ye2x=C.e2x1.

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