InterviewSolution
Saved Bookmarks
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.
| 501. |
If `(2+sinx)(dy)/(dx)+(y+1)cosx=0`and `y(0)=1,t h e ny(pi/2)`is equal to`-1/3`(2) `4/3`(3) `1/3`(4) `-2/3`A. (a) 1/3B. (b) -2/3C. (c) -1/3D. (d) 4/3 |
|
Answer» Correct Answer - (a) We have, `(2+sinx)dy/dx+(y+1)cos x =0` `rArr dy/dx+(cosx)/(2+sinx)y=(-cosx)/(2+sinx)` which is a linear differential equation. `therefore IF=e^(int(cosx)/(2+sinx)dx)=e^(log(2+sinx))=2 +sin x` `therefore` Required solution is given by `y cdot (2+sinx)=int(-cosx)/(2+sin x)cdot (2+sinx)dx+C` `rArr y(2+sinx) =-sinx+C` Also, `y(0)=1` `therefore 1(2+sin0)=-sin0+C` `rArr C =2` `therefore y=(2-sinx)/(2+sinx) rArr y(pi/2)=(2-sin frac{pi}{2})/(2+sin frac {pi}{2})=1/3` |
|
| 502. |
Integrating factor of the differential equation `(x+1)(dy)/(dx)-y=e^(3x)(x+1)^(2)`, isA. `-(x+1)`B. `log(x+1)`C. `e^(x+1)`D. `(1)/(x+1)` |
|
Answer» Correct Answer - D We have, `(dy)/(dx)-(1)/(x+1)y=e^(3x)(x+1)` `therefore" I.F."=e^(-int(1)/(x+1)dx)=e^(-log(x+1))=(1)/(x+1)` |
|
| 503. |
Q. The value of is equal `sum_(k=1)^13(1/(sin(pi/4+(k-1)pi/6)sin(pi/4+k pi/6))` is equalA. (a) `3-sqrt(3)`B. (b)`2(3-sqrt(3))`C. (c) `2(sqrt(3)-1)`D. (d) `2(2+sqrt3)` |
|
Answer» Correct Answer - (c) Here, `sum_(k=1)^(13)1/(sin{pi/4+((k-1)pi)/6}sin (pi/4+(kpi)/6))` Converting into differences, by multiplying and dividing by `sin [(pi/4+(kpi)/6)-{pi/4+((k-1)pi)/6}],i.e.sin (pi/6).` `therefore sum_(k=1)^(13)(sin[(pi/4=kpi/6)-{pi/4+(k-1)pi/6}])/(sin frac {pi}{6} {sin{pi/4+(k-1)pi/6}sin (pi/4+kpi/6)})` `=2sum_(k=1)^13 ([sin(pi/4+(kpi)/6)cos{pi/4+(k-1)pi/6} -sin{pi/4+(k-1)pi/6}cos(pi/4+(kpi)/6)])/(sin{pi/4+(k-1)pi/6}-cot (pi/4+kpi/6))` `=2sum_(k=1)^13 [cot{pi/4+(k-1)pi/6}-cot(pi/4+k pi/6)]` `=2 [{cot(pi/4)-cot(pi/4+ pi/6)}` `+ {cot(pi/4+pi/6)-cot(pi/4+ (2pi)/6)}` `+...+ {cot(pi/4+12 pi/6)-cot(pi/4+ 13pi/6)}]` `= 2 {cot frac{pi}{4}-cot(pi/4+ 13pi/6)}` `= 2 [1-cot ((29pi)/12)]=2 [1-cot(2pi+(5pi)/12)]` `=2[1-cot frac{5pi}{12}] [therefore cot frac {5pi}{12}=(2-sqrt3)]` `=2(1-2+sqrt(3))` `=2(sqrt(3)-1)` |
|
| 504. |
If : `(dy)/(dx)=(y+3)gt0 and y(0)=2,` then : `y(-log2)=`A. 5B. 13C. `-2`D. 7 |
| Answer» Correct Answer - D | |
| 505. |
The solution of differential equation ydx + (x+xy)dy = 0 isA. `y + log((y)/(x)) + x = 0`B. y + log xy + c = 0C. y - log xy + c = 0D. `x + log ((x)/(y)) + c = 0` |
| Answer» Correct Answer - B | |
| 506. |
The differential equation `(dy)/(dx)+x sin 2y=x^(3)cos^(2)y` when transformed to linear form becomesA. `(dz)/(dx)+(z)/(x^(2))=x`B. `(dz)/(dx)+zx=(x^(3))/(2)`C. `(dz)/(dx)+2xz=x^(3)`D. `(dz)/(dx)=(z)/(x)=x^(2)` |
