InterviewSolution
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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. |
` (1+ x) (dy)/(dx) - y = e ^(3x ) (1+ x)^2` |
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Answer» Correct Answer - `y = (1)/(3) ( 1 + x ) e ^( 3x ) + C (1+x)` |
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| 52. |
` x (dy)/(dx) + 2y = x ^(2)` |
| Answer» Correct Answer - `y = (x ^(2))/(4) + (C)/(x ^(2))` | |
| 53. |
`y dx - (x + 2y^(2)) dy = 0` |
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Answer» Correct Answer - ` x = 2y ^(2) +Cy` `y dx = (x + 2y ^(2)) dy rArr (dx)/(dy) = (x )/(y) + 2y ` `therefore (dx)/( dy )- (1)/(y) *x = 2y ` |
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| 54. |
The solution of the equation log `((dy)/(dx))=a x+b y`is(a)`( b ) (c) (d)(( e ) (f) e^(( g ) (h) b y (i))( j ))/( k ) b (l) (m)=( n )(( o ) (p) e^(( q ) (r) a x (s))( t ))/( u ) a (v) (w)+c (x)`(y)(b) `( z ) (aa) (bb)(( c c ) (dd) e^(( e e ) (ff)-b y (gg))( h h ))/( i i )(( j j )-b)( k k ) (ll)=( m m )(( n n ) (oo) e^(( p p ) (qq) a x (rr))( s s ))/( t t ) a (uu) (vv)+c (ww)`(xx)(c)`( d ) (e) (f)(( g ) (h) e^(( i ) (j)-b y (k))( l ))/( m ) a (n) (o)=( p )(( q ) (r) e^(( s ) (t) a x (u))( v ))/( w ) b (x) (y)+c (z)`(aa)(d) None of theseA. `(-e^(-by))/( b ) = ( e ^(ax))/( a ) + C `B. `e^(ax) - e ^(-by) = C `C. ` b e ^(ax ) + a e^(by) = C `D. none of these |
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Answer» Correct Answer - A `(dy)/(dx) = e ^( ax + by ) rArr int e ^(- by ) dy = int e ^(ax) dx rArr (e ^(-by))/( - b ) = (e ^(ax))/( a ) +C`. |
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| 55. |
Solve `x(dy)/(dx) +2y = x^2logx` |
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Answer» Correct Answer - `y = (x^(2))/(16) ( 4log x - 1) + (C)/(x^(2))` |
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| 56. |
The general solution of the DE `(dy)/(dx) = (sqrt (1 - x ^(2))) (sqrt (1 - y ^(2)))` isA. ` sin ^(-1) y - sin ^(-1) x = x sqrt(1 - x ^(2)) + C `B. `2sin ^(-1) y - sin ^(-1) x = x sqrt(1 - x ^(2)) + C `C. `2 sin ^(-1) y - sin ^(-1) x = C `D. none of these |
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Answer» Correct Answer - B ` int ( dy ) /( sqrt (1 - y ^(2))) = int sqrt (1 - x^(2)) dx rArr sin ^(-1) y = (x)/(2) sqrt (1 - x ^(2)) + (1)/(2) sin ^(-1) x + C`. ` rArr 2 sin ^(-1) y - sin ^(-1) x = x sqrt ( 1 - x ^(2)) + C`. |
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| 57. |
Solve the following differential equation: `(dy)/(dx)- ytanx=e^xsecx` |
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Answer» Correct Answer - ` y cos x = e ^(x) +C ` |
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| 58. |
The solution of `xsqrt(1+y^2)dx+ysqrt(1+x^2)dy=0`A. ` sin ^(-1) x + sin ^(-1) y = C `B. `sqrt (1 + x ^(2)) + sqrt (1 + y ^(2)) = C `C. ` tan ^(-1)x + tan ^(-1) y= C `D. none of these |
