Lakshya Education MCQs

Question: If xdydx=y(log ylog x+1), then the solution of the equation is
Options:
A.y log(xy)=cx
B.x log(yx)=cy
C.log(yx)=cx
D.log(xy)=cx
Answer: Option C
: C

dydx=yx(logyx+1)
Put y=vxdydx=v+xdvdx
v+xdvdx=vlogv+v1vlogvdv=1xdx1vlogvdv=1xdxlog(logv)=logx+logc
logyx=cx

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More Questions on This Topic :

Question 1. The degree and order of the differential equation of the family of all parabolas whose axis is x–axis, are respectively
  1.    1,2
  2.    3,2
  3.    2,3
  4.    2,1
Answer: Option A
: A

Equation of family of parabolas with x-axis as axis is y2=4a(x+α) where a,α are two arbitrary constants. So differential equation is of order 2 and degree 1.
Question 2. The solution of dydx=yx+tanyx is
  1.    y sin(yx)=cx
  2.    y sin(yx)=cy
  3.    sin(xy)=cx
  4.    sin(yx)=cx
Answer: Option D
: D

Put y=vx.Thendydx=v+xdvdx
Given equation is dydx=vx+tanyxv+xtanvcotvdv=dxxlogsinv=logx+logc
sinv=cxsin(yx=cx)
Question 3. If y1(x) is a solution of the differential equation dydxf(x)y=0, then a solution of the differential equation dydx+f(x)y=r(x)
  1.    1y1(x)∫r(x)y1(x)dx
  2.    y1(x)∫r(x)y1(x)dx
  3.    ∫r(x)y1(x)dx
  4.    None of these
Answer: Option A
: A

dydxf(x).y=0dyy=f(x)dx
ln y=f(x)dx
y1(x)=ef(x)dxThen for given equation I.F = ef(x)dx
Hence Solution y.y1(x)=r(x).y1(x)dx
y=1y1(x)r(x).y1(x)dx
Question 4. The solution of the differential equation (x2sin3yy2cosx)dx+(x3cosysin2y2ysinx)dy=0 is 
  1.    x3sin3y=3y2sinx+C
  2.    x3sin3y+3y2sinx=C  
  3.    x2sin3y+y3sinx=C  
  4.    2x2siny+y2sinx=C
Answer: Option A
: A

(x2sin3yy2cosx)dx+(x3cosysin2y2ysinx)dy=0dydx=y2cosxx2sin3yx3cosysin22ysinx(x3cosysin2y2ysinx)dy=(y2cosxx2sin3y)dx=0(x33dsin3ysindy2)sin3yd(x33)+y2dsinx=0
x33dsin2y+sin3yd(x33)(sindy2+y2dsinx)
d(x33sin3y)d(y2sinx)=0x33sin3yy2sinx=c
Question 5. The order of the differential equation whose general solution is given by y=C1e2x+C2+C3ex+C4sin(x+C5) is 
  1.    5
  2.    4
  3.    3
  4.    2
Answer: Option B
: B

y=C1e2x+C2+C3ex+C4sin(x+C5)=C1.eC2e2x+C3ex+C4(sinxcosC5+cosxsinC5)=Ae2x+C3ex+Bsinx+Dcosx
Here, A=C1eC2,B=C4cosC5,D=C4sinC5
(Since equation consists of four arbitrary constants)
order of differential equation = 4.
Question 6. The orthogonal trajectories of the family of curves an1y=xn are given by
  1.    xn+n2y= constant
  2.    ny2+x2= constant
  3.    n2x+yn= constant
  4.    n2x−yn= constant
Answer: Option B
: B

Differentiating, we have an1dydx=nxn1an1=nxn1dxdy
Putting this value in the given equation, we havenxn1dxdyy=xn
Replacing dydx by dxdy we have ny=xdxdy
nydy+xdx=0ny2+x2=constant. Which is the required family of orthogonal trajectories.

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