|
Answer» Correct Answer - C We have, `(dy)/(dx)+x sin 2y=x^(3)cos^(2)y` `rArr" "sin^(2)y.(dy)/(dx)+(2 tan y)x=x^(3)` Putting `tany=2 and sec^(2)y(dy)/(dx)=(dz)/(dx)`, we get `(dz)/(dx)+2xz=x^(3)` |
|
| 507. |
If `y=y(x)`satisfies the differential equation `8sqrt(x)(sqrt(9+sqrt(x)))dy=(sqrt(4+sqrt(9+sqrt(x))))^(-1)dx ,x >0a n dy(0)=sqrt(7,)`then `y(256)=`16 (b)80 (c) 3(d) 9A. (a) 16B. (b) 3C. (c) 9D. (d) 80 |
|
Answer» Correct Answer - (b) `dy/dx=1/(8sqrt(x)sqrt(9+sqrt(x))sqrt(4+sqrt(9+sqrt(x))))` `rArr y= sqrt(4+sqrt9+sqrt(x))+c` Now, `y(0)=sqrt(7)+c` `rArr c=0` `y(256)=sqrt(4+sqrt(9+16))=sqrt(4+5)=3` |
|
| 508. |
A curve passes through the point `(1,pi/6)`. Let the slope of the curve at each point `(x , y)`be `y/x+sec(y/x),x > 0.`Then theequation of the curve is(a) `( b ) (c)sin(( d ) (e) (f) y/( g ) x (h) (i) (j))=logx+( k )1/( l )2( m ) (n) (o)`(p)(q) `( r ) (s)"c o s e c"(( t ) (u) (v) y/( w ) x (x) (y) (z))=logx+2( a a )`(bb)(cc)`( d d ) (ee)sec(( f f ) (gg) (hh)(( i i )2y)/( j j ) x (kk) (ll) (mm))=logx+2( n n )`(oo)(pp)`( q q ) (rr)cos(( s s ) (tt) (uu)(( v v )2y)/( w w ) x (xx) (yy) (zz))=logx+( a a a )1/( b b b )2( c c c ) (ddd) (eee)`(fff)A. `sin((y)/(x))=logx+(1)/(2)`B. `"cosec"((y)/(x))=logx+2`C. `sec((2y)/(x))=logx+2`D. `cos((2y)/(x))=logx+(1)/(2)` |
|
Answer» Correct Answer - A It is given that `(dy)/(dx)=(y)/(x)+sec ((y)/(x)),x gt0" …(i)"` Let `y=vx.` Then, `(dy)/(dx)=v+x(dv)/(x)` Substituting these values in (i), we obtain `v+x(dv)/(dx)=v+sec v` `rArr" "x(dv)/(dx)=sec v` `rArr" "cos v dv =(1)/(x)dx` Integrating both sides, we get `sin v+log x+C` `rArr" "sin((y)/(x))=logx+C" ...(ii)"` `therefore" "sin.(pi)/(6)=log1+CrArr C=(1)/(2)` Putting `C=(1)/(2)` in (ii), we obtain the equation of the curve as `sin((y)/(x))=logx+(1)/(2)` |
|
| 509. |
The solution of the differential equation `(1+xy)xdy+(1-xy)ydx=0` ,isA. `(1)/(xy)+log((y)/(x))=C`B. `-xy+log((y)/(x))=C`C. `-(1)/(xy)+log((y)/(x))=C`D. `-(1)/(xy)+log((x)/(y))=C` |
|
Answer» Correct Answer - C The given equation can be written as `(xdy+ydx)+xy(xdy-ydx)=0` `rArr" "d(xy)+xy(xdy-ydx)=0` `rArr" "(d(xy))/((xy)^(2))+(xdy-ydx)/(xy)=0` `rArr" "(d(xy))/((xy)^(2))+(dy)/(y)-(dx)/(x)=0` `rArr" "(d(x))/((xy)^(2))+d(logy)-d(logx)=0` `rArr" "(d(xy))/((xy)^(2))+d(logy-logx)=0` `rArr" "(d(xy))/((xy)^(2))+dlog((y)/(x))=0` On integrating, we get `-(1)/(xy)+log((y)/(x))=C` as the required solution. |
|
| 510. |
Solution of the differential equation `dy/dx=(y^3)/(e^(2x)+y^2)` , isA. `e^(-2x) y^(2) + 2In|y|= c`B. `e^(2x)y^(2)=2In|y| = c`C. `e^(x)+In|y| = c`D. None of these |