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Answer» Correct Answer - B `int (2x )/( 2* sqrt ( 1+ x ^(2))) dx + int (2y )/( 2*sqrt ( 1 + y ^(2))) =C rArr (1)/(2) int (dt)/(sqrt t) + (1)/(2) int (du)/(sqrtu ) = C`, where ` 1 + x ^(2) = t and 1+ y^(2) =u` ` rArr sqrt t + sqrtu =C rArr sqrt (1+ x^(2)) + sqrt (1 + y ^(2)) =C`. |
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| 59. |
The solution of the DE ` (dy )/(dx) + sqrt ((1 -y ^(2))/( 1- x ^(2))) = 0 ` isA. `y + sin ^(-1) y = sin ^(-1) x + C `B. `sin ^(-1) y - sin ^(-1) x = C `C. ` sin ^(-1) y + sin ^(-1) x =C `D. none of these |
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Answer» Correct Answer - C ` int (1)/(sqrt(1 - y ^(2))) dy + int (1)/(sqrt (1 - x ^(2))) dx = C rArr sin ^(-1) y + sin ^(-1) x = C `. |
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| 60. |
Solve the equation`(x+y+1)((dy)/(dx))=1` |
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Answer» Correct Answer - ` x = Ce^(y) - (y + 2 )` ` (dx)/(dy) - x = ( y +1)` `IF = e^(-int dy ) = e ^(-y)` ` x xx e^(-y) = int (y + 1 ) e ^(-y) dy +C ` `" " = int underset ("I") y underset ("II") e ^(- y) dy + int (e ^(-y) dy +C `. ` " " = - y e ^(-y) + 2 int e ^(-y) dy +C ` ` " " = - y e ^(-y) -2 e ^(-y) +C`. ` therefore x = C e ^(y) - ( y + 2 )`. |
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| 61. |
Find the particular solution of the differential equation `(dy)/(dx)+ycotx=2x+x^2cotx(x!=0)`given that `y = 0`when `x=pi/2`. |
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Answer» Correct Answer - `y = x ^(2) +C cosec x ` `IF = e ^(int cot x dx ) = e ^(log sinx ) = sinx ` `yxx sinx = int (x ^(2) cot x + 2x) sinx dx + C ` ` " "= int x ^(2) cos x dx + int 2 x sinx dx +C` `= x ^(2) sinx - int 2x sin x dx + int 2x sinx dx +C = x ^(2) sin x +C ` |
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| 62. |
Find the particular solution of the differential equation ` (sqrt (1-y^(2)) dx = (sin ^(-1)y - x ) dy `, it being given that when `y=0`, then ` x =0`. |
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Answer» The given differential equation may be written as ` (dx)/(dy) = (sin ^(-1) y- x )/( sqrt (1- y^(2)))` `rArr (dx)/(dy ) + (1 )/(sqrt (1 -y ^(2))) *x = (sin ^(-1) y)/(sqrt(1- y^(2)))" " `... (i) This is of the form ` (dx)/(dy + Px = Q`, where ` P = (1)/(sqrt (1- y^(2))) and Q = (sin ^(-1) y )/( sqrt (1 - y^(2)))` Thus, the given equation is linear. `If = e ^(int Pdy) = e^(int (1)/(sqrt (1+ y^(2)))dy) = e ^(sin ^(-1)y)` So, the solution of the given differential equation is given by `x xx IF = int (Q xx IF )dy +C`, i.e., `x xx e ^(sin ^(-1)y) = int {(sin ^(-1)y)/(sqrt (1-y^(2))) * e ^( (sin ^(-1) y )) } dy + C ` ` " " = int underset ("I") t underset ("II") e ^(t) dt +C, ` where ` sin ^(-1)y =t and (1)/(sqrt (1-y^(2)))dy =dt ` ` " " = t e ^(t) - int 1*e ^(t) dt +C ` ` " " = t e ^(t) - int e ^(t) dt +C = ( te ^(t) - e ^(t)) + C ` ` = ( t- 1 ) e ^(t) +C ` ` = (sin ^(-1)y - 1 ) e ^(sin ^(-1) y )+C " " [because t = sin ^(-1 ) y ]` `therefore x = (sin^(-1) y - 1) + Ce^(-sin ^(-1) y )" "`...(ii) Putting y =0 and x =0 in (ii), we get C = 1 Hence, ` x= (sin ^(-1) y - 1 ) + e ^(-sin^(-1)y) ` is the required solution. |
