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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 51. |
The solution of the differential equation `(x^2y^2-1)dy+2xy^3dx=0`is(a)`( b ) (c)1+( d ) x^(( e )2( f ))( g ) (h) y^(( i )2( j ))( k )=c x (l)`(m) (b) `( n ) (o)1+( p ) x^(( q )2( r ))( s ) (t) y^(( u )2( v ))( w )=c y (x)`(y)(c)`( d ) (e) y=0( f )`(g)(d) `( h ) (i) y=-( j )1/( k )(( l ) (m) x^(( n )2( o ))( p ))( q ) (r) (s)`(t)A. `1+x^(2)y^(2)=cx`B. `1+x^(2)y^(2)=cy`C. `y=0`D. `y=-1/x^(2)` |
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Answer» Correct Answer - B `(x^(2)y^(2)-1)dy+2xy^(3)dx=0` or `x^(2)y^(2)dy+2xy^(3)dx=dy` or `x^(2)dy+2xydx=(dy)/y^(2)` or `intd(x^(2)y)=int(dy)/(y^(2))+c` or `x^(2)y=y^(-1)/-1+c` or `x^(2)y^(2)=-1+cy` i.e., `1+x^(2)y^(2)=cy` |
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| 52. |
The solution of the differential equation `(x+2y^(2))(dy)/(dx)=y`, isA. `x=y^(2)+C`B. `y=x^(2)+C`C. `x=y(y^(2)+C)`D. `y=x(x^(2)+C)` |
| Answer» Correct Answer - C | |
| 53. |
The solution of `(dy)/(dx)-y=1, y(0)=1` is given by y(x)=A. `-exp(x)`B. `-exp(-x)`C. `-1`D. `2exp(x)-1` |
| Answer» Correct Answer - D | |
| 54. |
The general solution of the differential equation `(dy)/(dx)+sin(x+y)/2=sin(x-y)/2`is(a) `( b ) (c)logtan(( d ) (e) (f) y/( g )2( h ) (i) (j))=c-2sinx (k)`(l)(m) `( n ) (o)logtan(( p ) (q) (r) y/( s )4( t ) (u) (v))=c-2sin(( w ) (x) (y) x/( z )2( a a ) (bb) (cc))( d d )`(ee)(ff) `( g g ) (hh)logtan(( i i ) (jj) (kk) y/( l l )2( m m ) (nn)+( o o )pi/( p p )4( q q ) (rr) (ss))=c-2sinx (tt)`(uu)(vv)`( w w ) (xx)logtan(( y y ) (zz) (aaa) y/( b b b )4( c c c ) (ddd)+( e e e )pi/( f f f )4( g g g ) (hhh) (iii))=c-2sin(( j j j ) (kkk) (lll) x/( m m m )2( n n n ) (ooo) (ppp))( q q q )`(rrr)A. `log tan(y/2)=c-2sinx`B. `log tan(y/2)=c-2sin(x/2)`C. `log tan(y/2+pi/4)=c-2sinx`D. `log tan(y/4+pi/4)=c-2sin(x/2)` |
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Answer» Correct Answer - B We have `(dy)/(dx) = sin(x-y)/2-sin(x+y)/2` `=-2cosx/2siny/2` or `1/2" cosec "y/2dy=-cosx/2dx` or `logtany/4-(sinx/2)/(1/2)+c` or `logtan(y/4)=c-2sinx/2` |
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| 55. |
The solution of the differential equation`(dy)/(dx)=(3x^2y^4+2x y)/(x^2-2x^3y^3)`is(a)`( b ) (c) (d)(( e ) (f) y^(( g )2( h ))( i ))/( j ) x (k) (l)-( m ) x^(( n )3( o ))( p ) (q) y^(( r )2( s ))( t )=c (u)`(v)(w) `( x ) (y) (z)(( a a ) (bb) x^(( c c )2( d d ))( e e ))/( f f )(( g g ) (hh) y^(( i i )2( j j ))( k k ))( l l ) (mm)+( n n ) x^(( o o )3( p p ))( q q ) (rr) y^(( s s )3( t t ))( u u )=c (vv)`(ww)(xx)`( y y ) (zz) (aaa)(( b b b ) (ccc) x^(( d d d )2( e e e ))( f f f ))/( g g g ) y (hhh) (iii)+( j j j ) x^(( k k k )3( l l l ))( m m m ) (nnn) y^(( o o o )2( p p p ))( q q q )=c (rrr)`(sss)(d) `( t t t ) (uuu) (vvv)(( w w w ) (xxx) x^(( y y y )2( z z z ))( a a a a ))/( b b b b )(( c c c c )3y)( d d d d ) (eeee)-2( f f f f ) x^(( g g g g )3( h h h h ))( i i i i ) (jjjj) y^(( k k k k )2( l l l l ))( m m m m )=c (nnnn)`(oooo)A. `y^(2)/x-x^(3)y^(2)=c`B. `x^(2)/y^(2)+x^(3)y^(3)=c`C. `x^(2)/y+x^(3)y^(2)=c`D. `x^(2)/(3y)-2x^(3)y^(2)=c` |
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Answer» Correct Answer - C Rewrite the differential equation as `(2xydx-x^(2)dy)+y^(2)(3x^(2)y^(2)dx+2x^(3)ydy)=0` Dividing by `y^(3)`, we get `(y^(2)xdx-x^(2)dy)/(y^(2))=y^(2)3x^(2)dx+x^(3)2ydy=0` or `d(x^(2)/y)+d(x^(3)y^(3))=0` Integrating, we get the solution `x^(2)/y+x^(3)y^(2)=c` |
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| 56. |
