Free Objective Test 02 Practice Test - 11th and 12th

The integrating factor of the differential equation dydx=y tan xy2 sec x is

[MP PET 1995; Pb. CET 2002]

A.

tan x

B.

secx

C.

-sec x

D.

cot x

SOLUTION

Solution : B

The differential equation
is dydxy tan x=y2 sec xI.F.=e tan x dx
This is Bernoulli's equation i.e. reducible to
linear equation.
Dividing the equation by y2, we get
1y2dydx1y tan x=sec x............(i)
Put 1y=y1y2dydx=dYdx
Equation (i) reduces to dydx=y tan x=sec xdYdx+Y tan x=sec x, Which is a linear equation
Hence I.F.=e tan x dx=sec x.

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

A. xn+n2y= constant
B. ny2+x2= constant
C. n2x+yn= constant
D. n2xyn= constant

SOLUTION

Solution : 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.

The order of the differential equation whose general solution is given by y=C1e2x+C2+C3ex+C4sin(x+C5) is

A. 5
B. 4
C. 3
D. 2

SOLUTION

Solution : B

y=C1e2x+C2+C3ex+C4sin(x+C5)=C1.eC2e2x+C3ex+C4(sin x cos C5+cos x sin C5)=Ae2x+C3ex+B sin x+D cos x
Here, A=C1eC2,B=C4cosC5,D=C4sinC5
(Since equation consists of four arbitrary constants)
order of differential equation = 4.

The degree and order of the differential equation of the family of all parabolas whose axis is x–axis, are respectively

A. 1,2
B. 3,2
C. 2,3
D. 2,1

SOLUTION

Solution : 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.

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)

A. 1y1(x)r(x)y1(x)dx
B. y1(x)r(x)y1(x)dx
C. r(x)y1(x)dx
D. None of these

SOLUTION

Solution : A

dydxf(x).y=0dyy=f(x)dx
ln y=f(x)dx
y1(x)=ef(x)dx Then 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

If xdydx=y(log ylog x+1), then the solution of the equation is

A. y log(xy)=cx
B. x log(yx)=cy
C. log(yx)=cx
D. log(xy)=cx

SOLUTION

Solution : C

dydx=yx(logyx+1)
Put y=vxdydx=v+xdvdx
v+xdvdx=v log v+v1vlogvdv=1xdx1v log vdv=1xdxlog(log v)=logx+log c
logyx=cx

The solution of dydx+xy=x2 is

A. 1y=cxxlog x
B. 1x=cyylog y
C. 1x=cxxlog y
D. 1y=cxylog x

SOLUTION

Solution : B

dxdy+xy=x2x2dxdy+x1y=1
Put x1=t.Thenx2dxdy=dtdyx2dxdy=dtdydtdy+ty=1dtdy1yt=1
It is linear in t. Here P=1y.Q=1
I.F=e(1y)dy=elogy=1y
Solution is t(1y)=(1)1ydy=logy+ct=y logy+cyx1cyy logy.

The solution of the differential equation (x2sin3yy2cosx)dx+(x3cosysin2y2ysinx)dy=0 is

A. x3sin3y=3y2sinx+C
B. x3sin3y+3y2sinx=C

C. x2sin3y+y3sinx=C

D. 2x2siny+y2sinx=C

SOLUTION

Solution : A

(x2sin3yy2cos x)dx+(x3 cos y sin2 y2y sin x)dy=0dydx=y2cosxx2sin3yx3 cos y sin22y sinx(x3 cos y sin2y2y sin x)dy=(y2 cos xx2 sin3y)dx=0(x33dsin3 ysin dy2)sin3yd(x33)+y2d sin x=0
x33d sin2y+sin3yd(x33)(sin dy2+y2 d sin x)
d(x33sin3y)d(y2sinx)=0x33sin3yy2sinx=c

The solution of dydx=yx+tanyx is

A. y sin(yx)=cx
B. y sin(yx)=cy
C. sin(xy)=cx
D. sin(yx)=cx

SOLUTION

Solution : D

Put y=vx.Thendydx=v+xdvdx
Given equation is dydx=vx+tanyxv+x tan vcot v dv=dxxlog sin v=log x+logc
sin v=cxsin(yx=cx)

The curves satisfying the differential equation (1x2)y1+xy=ax are

A. Ellipse and hyperbola
B. Ellipse and parabola
C. Ellipse and straight line
D. Circle and parabola

SOLUTION

Solution : A

The given equation is linear DE  and can be written as
dydx+x1x2y=ax1x2
Its integrating factor is ex1x2dx=e(12)ln(1x2)=11x2  [1<x<1] and if x2>1 then I.F.=1x21
ddx(y11x2)=ax(1x2)32=122ax(1x2)32
y11x2=a1x2+Cy=a+C1x2
(ya)2=C2(1x2)(ya)2+C2x2=C2
Thus, if 1<x<1 the given equation represents an ellipse.
If x2>1 then the solution is of the form (ya)2+C2x2=C2 which represents a hyperbola.

The solution of dydx+1=ex+y is

A. e(x+y)+x+c=0
B. e(x+y)x+c=0
C. ex+y+x+c=0
D. ex+yx+c=0

SOLUTION

Solution : A

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

Solution to the differential equation x+x33!+x55!+.....1+x22!+x44!+.....=dxdydx+dy is

A. 2ye2x=C.e2x+1

B. 2ye2x=C.e2x1

C. ye2x=C.e2x+2

D. 2xe2y=C.ex1

SOLUTION

Solution : B

Applying C and D, we get
dydx=exex=e2x2y=e2x+C
or 2ye2x=C.e2x1.

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

A. 2y=x2x
B. y=x2x
C. y=xx2
D. y=2(xx2)

SOLUTION

Solution : 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).

The family of curves passing through (0,0) and satisfying the differential equation y′′y=1 (where y=dydx and  y′′=d2ydx2) is

A. y=k
B. y=kx
C. y=k(ex+1)
D. y=k(ex1)

SOLUTION

Solution : D

dpdx=p(where p=dydx)
lnp=x+cp=ex+c
dydx=kexy=kex+λ
Satisfying (0,0), So λ=k
y=k(ex1)

S1: The differential equation of parabolas having their vertices at the origin and foci on the x-axis is an equation whose variables are separable
S2: The differential equation of the straight lines which are at a fixed distance p from the origin is an equation of degree 2
S3: The differential equation of all conics whose both axes coincide with the axes of coordinates is an equation of order 2

A. TTT
B. TFT
C. FFT
D. TTF

SOLUTION

Solution : A

S1 -Equation of parabola is y2=±4ax
2ydydx=±ra
D.E of parabola y2=2yxdydx
2dyy=dxx
Which is variable seperable
S2 -Equation of  line which is fixed distance. P from origin can be equation of tangent to circle
s2+y2=p2
Line is y=mx+p1+m2(m=dydx)
(yxdydx)2=P(1+(dydx)2)
So, degree is 2
S3 -Equation of conic whose both axis co-incide with co-ordinate axis is  ax2+by2=1
As there are two constants, so order of D.E is 2