|
Answer» Correct Answer - A `(dy)/(dx)=(y^(3))/(e^(2x)+y^(2))` Dividing by `e^(2x)` `therefore" "(dy)/(dx)=(y^(3)e^(-2x))/(1+y^(2)e^(-2x))` `therefore" "dy + y^(2)e^(-2x)dy = y^(3)e^(-2x)dx` `therefore" "int 2 (dy)/(y) + int 2 (ye^(-2x)dy - y^(2)e^(-2x)dx)= 0` `therefore" "int 2 (dy)/(y)+ int d(e^(-2x)y^(2))+c=0` `therefore" "2 "In"|y| + e^(-2x) y^(2) = c` |
|
| 511. |
`dy-e^(x-y)dx=x^(2)e^(-y)dx` A) `e^(x)=e^(y)+(x^(3))/(3)+c`B) `e^(x)=e^(y)+(y^(3))/(3)+c`C) `e^(y)=e^(x)+(x^(3))/(3)+c`D)`e^(x-y)=y+(x^(3))/(3)+c`A. `e^(x)=e^(y)+(x^(3))/(3)+c`B. `e^(x)=e^(y)+(y^(3))/(3)+c`C. `e^(y)=e^(x)+(x^(3))/(3)+c`D. `e^(x-y)=y+(x^(3))/(3)+c` |
| Answer» Correct Answer - C | |
| 512. |
The solution of the differential equation `ydx+ (x +x^2 y) dy =0` isA. log y = CxB. `-(1)/(xy) + log y = C`C. `(1)/(xy) + log y = C`D. `-(1)/(xy) = C` |
| Answer» Correct Answer - B | |
| 513. |
An integrating factor of the differential equation `ylogy(dx)/(dy)+x-logy=0`, isA. `log(logy)`B. `logy`C. `(1)/(logy)`D. `(1)/(log(logy))` |
|
Answer» Correct Answer - B We have, `(dx)/(dy)+(1)/(y log y)x=(1)/(y)` `"I.F."=e^(int(1)/(y log y))dy=e^(log(logy))=logy` |
|
| 514. |
Let y(x) be a solution of `xdy+ydx+y^(2)(xdy-ydx)=0` satisfying y(1)=1. Statement -1 : The range of y(x) has exactly two points. Statement-2 : The constant of integration is zero.A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1.B. Statement-1 is True, Statement-2 is True, Statement-2 is not a correct explanation for Statement-1.C. Statement-1 is True, Statement-2 is False.D. Statement-1 is False, Statement-2 is True. |
|
Answer» Correct Answer - C We have, `xdy+ydx+y^(2)(xdy-ydx)=0` `rArr" "d(xy)+x^(2)y^(2){(xdy-ydx)/(x^(2))}=0` `rArr" "(1)/((xy)^(2))d(xy)+d((y)/(x))=0` On integrating, we get `-(1)/(xy)+(y)/(x)=C" …(i)"` It is given that y = 1 when x = 1 `therefore" "-1+1=C rArr C=0` Putting C = 0 in (i), we get `-(1)/(xy)+(y)/(x)=0 rArr y^(2)-1=0 rArr y = pm1` Clearly, statement-1 and statement-2 are true. Also, statement-2 is not a correct explanation for statement-1. |
|
| 515. |
A curve passes through the point `(1,pi/6)`. Let the slope of the curve at each point `(x , y)`be `y/x+sec(y/x),x > 0.`Then theequation of the curve is(a) `( b ) (c)sin(( d ) (e) (f) y/( g ) x (h) (i) (j))=logx+( k )1/( l )2( m ) (n) (o)`(p)(q) `( r ) (s)"c o s e c"(( t ) (u) (v) y/( w ) x (x) (y) (z))=logx+2( a a )`(bb)(cc)`( d d ) (ee)sec(( f f ) (gg) (hh)(( i i )2y)/( j j ) x (kk) (ll) (mm))=logx+2( n n )`(oo)(pp)`( q q ) (rr)cos(( s s ) (tt) (uu)(( v v )2y)/( w w ) x (xx) (yy) (zz))=logx+( a a a )1/( b b b )2( c c c ) (ddd) (eee)`(fff)A. `sin(y/x)=logx+1/2`B. `"cosec "(y/x)=logx+2`C. `sec(2y)/(x)=logx+2`D. `cos(2y)/(x)=logx+1/2` |