The general solution of the differential equation `(dy)/(dx)+sin(x+y)/2=sin(x-y)/2`is(a) `( b ) (c)logtan(( d ) (e) (f) y/( g )2( h ) (i) (j))=c-2sinx (k)`(l)(m) `( n ) (o)logtan(( p ) (q) (r) y/( s )4( t ) (u) (v))=c-2sin(( w ) (x) (y) x/( z )2( a a ) (bb) (cc))( d d )`(ee)(ff) `( g g ) (hh)logtan(( i i ) (jj) (kk) y/( l l )2( m m ) (nn)+( o o )pi/( p p )4( q q ) (rr) (ss))=c-2sinx (tt)`(uu)(vv)`( w w ) (xx)logtan(( y y ) (zz) (aaa) y/( b b b )4( c c c ) (ddd)+( e e e )pi/( f f f )4( g g g ) (hhh) (iii))=c-2sin(( j j j ) (kkk) (lll) x/( m m m )2( n n n ) (ooo) (ppp))( q q q )`(rrr)A. `log tan((y)/(2))=C-2sinx`B. `log tan((y)/(4))=C-2sin((x)/(2))`C. `log tan ((y)/(2)+(pi)/(4))=C-2 sinx`D. `log tan ((y)/(2)+(pi)/(4))=C-2 sin (x)/(2)` |
| Answer» Correct Answer - B | |
| 57. |
The solution of differential equation `(2y+x y^3)dx+(x+x^2y^2)dy=0`is(a)`( b ) (c) (d) x^(( e )2( f ))( g ) y+( h )(( i ) (j) x^(( k )3( l ))( m ) (n) y^(( o )3( p ))( q ))/( r )3( s ) (t)=c (u)`(v)(b) `( w ) (x) x (y) y^(( z )2( a a ))( b b )+( c c )(( d d ) (ee) x^(( f f )3( g g ))( h h ) (ii) y^(( j j )3( k k ))( l l ))/( m m )3( n n ) (oo)=c (pp)`(qq)(c)`( d ) (e) (f) x^(( g )2( h ))( i ) y+( j )(( k ) (l) x^(( m )4( n ))( o ) (p) y^(( q )4( r ))( s ))/( t )4( u ) (v)=c (w)`(x) (d) None of theseA. `x^(2)y+(x^(3)y^(3))/3=c`B. `xy^(2)+(x^(3)y^(3))/3=c`C. `x^(2)y+(x^(4)y^(4))/4=c`D. None of these |
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Answer» Correct Answer - A `(2y+xy^(3))dx+(x+x^(2)y^(2))dy=0` or `(2ydx+xdy)+(xy^(3)dx+x^(3)y^(2)dy)=0` or `d(x^(2)y)+1/3d(x^(3)y^(3))=0` Integrating, we get `x^(2)y+(x^(3)y^(3))/3+c` |
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| 58. |
The solution of the differential equation`(dy)/(dx)=1/(x y[x^2siny^2+1])`is(a) `( b ) (c) (d) x^(( e )2( f ))( g )(cos( h ) y^(( i )2( j ))( k )-sin( l ) y^(( m )2( n ))( o )-2C (p) e^( q ) (r)-( s ) y^((( t )2( u ))( v ) (w))( x ))=2( y )`(z)(aa)`( b b ) (cc) (dd) y^(( e e )2( f f ))( g g )(cos( h h ) y^(( i i )2( j j ))( k k )-sin( l l ) y^(( m m )2( n n ))( o o )-2C (pp) e^( q q ) (rr)-( s s ) y^((( t t )2( u u ))( v v ) (ww))( x x ))=2( y y )`(zz)(aaa)`( b b b ) (ccc) (ddd) x^(( e e e )2( f f f ))( g g g )(cos( h h h ) y^(( i i i )2( j j j ))( k k k )-sin( l l l ) y^(( m m m )2( n n n ))( o o o )-( p p p ) e^( q q q ) (rrr)-( s s s ) y^((( t t t )2( u u u ))( v v v ) (www))( x x x ))=4C (yyy)`(zzz)(aaaa)None of these |
| Answer» `- (1)/(x^(2)) = (1)/(2)(sin y^(2) - cos y^(2)) + ke^(-y^(2))` | |
| 59. |
The slope of a curve, passing through (3,4) at any point is the reciprocal of twice the ordinate of that point. Show that it is parabola. |
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Answer» It is given that `(dy)/(dx) = 1/(2y)` `2ydy=dx` Integrating, we get `y^(2)=x+c` Now when `x=3, y=4` which gives `c=13` Hence the equation of the required curve is `y^(2)=x+13`, which is a parabola. |
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| 60. |
A family of curves is such that the slope of normal at any point (x, y) is 2(1-y). The area bounded by the curve y = f(x) of question number 1 and the line x+2y = 0 isA. `(10)/(3)` sq. unitsB. `(4)/(3)` sq. unitsC. `(28)/(3)` sq. unitsD. `(16)/(3)` sq. units |
| Answer» Correct Answer - B | |
| 61. |
The slope of a curve at any point is the reciprocal of twice the ordinate at that point and it passes through the point(4,3). The equation of the curve is:A. `x^(2)=y+5`B. `y^(2)=x-5`C. `y^(2)=x+5`D. `x^(2)=y-5` |
| Answer» Correct Answer - C | |
| 62. |