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| 63. |
Find the particular solution of the differential equation(`tan^(-1)y-x)dy=(1+y^2)dx ,`given thatwhen `x=0, y=0.` |
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Answer» The given differential equation may be written as ` (dx)/(dy) = ((tan ^(-1) y - x ))/( (1 + y^(2))` `rArr (dx)/(dy) = (1)/(( 1 + y ^(2))) *x = (tan ^(-1) y )/( ( 1 + y ^(2)))" "`...(i) This is of the form ` (dx)/(dy) + Px =Q ` ,where `P= (1)/((1+ y ^(2))) and Q = (tan ^(-1) y)/(( 1+y ^(2)))`. Thus, the given differential equation is linear. `IF = e ^(int Pdy) = e ^(int (1)/(( 1 +y^(2))) dy) = e ^( tan ^(-1) y)`. So, the solution of the given differential equation is given by ` x xx IF = int {Qxx IF } dy + C`, where C is an arbitary constant, i.e., `x xx e ^( tan ^(-1) y ) = int {(tan ^(-) y )/( (1 + y ^(2))) xx e ^(tan ^(-1) y ) }dy + C ` `" " int underset("I") t underset("II")e^(t) dt + C`, where ` tan ^(-1) y = t and (1)/(( 1+y ^(2))) dy = dt ` ` = t e ^(t) - int 1 * e ^(t) dt + C " " ` [ integrating by parts ] `" " = t e ^(t)- e ^(t) +C ` ` " "= (t- 1 ) e ^(t) +C ` ` = (tan ^(-1) y- 1) e^(tan^(-1) y) + C`. ` therefore x e ^(tan ^(-1) y ) = (tan ^(-1) y - 1 ) e ^( tan ^(-1) y) +C " "`...(ii) Putting ` x =0 and y =0 `in (ii), we get C = 1. `therefore xe^(tan^(-1)y) = (tan ^(-1) y - 1 ) e ^(tan ^(-1) y) + 1 ` Hence, `x = (tan ^(-1) y -1 ) + e ^(-tan ^(-1) y )` is the required solution. |
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| 64. |
` ( y^(3) - x) (dy)/(dx) = y ` |
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Answer» Correct Answer - ` x = (y^(3))/( 4) + (C )/(y)` `(y ^(3) - x ) (dy)/(dx) = y rArr (dx)/(dy) = ((y ^(3) - x ))/(y)` ` therefore (dx)/(dy) + (1)/(y) * x = y ^(2)` |
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| 65. |
The general solution of the DE ` (1+ x ^(2)) dy - xy dx = 0 ` isA. `y = C (1 + x ^(2)) `B. `y ^(2) = c (1 + x ^(2)) `C. `y sqrt (1 + x ^(2)) = C `D. none of these |
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Answer» Correct Answer - B ` int (1)/(y) dy = (1)/(2) int (2x)/((1+x ^(2))) dx rArr log y = (1)/(2) log (1 + x ^(2)) + log C` `rArr log y = log (C sqrt ( 1+ x^(2))) rArr y = C sqrt (1 + x ^(2))` . |
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| 66. |
The solution of the DE ` (dy)/(dx) = (- 2 xy )/( ( x ^(2) + 1))` isA. `y ^(2) (x + 1) = C `B. ` y (x ^(2) + 1 ) =C `C. `x ^(2) ( y + 1 ) = C `D. none of these |
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Answer» Correct Answer - B `int (1)/(y) dy = int (- 2x )/(( x ^(2) + 1 )) dx rArr log y + log (x ^(2) + 1 ) = log C`. ` therefore y (x^(2) + 1 ) =C`. |
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| 67. |
The solution of the DE ` (dy)/(dx) = 1 - x + y - xy ` isA. `log (1 + y ) = x - (x^(2))/( 2) + C `B. `e^(1+ y ) = x - (x ^(2))/(2) +C `C. ` e^(y)= x - (x^(2))/(2) +C `D. none of these |
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Answer» Correct Answer - A ` int (dy)/(( 1+y )) = int (1 -x ) dx rArr log ( 1 + y ) = x - (x ^(2))/(2) + C`. |