|
Answer» Correct Answer - A `(dy)/(dx) =y/x+secy/x`. Let `y=vx`. Then given equation reduces to `(dv)/(secv) = (dv)/x` or `intcosvdv=int(dx)/x` or `sinv="ln"x+c` or `sin(y/x)=logx+c` The curve passes through `(1,pi6)`. Thus, `sin(y/x)=logx+1/2`. |
|
| 516. |
`(1 + xy) ydx + (1-xy)xdy = 0`A. `(x)/(y)+(1)/(xy)=k`B. `log((x)/(y))=(1)/(xy)+k`C. `(x)/(y)+(1)/(xy)=k`D. `log((x)/(y))= xy + k` |
|
Answer» Correct Answer - C `y dx + x dy + xy^(2) dx - x^(2) y dy = 0` `rArr" "(y dx + x dy)/(x^(2)y^(2))+(dx)/(x)-(dy)/(y)=0` On integrating, we get `-(1)/(xy)+log x - log y = k rArr "log"(x)/(y)=(1)/(xy)+k` |
|
| 517. |
`(dy)/(dx)=(x y+y)/(x y+x)`A. `(y-x)+log(y-x)=c`B. `log((y)/(x))=(x-y)+c`C. `ylogy=xlogx+c`D. `log(xy)=(x-y)+c` |
| Answer» Correct Answer - B | |
| 518. |
Consider the differential equation `ydx-(x+y^(2))dy=0`. If for `y=1, x` takes value 1, then value of x when y = 4, isA. 64B. 9C. 16D. 36 |
|
Answer» Correct Answer - C We have, `ydx-(x+y)^(2)dy=0` `rArr" (dx)/(dy)+(-(1)/(y))x=y" ...(i)"` This is a linear differential equation with `"I.F."=e^(int(1)/(y)dy)=(1)/(y)` Multiplying both sides of (i) by `"I.F."=(1)/(y)` and integrating with respect to y, we get `(x)/(y)=int yxx(1)/(y)dy+C` `rArr" "(x)/(y)=y+C" ...(ii)"` It is given that y=1 when x = 1. Putting x = 1, y = 1 in (ii), we get C = 0. Putting C = 0 in (ii), we get `x=y^(2)`. When y = 4 this equation gives x = 16. |
|
| 519. |
The function `y=f(x)`is thesolution of the differential equation `(dy)/(dx)+(x y)/(x^2-1)=(x^4+2x)/(sqrt(1-x^2))`in `(-1,1)`satisfying`f(0)=0.`Then `int_((sqrt(3))/2)^((sqrt(3))/2)f(x)dx`is(a)`( b ) (c) (d)pi/( e )3( f ) (g)-( h )(( i )sqrt(( j )3( k ))( l ))/( m )2( n ) (o) (p)`(q)(b) `( r ) (s) (t)pi/( u )3( v ) (w)-( x )(( y )sqrt(( z )3( a a ))( b b ))/( c c )4( d d ) (ee) (ff)`(gg)(c)`( d ) (e) (f)pi/( g )6( h ) (i)-( j )(( k )sqrt(( l )3( m ))( n ))/( o )4( p ) (q) (r)`(s)(d) `( t ) (u) (v)pi/( w )6( x ) (y)-( z )(( a a )sqrt(( b b )3( c c ))( d d ))/( e e )2( f f ) (gg) (hh)`(ii)A. `pi/3-sqrt(3)/2`B. `pi/3-sqrt(3)/4`C. `pi/6-sqrt(3)/4`D. `pi/6-sqrt(3)/2` |
|