The equation of the curve through the point `(1,0)`, whose slope is `(y-1)/(x^2+x)`A. `(y-1)(x+1)+2x=0`B. `2x(y-1)+x+1=0`C. `x(y-1)(x+1)+2=0`D. none of these |
| Answer» Correct Answer - A | |
| 63. |
The equation of a curve passing through `(2,7/2)`and havinggradient `1-1/(x^2)`at `(x , y)`is(a)`( b ) (c) y=( d ) x^(( e )2( f ))( g )+x+1( h )`(i)(b) `( j ) (k) x y=( l ) x^(( m )2( n ))( o )+x+1( p )`(q)(c)`( d ) (e) x y=x+1( f )`(g) (d) None of theseA. `y=x^(2)+x+1`B. `xy=x^(2)+x+1`C. `xy=x+1`D. none of these |
| Answer» Correct Answer - B | |
| 64. |
Find the equation of a curve passing through `(0,1)`and having gradient `(1(y+y^3))/(1+x+x y)a t(x , y)` |
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Answer» Correct Answer - `xy+tan^(-1)y=pi/4` We have `(dy)/(dx)=(-y+y^(3))/(1+x(1+y^(2))` or `(dx)/(dy) = -(1+x(1+y^(3)))/(1+x(1+y^(2))` `=-1/(y(y^(2)+1))+x(1+y^(2))/(y(1+y^(2))]` `therefore (dx)/(dy) + x/y=-1/(y(1+y^(2))` I.F. `=e^(int(dy)/y)=e^(logy)=y`. Hence, the solution is `x.y=-int1/(y(1+y^(2)).ydy+C` `=-int(dy)/(1+y^(2))+C=-tan^(-1)y+C` or `xy+tan^(-1)y=C`. This passes through (0,1). Therefore, `tan^(-1)1=C, i.e., C=pi/4`. Thus, the equation of the curve is `xy+tan^(-1)y=pi/4` |
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| 65. |
`[ x tan(y/x) - y sec^2(y/x) ] dx + x sec^2 (y/x)] dy= 0` |
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Answer» Let `y = vx` Then, `dy/dx = v+x(dv)/dx` Our given equation becomes, `[xtanv-vxsec^2 v]dx + xsec^2v dy = 0` `=>-dy/dx = [xtanv-vxsec^2 v]/[xsec^2v]` `=>-v-x(dv)/dx = tanv/(sec^2v) - v` `=>- dx/x = (sec^2v dv)/tanv` Now, integrating both sides, `=> int - dx/x = int (sec^2v dv)/tanv` Let `tanv = u => sec^2vdv = du` `=> int - dx/x = int (du)/u` `=>-ln(x)+c = ln(u)` `=>c = ln(ux) ` `=>ln(xtanv) = c` `=>ln(xtan(y/x)) = c` |
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| 66. |
Find the co-ordinates of the point on curve `y=(x^2-1)/(x^2+1), (x>0)` where the gradient of the tangent to the curve is maximum. |
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Answer» Here, equaton of the curve, `y = (x^2-1)/(x^2+1)` `:. dy/dx = ((x^2+1)(2x)-(x^2-1)(2x))/(x^2+1)^2` `=>dy/dx = (4x)/(x^2+1)^2` We have to find a point, where slope is maximum. `:. (d^2y)/dx^2` should be `0`. `=>(d^2y)/dx^2 = ((x^2+1)^2 4-4x(2(x^2+1))(2x) )/(x^2+1)^4 ` `=>((x^2+1)^2 4-4x(2(x^2+1))(2x) )/(x^2+1)^4 =0` `=>(x^2+1)^2(4-16x^2) = 0` As `x^2+1` can not be `0`, `:. 4 - 16x^2 = 0` `=> x = +-1/2` We will take `x = 1/2` as `x gt 0`. `:. y = (x^2-1)/(x^2+1) = (1/4-1)/(1/4+1) = 3/5` So, the point at which slope is maximum is `(1/2,3/5)`. |
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| 67. |
Find the degree of the following : `(1+(ds)/(dt))^((3)/(2)) = 5(d^(2)s)/(dt^(2))` |
| Answer» Correct Answer - 2 | |
| 68. |
`(dy)/(dx)=3x-2y+5,` where `3x-2y+5=u` A)`6x-4y+10=ce^(2x)` B)`-6x+4y-7=c.e^(-2x)` C)`logu=2x+c` D)`logu=2u+c`A. `6x-4y+10=ce^(2x)`B. `-6x+4y-7=c.e^(-2x)`C. `logu=2x+c`D. `logu=2u+c` |
| Answer» Correct Answer - B | |
| 69. |
Show that `y^(2)dx + (xy + x^(2))dy = 0` is a homogeneous differential equation. Also find its general solution. |
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Answer» `(dy)/(dx) = -(y^(2))/(xy + x^(2))` Let `f(x,y) = -(y^(2))/(xy + x^(2))` Now `f(lambda x, lambda y) = -(lambda^(2) y^(2))/(lambda x.lambda y + lambda^(2) x^(2))` `= - lambda^(0)(y^(2))/(xy + x^(2))` `= lambda^(0)f(x, y)` `:.` The given differential equation is the homogeneous differential equation. Putting y = vx `rArr (dy)/(dx) = v + x(dv)/(dx)`, we get `v + x(dv)/(dx) = - (v^(2)x^(2))/(x.vx + x^(2))` `rArr x(dv)/(dx) = (-v^(2) - v^(2) - v)/(v+1)` `rArr (v+1)/(v(1+2v)) dv = -(dx)/(x)` Integrating both sides, we get `int ((1)/(v)-(1)/(1+2v))dv = -int (dx)/(x)` `rArr ln |v| - (1)/(2)ln|(1+2v)| = -ln|x| + lnc` `rArr (|vx|)/(sqrt(1+2v)) = c` `rArr xy^(2) = c^(2)(x+2y)` or `xy^(2) = k(x+2y)` |
