## Lakshya Education MCQs

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

Options:
 A. xn+n2y =constant B. ny2+x2 =constant C. n2x+yn=constant D. n2x−yn =constant
: 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
: A

x+y=z1+dydx=dzdxdydx+1=ex+ydydx=ezdz=dxezdz=dxez=x+ce(x+y)+x+c=0.
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)
: D

x(1x2)dy+(2x2yyax3)dx=0dydx+(2x21)x(1x2)y=ax2(1x2),P=2x21x(1x2).
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
: A

dxdy+11+y2x=11+y2etan1y
I.F=e11+y2dy=etan1y
Solution is x.etan1y
=etan1y.11+y2etan1dy=12e2tan1y+12k2xetan1y=e2tan1y+k
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)
: C

The point on y-axis is (0,yxdydx)
According to given condition,
x2=yx2dydxdydx=2yx1
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