Answer» Correct Answer - B `(dy)/(dx) + x/(x^(2)-1)y=(x^(4)+2x)/sqrt(1-x^(2))` This is a linear differential equation. `I.F. =e^(intx/(x^(2)-1)dx)` `e^(1/2"ln "|x^(2)-1|)=sqrt(1-x^(2))` (`therefore x in (-1,1))` Therefore, solution is: `ysqrt(1-x^(2))=int(x(x^(3)+2))/sqrt(1-x^(2)).sqrt(1-x^(2))dx` or `ysqrt(1-x^(2))=int(x^(4)+2x)dx=x^(5)/5+x^(2)+C` Since, `f(0)=0,C=0` `therefore f(x) sqrt(1-x^(2))=x^(5)/5+x^(2)` `rArr f(x) = x^(5)/(5sqrt(1-x^(2))+x^(2)/sqrt(1-x^(2))` `therefore int_(-sqrt(3//2))^(sqrt(3)/2) f(x)dx= int_(sqrt(3)/2)^(sqrt(3)/2) (x^(5)/(5sqrt(1-x^(2))+x^(2)/sqrt(1-x^(2))))dx` `=2int_(0)^(sqrt(3)//2)(x^(2)/sqrt(1-x^(2)))dx` `rArr 2int_(0)^(pi//3) sin^(2)theta(d)theta` `=2int_(0)^(pi//3)(1-cos2theta)/(2)(d)theta` `=2[theta/2-(sin2theta)/4]_(0)^(pi//3)` `=2(pi/6) -2(sqrt(3))/8 = pi/3-sqrt(3)/4` |
|
| 520. |
`(dy)/(dx)=(xy)/((1-x)(1+y))` A) `x+y+log[y(1-x)]=c`B) `x+y+log[x(1-y)]=c`C) `x+y+log(x+y)=c`D)`y-x+log[x(1-y)]=c`A. `x+y+log[y(1-x)]=c`B. `x+y+log[x(1-y)]=c`C. `x+y+log(x+y)=c`D. `y-x+log[x(1-y)]=c` |
| Answer» Correct Answer - A | |
| 521. |
The differential eqaution of the family of curve `y^(2)=4a(x+1)`, isA. `y^(2)=4(dy)/(dx)(x+(dy)/(dx))`B. `2y=(dy)/(dx)+4a`C. `y^(2)((dy)/(dx))^(2)+2xy(dy)/(dx)-y^(2)=0`D. `y^(2)(dy)/(dx)+4y=0` |
| Answer» Correct Answer - C | |
| 522. |
The function y=f(x) is the solution of the differential equation `[dy]/[dx]+[xy]/[x^2-1]=[x^4+2x]/sqrt[1-x^2]` in (-1, 1), satisfying `f(0)=0`. Then `int_[-sqrt3/2]^[sqrt3/2] f(x)dx` is (A) ` pi/3 - sqrt3/2` (B) ` pi/3 - sqrt3/4` (C) ` pi/6 - sqrt3/4` (D) ` pi/6 - sqrt3/2`A. `(pi)/(3)-(sqrt3)/(2)`B. `(pi)/(3)-(sqrt3)/(4)`C. `(pi)/(6)-(sqrt3)/(4)`D. `(pi)/(6)-(sqrt3)/(2)` |
|
Answer» Correct Answer - B We have, `(dy)/(dx)+(xy)/(x^(2)-1)=(x^(4)+2x)/(sqrt(1-x^(2)))" ...(i)"` This is a linear differential equation with `"I.F."=e^(int(x)/(x^(2)-1)dx)=e^((1)/(2)log|x^(2)-1|)=e^((1)/(2)log(1-x^(2)))=sqrt(1-x^(2))` Multiplying both sides of (i) by I.F. and integrating with respect to x, we obtain `ysqrt(1-x^(2))=int(x^(4)+2x)dx+C` or, `ysqrt(1-x^(2))=(x^(5))/(5)+x^(2)+C" ...(ii)"` Putting x = 0 and y = f (0) = 0 in (ii), we get C = 0 Putting C = 0 in (ii), we get `ysqrt(1-x^(2))=(x^(5))/(5)+x^(2)` `rArr" "f(x)-(x^(5))/(5sqrt(1-x^(2)))+(x^(2))/(sqrt(1-x^(2)))` `therefore" "int_(-sqrt3//2)^(sqrt3//2)f(x)dx=int_(-sqrt3//2)^(sqrt3//2)(x^(5))/(sqrt(1-x^(2)))dx+int_(-sqrt3//2)^(sqrt3//2)(x^(2))/(sqrt(1-x^(2)))dx` `=2int_(0)^(sqrt3//2)(x^(2))/(sqrt(1-x^(2)))dx` `=2int_(0)^(pi//3)sin^(2)d theta," where x "=sin theta` `=int_(0)^(pi//3)(1-cos2 theta)d theta` `[theta-(1)/(2)sin2 theta]_(0)^(pi//3)=(pi)/(3)-(sqrt3)/(4)` |
|
| 523. |
Find the differential equation of `xy = ae^x + be^(-x)`.A. `xy_(1)+2y_(2)=xy`B. `xy_(2)+2y_(1)=xy`C. `xy_(2)=y`D. `xy_(1)+y=xy` |
| Answer» Correct Answer - B | |
| 524. |