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| 70. |
Solution of the differential equation sin x. cos y dy + cos x. sin y dx = 0 isA. sin x + sin y = CB. cos x + cos y = CC. sin x . Sin y = CD. `(sin x)/(sin y) = C` |
| Answer» Correct Answer - C | |
| 71. |
Solve the differential equation : `(dy)/(dx)=(x^(2)-y^(2))/(xy)`. |
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Answer» `(dy)/(dx)=(x^(2)-y^(2))/(xy)`……….`(1)` It is a homogenous differential equation. Let `y=vx` `implies (dy)/(dx)=v+x(dv)/(dx)` Put these values in eq. `(1)` `v+x(dv)/(dx)=(x^(2)-v^(2)x^(2))/(x^(2)v)=(1-v^(2))/(v)` `implies x(dv)/(dx)=(1-v^(2))/(v)-v=(1-2v^(2))/(v)` `implies (v)/(1-2v^(2))dv=(dx)/(x)` `implies int(v)/(1-2v^(2))dv=int(dx)/(x)` Let `1-2v^(2)=t` `implies int(dt)/(-4t)=int(dx)/(x)impliesvdv=(dt)/(-4)` `implies -(1)/(4)logt+logc=logx` `implies logc=logx+logt^(1//4)` `implies c=x*t^(1//4)` `implies c^(4)=x^(4)*t=x^(4)*(1-2v^(2))` `=x^(4)(1-(2y^(2))/(x^(2)))` `implies c_(1)=x^(2)(x^(2)-2y^(2))`. |
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| 72. |
Solve the following differential equation `x^2(dy)/(dx)-xy=1+cos(y/x), x!=0` |
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Answer» `x^2dy/dx - xy = 1+cos(y/x)` `=>dy/dx -y/x = 1/x^2(1+cos(y/x))` Let `y = vx`, then `dy/dx = v+x(dv)/dx` Now, our equation becomes, `v+x(dv)/dx -v = 1/x^2(1+cosv)` `=>(dv)/(1+cosv) = 1/x^3dx` `=>1/2sec^2(v/2) dv = dx/x^3` Integrating both sides, `=>1/2 int sec^2(v/2) dv = int dx/x^3` `=>1/2(2)tan(v/2) = -1/2 x^-2+c` `=>tan(v/2) = -1/(2x^2)+c` `=>tan(y/(2x)) = -1/(2x^2)+c` |
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| 73. |
show that the given differential equation is homogeneous and solve each of them `(x^2+xy)dy=(x^2+y^2)dx` |
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Answer» `(dy)/(dx)=(x^(2)+y^(2))/(x^(2)+xy)` `implies v+x(dv)/(dx)=(x^(2)+x^(2)v^(2))/(x^(2)+x^(2)v)` Let `y=vx` `implies (dv)/(dx)=v+x(dv)/(dx)` `implies x(dv)/(dx)=(1+v^(2))/(1+v)-v=(1-v)/(1+v)` `implies (1+v)/(1-v)dv=(dx)/(x)` `implies int((2)/(1-v)-1)dv=int(dx)/(x)+logc` `implies -2log(1-v)-v=logx+logc` `implies -v=logcx+2log(1-v)=log{cx(1-v)^(2)}` `implies cx(1-v)^(2)=e^(-v)` `implies cx(1-(y)/(x))^(2)=e^(-y//x)` `implies c(x-y)^(2)=x*e^(-y//x)` `implies c(x-y)^(2)*e^(y//x)=x` |
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| 74. |
Solution of differential equation `(dy)/(dx) = (2)/(x+y)` isA. `x+y+2 = ke^(y//2)`B. `x-y + 2 = ke^(y//2)`C. `x + y +2 = ke^(-y//2)`D. `x-y+2 = ke^(-y//2)` |
| Answer» Correct Answer - A | |
| 75. |
Solution of the differential equation `(dx)/(x)+(dy)/(y)=0` isA. `(1)/(x)+(1)/(y)=C`B. `logxlogy=C`C. `xy=C`D. `x+y=C` |
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Answer» Correct Answer - C We have, `(1)/(x)dx+(1)/(y)dy=0` On integrating, we get `logx+logy =logCrArr xy =C`. |
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| 76. |
Find the differential equation of system of cocentric circles with centre (1,2) |
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Answer» The family of concentric circles with centre (1, 2) and radius a is given by `" "(x-1)^(2)+(y-2)^(2)=a^(2)` `rArr" "x^(2)+1-2x+y^(2)+4-4y=a^(2)` `rArr" "x^(2)+y^(2)-2x-4y+5=a^(2)` On differentiating Eq. (i) w.r.t. x, we get `" "2x+2y(dy)/(dx)-2-4(dy)/(dx)=0` `rArr" "(2y-4)(dy)/(dx)+2x-2=0` `rArr" "(y-2)(dy)/(dx)+(x-1)=0` |
