# Free Objective Test 01 Practice Test - 11th and 12th

The general solution of the differential equation dydx=y tan xy2sec x is

A. tan x = (c + sec x)y
B. sec y = (c + tan y )x
C. sec x = (c + tan x)y
D. None of these

#### SOLUTION

Solution : C

We have dydx=y tan xy2sec x1y2dydx1ytanx=secx
Putting 1y=v1y2dydx=dvdx, we obtain
dvdx+tan x.v=secxwhich is linear
I.F=etanxdx=elogsecx=secx
The solution is
v secx=sec2xdx+c1ysecx=tanx+c
secx=y(c+tanx)

If integrating factor of x(1x2)dy+(2x2yyax3)dx=0 is ePdx, then P is equal to

A. 2x2ax3x(1x2)

B. (2x21)

C. 2x21ax3

D. (2x21)x(1x2)

#### SOLUTION

Solution : D

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

Solution of the equation xdy=(y+xf(yx)f(yx))dx

A. f(xy)=cy
B. f(yx)=cx
C. f(yx)=cxy
D. f(yx)=0

#### SOLUTION

Solution : B

We have, xdy=(y+xf(yx)f(yx))dx
dydx=yx+f(yx)f(yx) which is homogenous
Putting y=vxdydx=v+xdvdx, we obtain
v+xdvdx=v+f(v)f(v)f(v)f(v)dv=dxx
Integrating, we get log f(v)=logx+logc
logf(v)=logcxf(yx)=cx

The solution of the differential equation (1+y2)+(xetan1y)dydx=0, is

A. 2x etan1y,=e2tan1y+k
B. x etan1y,=etan1y+k
C. x e2tan1y,=etan1y+k
D. (x2)k etan1y

#### SOLUTION

Solution : A

dxdy+11+y2x=11+y2etan1y
I.F=e11+y2dy=etan1y
Solution is x.etan1y
=etan1y.11+y2etan1dy=12e2 tan1y+12k2x etan1y=e2 tan1y+k

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 have nxn1dxdyy=xn
Replacing dydxby dydy, we have ny=xdxdy
nydy+xdx=0ny2+x2 =constant. Which is the required family of orthogonal trajectories.

The solution of (y(1+x1)+siny)dx+(x+logx+x cosy)dy=0 is

A. (1+y1siny)+x1logx=C

B. (y+siny)+xy log x=C

C. xy+ylogx+x sin y=C
D. None of these

#### SOLUTION

Solution : C

The given equation can be written as y(1+x1)dx+(x+logx)dy+sin ydx+xcos ydy=0
d(y(x+logx))+d(xsiny)=0y(x+logx)+xsiny=C

The solution of y27y1+12y=0 is

A. y=C1e3x+C2e4x

B. y=C1xe3x+C2e4x

C. y=C1e3x+C2xe4x
D. None of these

#### SOLUTION

Solution : A

The given equation can be written as (ddx3)(dydx4y)=0....(i)
If dydx4y=u then (I) reduces to dudx3u=0
duu=3dxu=C1e3x. Therefore, we have dydx4y=C1e3x which is a linear equation whose I.F.is e4x. So ddx(ye4x)=C1ex
ye4x=C1ex+C2y=C1e3x+C2e4x

The solution lof dydxx tan(yx)=1

A. c eex22=sin(yx)
B. c eex22=sin(y+x)
C. c eex22=sin(yx)2
D. c eex22=cos(yx)

#### SOLUTION

Solution : A

Put y-x = z. Then dydx1=dzdxdydx=1+dzdx
Given dydxx tan(yx)=11+dzdxx tan z=1dzdx=x tan z1tanzdz=x dxcotz dz=x dx
log|sin z|logc=x22log|sin zc|=x22=sinzc=ex22sin(yx)=c ex22
The solution is sin(yx)=cex22, where c is arbitrary constant.

Solution of the differential equation : dydx=3x2y4+2xyx22x3y3 is

A. x2y2+x2y=c

B. x3y2+x2y=c

C. x3y2+y2x=c

D. x2y3+x2y=c

#### SOLUTION

Solution : B

x2dy2x3y3dy=3x2y4dx+2xydxx2dy2xydx=3x2y4dx+2x3y3dy2xydxx2dyy2+3x2y2dx+2x3ydy=0d(x2y)+d(x3y2)=0x2y+x3y2=C

The degree of the differential equation satisfying the relation 1+x2+1+y2=λ(x1+y2y1+x2) is

A. 1
B. 2
C. 3
D. none of these

#### SOLUTION

Solution : A

On Putting x=tanA,y=tanB we get
secA+secB=λ(tanA secBtanB secA)cosA+cosB=λ(sinAsinB)tan(AB2)=1λtan1xtan1y=2tan11λ
On differentiating 11+x211+y2dydx=0

The order of the differential equation of all tangent lines to the parabola y=x2 is

A. 1
B. 2
C. 3
D. 4

#### SOLUTION

Solution : A

The parametric form of the given equation is  x=t,y=t2.The equation of any tangent at t is 2xt=y+t2, Differentiating we get 2t=y1(=dydx) putting this value in the above equation, we have 2xy12=y+(y12)24xy1=4y+y21
The order of this equation is 1
Hence (A) is the correct answer

A function y =f(x) has a second order derivative f"=6(x-1). If its graph passes through the point(2,1) and at that point the tangent to the graph is y =3x -5, then the function is

A. (x1)2

B. (x1)3

C. (x+1)2

D. (x+1)3

#### SOLUTION

Solution : B

Since f"(x)=6(x-1)
f(x)=3(x1)2+c (integrating)        ----(i)
Also, at the point (2,1), the tangent to the graph is y =3x-5and slope of thetangent = 3
f(2)=3
3(21)3+c=3 [from eq(i)
3+c=3c=0
From Eq (i) we have
f(x)=3(x1)2
f(x)=(x1)3+k (Integrating)----(ii)
1=(21)3+kk=0
Hence the equation of the function is f(x)=(x1)3.

The order and degree of the differential equation [4+(dydx)2]2/3=d2ydx2 are

A.

2,2

B.

3,3

C.

2,3

D.

3,2

#### SOLUTION

Solution : C

Here power on the differential coefficient is fractional, therefore change it into positive integer, so
[4+(dydx)2]2/3=d2ydx2[4+(dydx)2]2=[d2ydx2]3
Hence order is 2 and degree is 3.

The order and degree of the differential equation y=xdydx+a2(dydx)2+b2 are

A.

1,2

B.

2,1

C.

1,1

D.

2,2

#### SOLUTION

Solution : A

Given differential equation can be written as
y2=x2(dydx)22xy.dydx=a2(dydx)2+b2.
Hence it is of 1st order and 2nddegree differential equation.

The general solution of  y2 dx+(x2xy+y2)dy=0 is

[EAMCET 2003]

A.

tan1(xy)+log y+c=0

B.

2 tan1(xy)+log x+c=0

C.

log(y+x2+y2)+log y+c=0

D.

sin h1(xy)+log y+c=0

#### SOLUTION

Solution : A

dxdy+x2xy+y2y2=0dxdy+(xy)2(xy)+1=0
Put v=x/yx=vydxdy=v+ydvdyv+ydvdy+v2v+1=0dvv2+1+dyy=0dvv2+1+dyy=0tan1(v)+log y+C=0tan1(x/y)+log y+c=0.