`(dy)/(dx)=(1+x)(1+y^(2))`A)`(1)/(2)log[(1+y)/(1-y)]=x+(x^(2))/(2)+c` B)`log(1+y^(2))=x+(x^(2))/(2)+c` C)`tan^(-1)y=x+(x^(2))/(2)+c` D)`tan^(-1)x=y+(y^(2))/(2)+c`A. `(1)/(2)log[(1+y)/(1-y)]=x+(x^(2))/(2)+c`B. `log(1+y^(2))=x+(x^(2))/(2)+c`C. `tan^(-1)y=x+(x^(2))/(2)+c`D. `tan^(-1)x=y+(y^(2))/(2)+c` |
| Answer» Correct Answer - C | |
| 525. |
Form the Differential equation for `y=ae^(2x-1)+be^(-2x-1)-1`A. `y_(2)=4y`B. `y_(2)=4+y`C. `y_(2)=4(y+1)`D. `y_(2)=4(x+1)` |
| Answer» Correct Answer - C | |
| 526. |
The differential equation obtained by eliminating the constants a and b from `xy = ae^x +be^(-x) + x^2` isA. `xy_(2)+2y_(1)+x^(2)-xy-2=0`B. `xy_(2)+2y_(1)=x^(2)-xy+2`C. `xy_(1)+2y_(2)+x^(2)-xy+2=0`D. `y_(2)=2xy` |
| Answer» Correct Answer - A | |
| 527. |
`sqrt(1-y^(2))dx-sqrt(1-x^(2))dy=0`A)`sin^(-1)x-cos^(-1)y=c` B)`sin^(-1)x-sin^(-1)y=c` C)`log(x+sqrt(1-x^(2)))=log(y+sqrt(1-y^(2)))+c` D)`x-y=c(1+xy)`A. `sin^(-1)x-cos^(-1)y=c`B. `sin^(-1)x-sin^(-1)y=c`C. `log(x+sqrt(1-x^(2)))=log(y+sqrt(1-y^(2)))+c`D. `x-y=c(1+xy)` |
| Answer» Correct Answer - B | |
| 528. |
`ysqrt(1-x^(2)) dy + x sqrt(1-y^(2)) dx=0`A. `sin^(-1)x+sin^(-1)y=c`B. `sec^(-1)x+sec^(-1)y=c`C. `sin^(-1)x+cos^(-1)y=c`D. `sqrt(1+x^(2))+sqrt(1-y^(2))=c` |
| Answer» Correct Answer - D | |
| 529. |
Solve `(x-y^(2)x)dx=(y-x^(2)y)dy`. |
|
Answer» Correct Answer - `(x^(2)-1)=C(y^(2)-1)` We have `x(1-y^(2))dx=y(1-x^(2))dy` `therefore (2x)/(x^(2)-1)dx=(2y)/(y^(2)-1)dy` Integrating both sides, we get `log_(e)(x^(2)-1)=log_(e)(y^(2)-1)log_(e)C` `therefore (x^(2)-1)=C(y^(2)-1)` |
|
| 530. |
If `y=ae^(2x)+b cos 2x+c sin 2x, ` thenA. `y_(3)=8y`B. `y_(3)-2y_(2)+4y_(1)-8y=0`C. `y_(3)+2y_(2)-4y_(1)+8y=0`D. `y_(3)+8y=0` |
| Answer» Correct Answer - B | |
| 531. |
The order of differential equation of family of circles passing through intersection of `3x+4y-7=0` and `S=-x^(2)+y^(2)-2x+1=0` isA. 1B. 2C. 3D. 4 |
|
Answer» Correct Answer - A Equation of family of circles is `(x^(2)+y^(2)-2x-2y+1)+lambda(3x+4y-7)=0` Arbitrary constants is 1 So, order is 1. |
|
| 532. |
The degree of the differential equation satisfying the relation `sqrt(1+x^2) + sqrt(1+y^2) = lambda (x sqrt(1+y^2)- ysqrt(1+x^2))` is |
|
Answer» Correct Answer - `(dy)/(dx) = (1+y^(2))/(1+x^(2))` Putting x=tanA and y=tanB in the given relation, we get cosA + cosB=`lambda`(sinA-sinB) or `tan^(-1)x-tan^(-1)y=2tan^(-1)(1/lambda)` Differentiating w.r.t. to x, we get `1/(1+x^(2))-1/(1+y^(2))(dy)/(dx)=0` or `(dy)/(dx)=(1+y^(2))/(1+x^(2))` |
|
| 533. |
Solve `e^((dy)/(dx))=x+1,`given that when `x=0,y=3.` |
|