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| 77. |
If `y+d/(dx)(xy)=x(sinx+logx)`, find `y(x)`. |
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Answer» The given differential equation is `y+d/(dx)(xy) = x(sinx+logx)` i.e., `x(dy)/(dx)+2/xy=sinx+logx`.............(1) This is a linear different equation I.F.`=e^(2int1/xdx)= e^(2logx)=x^(2)`............(2) Thus, solution is given by `yx^(2)=intx^(2)(sinx+logx)dx+c` `=-x^(2)cosx+2xsinx+2cosx+x^(3)/(3)logx-x^(3)/9+c` or `y=-cosx+2/xsinx+2/x^(2)cosx+x/3logx-x/3+c/x^(2)` |
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| 78. |
Solve: `dy/dx = sin(x+y) + cos(x+y)` |
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Answer» Given, `" "(dy)/(dx)=cos(x+y)+sin(x+y)" "`...(i) Put `" "x+y=z` `rArr" "1+(dy)/(dx)=(dz)/(dx)` On substituting these values in Eq. (i), we get `" "((dz)/(dx)-1)=cosz+sinz` `rArr" "(dz)/(dx)=(cosz+sinz+1)` `rArr" "(dz)/(cosz+sinz+1)=dx` On integrating both sides, we get `" "int(dz)/(cosz+sinz+1)=int1dx` `rArr" "int(dz)/((1-tan^(2)z//2)/(1+tan^(2)z//2)+(2tanz//2)/(1+tan^(2)z//2)+1)=intdx` `rArr" "int(dz)/((1-tan^(2)z//2+2tanz//2+1+tan^(2)z//2)/((1+tan^(2)z//2)))=intdx` `rArr" "int((1+tan^(2)z//2)dz)/(2+2tan^(2)z//2)=intdx` `rArr" "int(sec^(2)z//2dz)/(2(1+tanz//2))=intdx` Put `1+tanz//2 =t` `rArr" "((1)/(2)sec^(2)z//2)dz=dt` `rArr" "int(dt)/(t)=intdx` `rArr" "lot|t|=x+C` `rArr" "log|1+tanz//2|=x+C` `rArr" "log|1+tan""((x+y))/(2)|=x+C` |
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| 79. |
If `(2+sinx)(dy)/(dx)+(y+1)cosx=0` and `y(0)=1`, then `y((pi)/(2))` is equal toA. `+(1)/(3)`B. `(4)/(3)`C. `(1)/(2)`D. `-(2)/(3)` |
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Answer» Correct Answer - C The given differential equation is `(2+sinx)(dy)/(dx)+(y+1)cosx=0` `rArr" "(2+sinx)dy(y+1)cosx dx =0` `rArr" "(1)/(y+1)dy+(cosx)/(2+sinx)dx=0` Integrating both sides, we get `log(y+1)+log(2+sinx)=logC` `rArr" "(y+1)(2+sinx)=C" ...(i)"` It is given that `y(0)=1` i.e. y = 1 when x = 0. Putting x = 0, y = 1 in (i), we get `2(2+0)=CrArr C=4` Putting C = 4 in (i), we obtain `(y+1)(2+sinx)=4" ...(ii)"` Putting `x=(pi)/(2)` in (ii), we get `y=(1)/(3)`. |
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| 80. |
Solve the differential equation `(dy)/(dx)=1+x+y^(2)+xy^(2)`, when y=0 and x=0. |
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Answer» Given that, `(dy)/(dx)=1+x+y^(2)+xy^(2)` `Rightarrow (dy)/(dx)=(1+x)+y^(2)(1+x)` `Rightarrow (dy)/(dx)=(1+y^(2))(1+x)` `Rightarrow (dy)/(1+y^(2))=(1+x)(dx)` On integrating both sides, we get `tan^(-1)y=x+(x^(2))/(2)+k..(i)` When y=0 and x=0, then substituting these values in Eq. (i) we get `tan^(-1)(0)=0+0+K` `Rightarrow K=0` `Rightarrow tan^(-1) y=x+(x^(2))/(2)` `Rightarrow y=tan(x+(x^(2))/(2))` |
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| 81. |
The differential equation `(dy)/(dx)=(sqrt(1-y^2))/y`determinesa family of circle with(a) variable radii and a fixed centre at (0, 1)(b)variable radii and a fixed centre at `( c ) (d)(( e ) (f)0,-1( g ))( h )`(i)(j) Fixed radius 1 and variable centres along thex-axis.(k) Fixed radius 1 and variable centres along they-axis.A. Variable radii and a fixed centre at (0, 1)B. Variable radii and a fixed centre at (0, -1)C. Fixed radius 1 and variable centre along the x-axisD. Fixed radius 1 and variable centre along the y-axis |
| Answer» Correct Answer - C | |
| 82. |