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| 68. |
The solution of the `DE (dy)/(dx) = e ^( x+ y )` isA. ` e ^(x) + e^(y) =C `B. ` e ^(x) - e ^(-y) = C `C. `e ^(x) + e ^(-y) = C `D. none of these |
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Answer» Correct Answer - C The given DE is ` e ^(x)dx = e ^(-y) dy `. `therefore int e ^(x) dx = int e ^(-y) dy rArr e ^(x) = - e^(-y) + C rArr e ^(x) + e ^(-y) = C `. |
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| 69. |
The general solution of the DE ` 2x y dy + (x ^(2) - y ^(2)) dx = 0 ` isA. ` x ^(2) + y ^(2) = C x `B. ` x ^(2) + y ^(2) = Cy`C. ` x ^(2) + y ^(2) = C `D. none of these |
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Answer» Correct Answer - A `(dy)/(dx) = (y ^(2) -x ^(2))/( 2x y ) `, which is homogeneous. Put ` y = vx and (dy)/(dx) = v + x (dv)/(dx)`. Then , `int (2v)/(( 1 + v ^(2))) dv = - int (1)/(x) dx rArr log (1 + v ^(2)) = -log x + log C`. `rArr x (1 + v^(2)) = C rArr x ^(2) + y ^(2) =Cx` |
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| 70. |
The general solution of the DE ` x ^(2) (dy)/(dx) = x ^(2) + xy + y ^(2) ` isA. ` tan ^(-1) ""(y)/(x) = log x + C `B. ` tan ^(-1)""(x)/(y) =log x + C `C. ` tan ^(-1) ""(y)/(x)=log y +C `D. none of these |
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Answer» Correct Answer - A ` (dy)/(dx) = d ( x ^(2) + xy + y ^(2))/( x ^(2))`, which is homogeneous. Put `y = vx and (dy)/(dx) = v + ( dv)/(dx) ` to get ` v + x (dv)/(dx) = (1 + v + v ^(2))` ` therefore int (dv)/((1 + v ^(2))) = int (1)/(x) dx rArr tan ^(-1) v = log x + C rArr tan ^(-1) ""(y)/(x) = log x +C`. |
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| 71. |
Find the general solution of the differential equation `(dy)/(dx)-y=cosx` |
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Answer» The given differential equation is ` (dy)/(dx) -y = cos x " " ` ... (i) This is of the form ` (dy)/(dx) + P y = Q, ` where ` P =- 1 and Q = cos x ` Thus, the given differential equation is linear. `IF= e ^(int Pdx)= e ^( int - dx) = e ^(-x )` So, the required solution is `y xx IF = int { Q xxIF} dx + C, ` i.e., `y xx e ^(-x) = int (cos x ) e ^(-x) dx + C " "` ... (ii) Let `I =int (cos x ) e ^(-x) dx ` ` =( cos x ) (- e ^(-x)) - int (-sin x ) (-e ^(- x )) dx ` [ integrating by parts] `" " = - (cos x ) e ^(- x) - int underset ("I")((sinx )) underset("II")((e^(-x)))dx` ` " " = - (cos x ) e ^(-x) - { (sin x ) (-e^(-x)) } - int (cos x ) (-e^(-x)) dx ` ` = - ( cos x ) e ^(-x) + (sin x ) (e ^(-x)) -I.` `therefore 2 I = (sin x - cos x ) e^(-x) rArr I = (1)/(2) (sinx - cos x ) e ^(-x)`. Putting the value in (ii), we get ` y xx e ^(-x) = (1)/(2) ( sinx - cos x ) e ^(-x) + C rArr y = (1)/(2)(sin x - cos x ) + Ce^(x)` Hence, `y = (1)/(2)( sinx - cos x ) + Ce ^(x) ` is the required solution. |
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| 72. |
Solve the differential equation:(i) `(1+y^2) + ( x - e^(tan^-1 y) ) dy/dx = 0`(ii) ` x dy/dx + cos^2 y = tan y dy/dx` |