Answer» Correct Answer - `y=(x+1)(log_(e)(x+1))-x+3` `e^((dy)/(dx))=x+1` or `(dy)(dx) = log(x+1)` or `intdy=intlog(x+1)dx` or `y=(x+1)log(x+1)-x+c` when x=0, y=3 is c=3 Hence, the solution is `y=(x+1)log(x+1)-x+3` |
|
| 534. |
Obtain the differential equation of the family of circles passing through the point (a,0) and (-a,0).A. `y_(1)(y^(2)-x^(2))+2xy+a^(2)`B. `y_(1)y^(2)+xy+a^(2)x^(2)=0`C. `y_(1)(y^(2)-x^(2)+a^(2))+2xy=0`D. none of these |
| Answer» Correct Answer - C | |
| 535. |
Find the differential equation whose general solution is given by `y=(c_(1)+c_(2))cos(x+c_(3))-c_(4)e^(x+c)`, where `c_(1),c_(2), c_(3), c_(4), c_(5)` are arbitary constants. |
|
Answer» Correct Answer - `(d^(3)y)/(dx^(3))-(d^(2)y)/(dx^(2))+(dy)/(dx)-y=0` `y=(c_(1)+c_(2))cosx+c_(3)-c_(4)e^(x+c_(5))` or `y=Acos(x+B)-Ce^(x)`, where `A=c_(1)+c_(2),B=c_(3)` and `C=c_(4)e^(c_(5))` `rArr (dy)/(dx)=-Asin(x+B)=Ce^(x)` `rArr (d^(2)y)/(dx^(2))=-Acos(x+B)-Ce^(x)` `rArr (d^(2)y)/(dx^(2))+y-2Ce^(x)` `rArr (d^(2)y)/(dx^(2))+y=-2Ce^(x)` `rArr (d^(3)y)/(dx^(3))-(d^(2)y)/(dx^(2))+(dy)/(dx)-y=0` |
|
| 536. |
The differential equation of all parabolas whoseaxis are parallel to the y-axis is(a)`( b ) (c) (d)(( e ) (f) d^(( g )3( h ))( i ) y)/( j )(( k ) d (l) x^(( m )3( n ))( o ))( p ) (q)=0( r )`(s)(b) `( t ) (u) (v)(( w ) (x) d^(( y )2( z ))( a a ) x)/( b b )(( c c ) d (dd) y^(( e e )2( f f ))( g g ))( h h ) (ii)=C (jj)`(kk)(c)`( d ) (e) (f)(( g ) (h) d^(( i )3( j ))( k ) y)/( l )(( m ) d (n) x^(( o )3( p ))( q ))( r ) (s)+( t )(( u ) (v) d^(( w )2( x ))( y ) x)/( z )(( a a ) d (bb) y^(( c c )2( d d ))( e e ))( f f ) (gg)=0( h h )`(ii)(d) `( j j ) (kk) (ll)(( m m ) (nn) d^(( o o )2( p p ))( q q ) y)/( r r )(( s s ) d (tt) x^(( u u )2( v v ))( w w ))( x x ) (yy)+2( z z )(( a a a ) dy)/( b b b )(( c c c ) dx)( d d d ) (eee)=C (fff)`(ggg)A. `y_(2)=2y_(1)+x`B. `y_(3)=2y_(1)`C. `y_(2)^(3)=y_(1)`D. none of these |
| Answer» Correct Answer - D | |
| 537. |
The solution of the differential equation `y_(1)y_(3)=3y_(2)^(2)`, isA. `x=A_(1)y^(2)+A_(2)y+A_(3)`B. `x=A_(1)y+A_(2)`C. `x=A_(1)y^(2)+A_(2)y`D. none of these |
| Answer» Correct Answer - A | |
| 538. |
The solution of differential equation `(1+y^(2))+(x-e^(tan^(-1)y))(dy)/(dx)=0`, isA. `(x-2)=Ce^(-2tan^(-1)y)`B. `2x^(tan^(-1)y)=e^(2tan^(-1)y)+C`C. `xe^(tan^(-1)y)=tan^(-1)y+C`D. `xe^(2tan^(-1)y)=e^(tan^(-1)y)+C` |
| Answer» Correct Answer - B | |
| 539. |
The order of the differential equation whose general solution is given by `y=(C_(1)+C_(2))sin(x+C_(3))-C_(4)e^(x+(C_(5))`, isA. 5B. 4C. 2D. 3 |
|