if `y=y(x)`and `(2+sinx)/(y+1)((dy)/(dx))=-cosx ,y(0)=1,`then `y(pi/2)=`(a)`( b ) (c) (d)1/( e )3( f ) (g) (h)`(i)(b) `( j ) (k) (l)2/( m )3( n ) (o) (p)`(q)(c) `( r ) (s)-( t )1/( u )3( v ) (w) (x)`(y) (d)1A. (a) 1/3B. (b) 2/3C. (c) -1/3D. (d) 1 |
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Answer» Correct Answer - (a) Given, `dy/dx=(-cosx(y+1))/(2+sin x)` `rArr dy/(y+1)=(-cosx)/(2+sinx)dx` On integrating both sides `int dy/(y+1)=-int (cosx)/(2+sinx)dx` `rArr log (y+1)=-log(2+sin x) +log e` When `x =0, y=1 rArr c = 4` `rArr y+1=4/(2+sin x` `therefore y(pi/2)=4/3-1` `rArr y(pi/2)=1/3` |
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| 83. |
The differental equation `y(dy)/(dx)+x=C` representsA. a family of hyperbolaB. a family of circles whose centres are on y-axisC. a family of circle of parabolasD. a family of circles whose centres are on x-axis. |
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Answer» Correct Answer - D We have, `y(dy)/(dx)+x=C` `rArr" "ydy+(x-C)dx=0` `rArr" "(y^(2))/(2)+((x-C)^(2))/(2)=C_(1) rArr (x-C)^(2)+y^(2)=2C_(1)` Clearly, it represents a family of circles having their centres on x-axis. |
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| 84. |
The differential equation `(dy)/(dx)=(sqrt(1-y^2))/y`determinesa family of circle with(a) variable radii and a fixed centre at (0, 1)(b)variable radii and a fixed centre at `( c ) (d)(( e ) (f)0,-1( g ))( h )`(i)(j) Fixed radius 1 and variable centres along thex-axis.(k) Fixed radius 1 and variable centres along they-axis.A. variable radii and fixed centre at (0, 1)B. variable radii and a fixed centre at `(0, -1)`C. fixed radius 1 and variable centre along the x-axisD. fixed radius 1 and variable centre along the y-axis |
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Answer» Correct Answer - C The given differential equation is |
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| 85. |
If `y+d/(dx)(x y)=x(sinx+logx),fin dy(x)dot` |
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Answer» Given differential equation is `" "y+(d)/(dx)(xy)=x(sinx+logx)` `rArr" "y+x(dy)/(dx)+y=x(sinx+logx)` `rArr " "x(dy)/(dx)+2y=x(sinx+logx)` `rArr" "(dy)/(dx)+(2)/(x)y=sinx+logx` which is a linear differential equation. On comparing it with `" "(dy)/(dx)+Py=Q,` we get `" "P=(2)/(x),Q=sinx+logx` `" "IF=e^(int(2)/(x)dx)=e^(2logx)=x^(2)` The general solution is `" "y*x^(2)=int(sinx+logx)x^(2)dx+C` `rArr" "y*x^(2)=int(x^(2)sinx+x^(2)logx)dx+C` `rArr" "y*x^(2)=intx^(2)sinxdx+int x^(2)logxdx+C` `rArr" "y*x^(2)=I_(1)+I_(2)+C" "`...(i) Now, `" "I_(1)=intx^(2)sinxdx` `" "=x^(2)(-cosx)+int2xcosxdx` `" "=-x^(2)cosx+[2x(sinx)-int2sinxdx]` `" "I_(1)=-x^(2)cosx+2xsinx+2cosx" "`...(ii) and `" "I_(2)=intx^(2)logxdx` `" "=logx*(x^(3))/(3)-int(1)/(x)*(x^(3))/(3)dx` `" "=logx*(x^(3))/(3)-(1)/(3)intx^(2)dx` `" "=logx*(x^(3))/(3)-(1)/(3)*(x^(3))/(3)" "`...(iii) On substituting the value of `I_(1) and I_(2)` in Eq. (i), we get `" "y*x^()=-x^(2)cosx+2xsinx+2cosx+(x^(3))/(3)logx-(1)/(9)x^(3) +C` `therefore " "y=-cosx+(2sinx)/(x)+(2cosx)/(x)+(x)/(3)logx-(x)/(9)+Cx^(-2)` |
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| 86. |
Find the general solution of `(x+2y^(3))(dy)/(dx)=y` |
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Answer» Given that, `(x+2y^(3))(dy)/(dx)=y` `y(dy)/(dx)=x+2y^(3)` `Rightarrow (dy)/(dx)=(x)/(y)+2y^(2)["dividing throughout by y"]` Which linear different equation. On comparing it with `(dx)/(dy)+Px=0` , we get `P=-(1)/(y),Q=2y^(2)` `IF=e^(int-(1)/(y)dy)=e^(-1(1)/(y)dy)` `=e^(-logy)=(1)/(y)` The genergal solution is `x.(1)/(y)=int2y^(2).(1)/(y)dy+C` `Rightarrow (x)/(y)=(2y^(2))/(2)+C` `Rightarrow (x)/(y)=y^(2)+C` `Rightarrow x=y^(3)+Cy` |