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Answer» Correct Answer - ` x e ^(tan^(-1)y) = tan ^(-1) y ` ` (1+ y ^(2)) dx = ( e ^(-tan ^(-1)y ) - x ) dy rArr (dx)/(dy)= ( e^(-tan ^(-1) y) - x )/( ( 1+ y ^(2)))`. ` therefore (dx )/(dy) + (1)/((1 + y ^(2))) * x = ( e ^(-tan ^(-1) y ))/( ( 1 + y ^(2)))` `IF = e ^(int (1)/((1 + y ^(2))) dy) = e ^(tan ^(-1) y ). ` ` therefore x xx e ^(tan^(-1) y ) = int {( e ^(-tan ^(-1) y ))/( ( 1+y ^(2))) xx e ^(tan ^(-1) y )} dy = C = int (1)/(( 1+y ^(2))) dy +C ` ` rArr x xx e^(tan ^(-1)y ) = tan ^(-1) y +C `. Putting ` x = 0 and y = 0` in (i), we get C =0. `" " ` ... (i) |
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| 73. |
The general solution of the DE ` ( y - x ) dy + (x + y ) dx = 0 ` isA. ` tan ^(-1)""(y)/(x) = C sqrt (x ^(2) + y ^(2))`B. ` e ^(tan ^(-1) ( y//x)) = C sqrt (x ^(2) + y ^(2))`C. `tan ^(-1) ((y)/(x))= x ^(2) + y ^(2) + C `D. none of these |
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Answer» Correct Answer - B ` (dy)/(dx) = ( x+ y )/( x - y ) `, which is clearly homogeneous. Put ` y= vx and (dy)/(dx) = v + x ( dv )/(dx) rArr int ((1 - v ))/(( 1+ v ^(2))) dv = int (1)/(x) dx ` `rArr int (dv)/((1 + v ^(2))) - (1)/(2) int (2v)/((1+ v^(2))) dv = int (1)/(x) dx ` `rArr tan ^(-1) v - (1)/(2) log (1 + v ^(2)) = log x + log C rArr tan ^(-1) v = log (Cx sqrt (1 + v ^(2)))` ` rArr tan ^(-1) "" ( y )/(x) = log (C sqrt ( x ^(2) + y ^(2))) rArr e ^( tan ^(-1) y//x) = C sqrt (x ^(2) + y ^(2))`. |
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| 74. |
The general solution of the DE `(dy)/(dx) + y cot x = 2 cos x ` isA. `(y + sinx ) sin x = C `B. `(y + cos x ) sin x = C `C. `(y- sinx ) sinx = C `D. none of these |
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Answer» Correct Answer - C Given DE is linear. `IF = e ^(int cot x dx ) = e ^( log sin x ) = sinx `. ` therefore y (sinx ) = int 2 sin x cos x dx rArr y sinx = sin ^(2) x +C ` `therefore ( y - sinx ) sinx = C `. |
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| 75. |
The general solution of the DE ` (dy)/(dx) + (y)/(x) = x ^(2) ` isA. `xy = x ^(4) + C `B. `4xy = x ^(4) + C `C. ` 3xy = x ^(3) + C `D. none of these |
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Answer» Correct Answer - B Given DE is linear. `IF = e ( int (1)/(x) dx ) = e (log x ) = x ` ` therefore y xx x = int ( x xx x ^(2)) dx = int x ^(3) dx = (x ^(4))/( 4) + C ` ` rArr 4xy = x ^(4) +C`. |
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| 76. |
` xdy + (y - x ^(3)) dx =0` |
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Answer» Correct Answer - `y = (x^(3))/( 4) + (C )/(x)` ` (dy)/(dx) = (x ^(3) - y)/(x )rArr (dy)/(dx) + (1)/(x) *y = x ^(2)` |
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| 77. |
The general solution of the DE ` (dy)/(dx) + y tanx = sec x ` isA. ` y = sin x -C cos x `B. `y = sin x + C cos x `C. `y = cos x - C sin x `D. none of these |