Answer» Correct Answer - D We have, `y=(C_(1)+C_(2))sin(x+C_(2))-C_(4)e^(x+C_(5))` `rArr" "y=C_(6)sin(x+C_(3))-C_(4)e^(C_(5)).e^(x)," where "C_(6)=C_(1)+C_(2)` `rArr" "C_(6)sin(x+C_(3))-C_(7)e^(x)," where "C_(4)e^(C_(5))=C_(7)` Clearly, the above relation contains three arbitrary constants. So, the order of the differential equation satisfying it is 3. |
|
| 540. |
`c+4yx=0` is a solution of the D.E.A. `xy+y_(1)=0`B. `xy_(1)+y=0`C. `y_(1)=4xy`D. `xy_(1)+1=0` |
| Answer» Correct Answer - B | |
| 541. |
Derivative of`y=be^(-x)`A. `y_(1)=y`B. `y_(1)+ye^(x)=0`C. `y_(1)+y=0`D. `y_(1)+x=0` |
| Answer» Correct Answer - C | |
| 542. |
The degree and order of the differential equation of the family of all parabolas whose axis is x-axs are respectivelyA. 2, 1B. 1, 2C. 3, 2D. 2, 3 |
| Answer» Correct Answer - B | |
| 543. |
The degree and order of the differential equation of the family of all parabolas whose axis is x-axs are respectivelyA. 2B. 1,2C. 3,2D. none of these |
| Answer» Correct Answer - B | |
| 544. |
The degree and order of the differential equation of the family of all parabolas whose axis is x-axs are respectivelyA. 2,1B. 1,2C. 3,2D. none of these |
|
Answer» Correct Answer - B The equation of all parabolas whose axis is x-axis is given by `y^(2)=4a(x-h)` Differentiating w.r. to x, we get `2yy_(1)=4a rArr yy_(1)=2a rArr y_(1)^(2)+yy_(2)=0` This is a differential equation of order 2 and degree 1. |
|
| 545. |
Form differential equation for `y=ax^(x)`A. `y=y_(1)e^(x)`B. `y_(1)=ye^(x)`C. `yy_(1)=e^(x)`D. `y_(1)=y(1+logx)` |
| Answer» Correct Answer - D | |
| 546. |
Form differential equation for `y=Ax^(2)+b`A. `xy_(1)=y_(2)`B. `xy_(1)y_(2)=1`C. `y_(1)y_(2)=x`D. `y_(1)=xy_(2)` |
| Answer» Correct Answer - D | |
| 547. |
What is the differential equation of the curve ` y=ax^(2)+bx` ?A. `x^(2)y_(2)-2xy_(1)+2y=0`B. `x^(2)y_(1)-2xy_(2)+2y=0`C. `2xy_(2)-x^(2)y_(1)+2y=0`D. `2xy_(2)-x^(2)y_(1)+y=0` |
| Answer» Correct Answer - A | |
| 548. |
Form differential equation for `y=x+2ax^(2)`A. `2x=y(1+y_(1))`B. `2y=x(1+y_(1))`C. `2xy=1+y_(1)`D. `2y_(1)=x(1+y)` |
| Answer» Correct Answer - B | |
| 549. |
Form differential equation for `y=A log x+B`A. `xy_(2)+y_(1)=0`B. `xy_(1)=+y_(2)=0`C. `y_(1)+y_(2)=x`D. `y_(2)=x+y_(1)` |
| Answer» Correct Answer - A | |
| 550. |
Form differential equation for `(y-c)^(2)=cx` A)`y=2xy_(1)+4x(y_(1))^(2)` B)`y_(1)=2xy+4xy^(2)` C)`2yy_(1)=2x+4x^(2)` D)`y_(1)=x^(2)y+x^(4)y^(2)`A. `y=2xy_(1)+4x(y_(1))^(2)`B. `y_(1)=2xy+4xy^(2)`C. `2yy_(1)=2x+4x^(2)`D. `y_(1)=x^(2)y+x^(4)y^(2)` |
| Answer» Correct Answer - A | |