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| 87. |
if `y=y(x)`and `(2+sinx)/(y+1)((dy)/(dx))=-cosx ,y(0)=1,`then `y(pi/2)=`(a)`( b ) (c) (d)1/( e )3( f ) (g) (h)`(i)(b) `( j ) (k) (l)2/( m )3( n ) (o) (p)`(q)(c) `( r ) (s)-( t )1/( u )3( v ) (w) (x)`(y) (d)1A. `(1)/(3)`B. `(2)/(3)`C. `-(1)/(3)`D. 1 |
| Answer» Correct Answer - A | |
| 88. |
If `(2+sinx)(dy)/(dx)+(y+1)cosx=0` and `y(0)=1`, then `y((pi)/(2))` is equal toA. `4/3`B. `1/3`C. `-2/3`D. `-1/3` |
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Answer» Correct Answer - B We have `(2+sinx)(dy)/(dx) +(y+1)cosx=0` `rArr d/(dx)[(2sinx)(y+1)]=0` `rArr (2+sinx)(y+1)=c` Since `y(0)=1, c=4` So, `y+1=4/(2+sinx)` `therefore y(pi/2)=4/3-1=1/3` |
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| 89. |
If `y(x)`is a solution of thedifferential equation `((2+sinx)/(1+y))(dy)/(dx)=-cosx`and `y(0)=1`, then find the value of `y(pi/2)dot` |
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Answer» Given that, `((2+sin x)/(1+y))(dy)/(dx)=-cos x` `Rightarrow (dy)/(1+y)=-(cosx)/(2+sinx)dx` On integrating both sides, we get `int(1)/(1+y)dy=-int (cos x)/(2+sinx)dx` `Rightarrowlog(1+y)=-log(2+sinx)+logC` `Rightarrow log(1+y)+log(2+sinx)=logC` `Rightarrow log(1+y)(2+sinx)=logC` `Rightarrow (1+y)(2+sinx)=C` `Rightarrow 1+y=(C)/(2+sinx)` When, x=0 and y=1, then `1=(C)/(2+sin x)-1` `Rightarrow C=4` On putting C=4 in Eq (i) we get `y=(4)/(2+sinx)-1` `y((pi)/(2))=(4)/(2+"sin"(pi)/(2))-1=(4)/(2+1)-1` `(4)/(3)-1=(1)/(3)` |
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| 90. |
If `y=y(x)` and `((2+sinx)/(y+1))dy/dx=-cosx, y(0)=1`, then `y(pi/2)` equals (A) `1/3` (B) `2/3` (C) `-1/3` (D) `1`A. `(1)/(3)`B. `(2)/(3)`C. `-(1)/(3)`D. 1 |
| Answer» Correct Answer - A | |
| 91. |
Solution of differential equation `x^(2)y - x^(3) (dy)/(dx)=y^(4) cos x` isA. `x^(2)y^(-3) = 2 sin x + c`B. `x^(2)y^(-3) = 3 cos x + c`C. `x^(3)y^(-3) = 3 sin x + c`D. `x^(2)y^(3) = 3 sin x + c x^(2) y ` |
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Answer» Correct Answer - C `x^(2) y - x^(3)(dy)/(dx)=y^(4)cos x` `(1)/(y^(4))(dy)/(dx)+((-1)/(x))(1)/(y^(3))=(cos x)/(x^(3))` put `(1)/(y^(3))=v` `(1)/(y^(4))(dy)/(dx)=-(1)/(3)(dv)/(dx)` `(-1)/(3)(dv)/(dx)+(-(-1)/(x))v=-(cos x)/(x^(3))` `(dv)/(dx)+(3)/(x)v=3 cos x` `I.F.=e^(3 int (1/2dx)=x^(3)` solution `v*x^(3)=3int cos x dx + c` `(1)/(y^(3))x^(3)=3 sin x + c` `x6(3)y^(-3)=3 sin x + c` |
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| 92. |
Solve the following differential equation: `tany(dy)/(dx)=sin(x+y)+sin(-y)`A. sec y + 2 cos x = cB. sec y - 2 cos x = cC. cos y - 2 sin x = cD. tan y - 2 sec x = c |
| Answer» Correct Answer - A | |
| 93. |
`(dy)/(dx)=2(sinx-cosx)^(-2)`A. `y-(2)/(1-tanx)=c`B. `y=cosx+sinx+c`C. `y(1-tanx)=c`D. `y-2(1-cotx)=c` |
| Answer» Correct Answer - A | |
| 94. |
Find the general solution of the differential equation `(1+tany)(dx-dy)+2x dy=0`A. `x(sin y + cos y)=sin y + ce^(y)`B. `x(sin y + cos y) = sin y + ce^(-y)`C. `y(sin x + cos x)=sin x + ce^(x)`D. None of these |
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Answer» Correct Answer - B We have, `(1 + tan y)(dx - dy) + 2x dy = 0` `rArr" "(1+tan y)dx =(1+tan y - 2x)dy` `rArr" "(dx)/(dy)+(2)/(1+tan y)x=1`, which is linear differential equation. I.F.`=e^(2int(dy)/(1+tan y))=e^(int(2 cos y)/(sin y + cos y)dy)` `=e^(int(1+(cos y-sin y)/(sin y + cos y))dy)=e^(y+log(cos y + sin y))` `=(cos y + sin y)e^(y)` So, the solution is `Xe^(y)(sin y + cos y) = int e^(y) (sin y + cos y) dy + c` `i.e. xe^(y)(sin y + cos y)=e^(y) sin y + c`. `i.e. x(sin y + cos y)=sin y + ce^(-y)` |