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Answer» Correct Answer - B Given DE is linear `IF = e ^( int tan x dx ) = e ^( log (sec x)) = sec x ` `therefore ` its solution is `y (sec x ) = int sec ^(2) x dx rArr y ( secx ) = tan x + C rArr y = sin x + C cos x `. |
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| 78. |
Solve the differential equation ` (dy)/(dx) + y tan x = 2x + x ^(2)tanx ` |
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Answer» The given differential equation is of the form `(dy)/(dx) + Py = Q`, where` P = tan x and Q = 2 x + x ^(2) tanx.` Thus, the givebn differential equation is linear. `IF = e ^( int Pdx) = e ^(int tan x dx ) = e ^( log secx ) = secx` So, its solution is given by `y xx IF = int (Q xx IF )dx +C`, i.e, `y xx secx = int (2x + x ^(2) tanx ) sec x dx + C ` ` " "= int 2 x secx dx + int underset ("I") x^(2) underset("II") ( secx tanx ) dx +C ` ` " " = int 2x secx dx + {x ^(2) secx - int 2 x secx dx} + C ` ` " " = x ^(2) sec x + C ` ` therefore y = x ^(2) +C cos x is the required solution. |
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| 79. |
Solve: `(dy)/(dx)+y=cos x=sinx` |
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Answer» Correct Answer - ` y = cos x + Ce^(-x)` `IF = e ^(int dx ) = e ^(x)` ` y xx e ^(x) = int e ^(x) (cos x - sin x ) dx +C ` `= int (underset ("I")((cos x ) underset ( "II") (e ^(x)) dx - int (sinx ) e ^(x) dx + C ` ` = (cos x ) e ^(x) - int ( - sinx ) e ^(x) dx - int (sinx ) e ^(x) dx + C ` ` = (cos x )e ^(x) +C `. |
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| 80. |
` (dy)/(dx ) + 2y =sinx` |
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Answer» Correct Answer - `y = (1)/(5) (2 sin x - cos x ) + C en ^(-2x)` `IF = e ^(int 2 dx ) = e^( 2x )` `therefore y xx e ^( 2x ) = int e ^( 2x ) sin x dx + C ` ` " " = e ^( 2x ) { ( 2 sin x - cos x )/(( 2^(2) + 1 ^(2)))} + C [ int e ^(ax) sin bx dx = e^(ax){(asin bx + b cos bx)/((a ^(2) + b^(2)))} ]` ` rArr y = ( (2 sin x - cos x ))/( 5) + Ce^(-2x)` |
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| 81. |
` (1 + x^(2)) (dy)/(dx) + 2xy = 4x ^(2)`, given that `y = 0`, when `x =0` |
| Answer» Correct Answer - `y = ( 4x^(3))/( 3 ( 1+ x ^(2)))` | |
| 82. |
` secx (dy)/(dx) -y = sin x ` |
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Answer» Correct Answer - `y = Ce^(sin x ) - (1+ sin x )` ` (dy)/(dx ) - (cos x )y= sinx cos x ` `IF = e ^(-int cos xdx ) = e ^(-sin x )` ` therefory y xx e ^(-sinx ) = int (sin x cos x ) e ^(-sinx ) dx +C ` ` " " =int underset ("I")t underset ("II") e^(-t) dt +C `, where ` sin x = t `. |
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| 83. |
`(dy)/(dx) + y tan x = 2x + x ^(2) tan x `, given that ` y = 1 ` when ` x=0` |
| Answer» Correct Answer - `y = x ^(2) +cos x ` | |
| 84. |
` x (dy)/(dx) + y = x ^(3)`, given that ` y = 1` when ` x = 2 ` |
| Answer» Correct Answer - ` y = (x^(3))/(4) - (2)/(x)` | |
| 85. |
` x (dy)/(dx) - y = 2x ^(2) secx ` |
| Answer» Correct Answer - ` y = Cx + 2x log | secx + tan x |` | |
| 86. |
`x(dy)/(dx)+y=3x^2-2, x gt0` |
| Answer» Correct Answer - `y = x ^(2) - 2 + (C)/(x)` | |