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| 95. |
`(dy)/(dx)tany=sinx+cosx`A. `sinx+log(secy)=cosx+c`B. `cosx-log(cosy)=sinx+c`C. `cosx+log(cosy)=sinx+c`D. `cosy+log(cosx)=siny+c` |
| Answer» Correct Answer - B | |
| 96. |
General solution of `(tany)dy=dx` isA. `e^(x)siny=c`B. `e^(x)cosy=c`C. `e^(x)sec y=c`D. `e^(x)cosy=c` |
| Answer» Correct Answer - B | |
| 97. |
General solution of `coty cos^(2)xdy+cotxcos^(2)ydx=0` isA. `tanx+tany=c`B. `tanx-ctany`C. `tanxtany=c`D. `tanx-tany=c` |
| Answer» Correct Answer - C | |
| 98. |
A country has a food deficit of `10%`. Its population grows continuously at a rate of `3%` per year. Its annual food production every year is 4% more than that of the last year. Assuming that the average food requirement per person remains constant, prove that the country will become self-sufficient in food after `n` years, where `n` is the smallest integer bigger than or equal to `(ln10-ln9)/(ln(1.04)-0.03)` |
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Answer» Let `P_(0)` be the initial population of country and P be the population of country in year t. Then, `(dP)/(dt)`= rate of change of population =`3/100P=0.03P` `therefore` Population of P at the end of n years is given by `int_(P_(0))^(P) (dP)/(P)= int_(0)^(P)0.03dt` or `inP-"ln"P_(0)=(0.03)n` or `"ln"P="ln"P_(0)+(0.03)n`.............(1) If `F_(0)` is its initial food production and F is the food production in year n, then `F_(0) = 0.9P_(0)` and `F=(1.04)^(n)F_(0)` or `InF=n"ln"(1.04)+"ln"F_(0)`...............(2) The country will be self-sufficient if `FgeP` or `"ln"Fge"ln"P` or `n "ln"(1.04)+"ln"F_(0)ge"ln"P_(0)+(0.03)n` or `nge("ln"P_(0)-"ln"F_(0))/("ln"(1.04)-(0.3))=("ln"10-"ln"9)/("ln"(1.04)-0.03)` Hence, `nge("ln"10-"ln"9)/("ln"(1.04)-0.03)` Thus,the least integral values of the year n, when the country becomes self-suffcient, is the smallest integer greater than or equal to `("ln"10-"ln"9)/("ln"(1.04)-0.03)` |
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| 99. |
The general solution of the differential equation `(ydx-xdy)/(y)=0` is :A. `xy=C`B. `x=Cy^(2)`C. `y=Cx`D. `y=Cx^(2)` |
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Answer» Given differential equation is `(ydx-xdy)/(y)=0` `implies ydx-xdy=0 implies (1)/(x)dx-(1)/(y)dy=0` On integration, `log|x|-log|y|=logk` `implies log|(x)/(y)|=logkimplies(x)/(y)=k` ` y=(1)/k)ximpliesy=Cx` (Let `(1)/(k)=C`) |
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| 100. |
find the particular solution satisfying the given condition, for the following differential equation: `(x+1)(dy)/(dx)=2e^-y-1` given that `y=0` when `x=0` |
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Answer» Given differential equation is `(x+1)(dy)/(dx)=2e^(-y)-1`……`(1)` `implies (dy)/(2e^(-y)-1)=(dx)/(x+1)implies(e^(y)dy)/(2-e^(y))=(dx)/(x+1)` On integration, `int(e^(y)dy)/(2-e^(y))=int(dx)/(x+1)`……`(2)` Let `2-e^(y)=timplies-e^(y)=(dt)/(dy)impliese^(y)dy=-dt` From equation `(2)`, `int-(dt)/(t)=int(dx)/(x+1)` `implies -log|t|=log|x+1|+logC` `implies -log|2-e^(y)|=log|C(x+1)|` `implies (1)/(2-e^(y))=C(x+1)` `implies 2-e^(y)=(1)/(C(x+1))` Now, at `x=0` and `y=0`, `2-1=(1)/(C)impliesC=1` Put the value of `C` in equation `(3)`, `2-e^(y)=(1)/((x+1))impliese^(y)=2-(1)/(x+1)` `impliese^(y)=(2x+2-1)/(x+1)impliese^(y)=(2x+1)/(x+1)` `implies y=log|(2x+1)/(x+1)|` |
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