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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.
| 601. |
Thedifferential equation of all circles passing through the origin and havingtheir centres on the x-axis is(1) `x^2=""y^2+""x y(dy)/(dx)`(2) `x^2=""y^2+"3"x y(dy)/(dx)`(3) `y^2=x^2""+"2"x y(dy)/(dx)`(4) `y^2=x^2""-"2"x y(dy)/(dx)`A. `y^(2) = x^(2) - 2xy (dy)/(dx)`B. `x^(2) = y^(2) + xy(dy)/(dx)`C. `x^(2) = y^(2) + 3xy (dy)/(dx)`D. `y^(2) = x^(2) + 2xy (dy)/(dx)` |
| Answer» Correct Answer - D | |
| 602. |
The differential equation of the family of circles passing through the origin and having centres on the x-axis isA. `x^(2)=y^(2)+xy(dy)/(dx)`B. `x^(2)=y^(2)+3xy(dy)/(dx)`C. `y^(2)=x^(2)+2xy(dy)/(dx)`D. `y^(2)=x^(2)-2xy(dy)/(dx)` |
| Answer» Correct Answer - C | |
| 603. |
If the D.E. `(dy)/(dx)=(gx+3)/(2y+f)` represents a circle, then g = A)2 B)-2 C)3 D)-4A. 2B. `-2`C. 3D. `-4` |
| Answer» Correct Answer - B | |
| 604. |
Determine the order and degree of each of thefollowing differential equation. State also whether they are linear ornon-linear: `2(d^2y)/(dx^2)+3 sqrt(1-((dy)/(dx))^2)-y=0`A. 2, 3B. 3, 2C. 2, 4D. 2, 2 |
| Answer» Correct Answer - D | |
| 605. |
If a curve passes through the point (1,-2) and has slope of the tangent at any point (x, y) on it as `(x^2_2y)/x`, then the curve also passes through the point -(a) `(sqrt3,0)`(b) `(-1,2) `(c) `(-sqrt2,1)`(d) `(3,0) `A. (a) `(sqrt3,0)`B. (b) (-1,2)C. (c) `(-sqrt2,1)`D. (d) (3,0) |
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Answer» Correct Answer - (a) We know that, slope if the tangent at anypoint (x,y) on the curve is `dy/dx = (x^(2) -2y)/x` (given) `rArr dy/dx + 2/xy=x` shich is a linear differential equation of the form `dy/dx+P(x)cdot y=Q(x),` where ` P(x)=2/xand Q(x) = x` Now, integrating factor `(IF)=e^(int P(x)dx)=e^(int2/xdx)=e^(2log_(e)x` `=e^(2log_(e)x^(2) [therefore m log a =log a^(m)]` `=x^(2) [therefore e^(2log_(e)f(x))=f(x)]` and the solution of differential Eq.(i) is `y(IF) = int Q(x)(IF)dx+ CrArry(x^(2))= int xcdotx^(2)dx+C` `rArr yx^(2) =x^(4)/4+C …(ii)` `therefore` The curve (ii) passes through the point (1,-2), therrefore `-2=1/4+C rArr C= -9/4` `therefore` Equation of required curve is `4yx^(2)+x^(4)-9.` Now, checking all the option, we get only `(sqrt3,0)` satisfy the above equation. |
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| 606. |
`(x-1)dy-(1+y)dx=0, y=4" if " x=2` A)`5(x-1)(1+y)=1` B)`(x-1)(1+y)=5` C)`1+y=5(x-1)` D)`5(1+y)=x-1`A. `5(x-1)(1+y)=1`B. `(x-1)(1+y)=5`C. `1+y=5(x-1)`D. `5(1+y)=x-1` |
| Answer» Correct Answer - C | |
| 607. |
D.E. of lines, passing through the origin, isA. `xy_(1)=y`B. `xy=y_(1)`C. `xy+y_(1)=0`D. `y_(2)=y_(1)+y` |
| Answer» Correct Answer - A | |
| 608. |
Solve the following differential equation: `(1+y^2)tan^(-1)dx+2y(1+x^2)dy=0`A. `(1)/(2)(tan^(-1)x)^(2)+log(1+y^(2))=c`B. `(tan^(-1)x)^(2)+logsqrt(1+y^(2))=c`C. `(1+y^(2))tan^(-1)x=c`D. `(tan^(-1)x)(tan^(-1)y)=c` |
| Answer» Correct Answer - A | |
| 609. |
Solve : `3e^xtany dx+(1-e^x)sec^2y dy=0`A. `3tany=c(1-e^(x))`B. `tany=c(1-e^(x))^(3)`C. `tany=c(1-e^(3x))`D. `tanx=c(1-e^(x))` |
| Answer» Correct Answer - B | |
| 610. |
`3x^(2)y+(dy)/(dx)=0`A. `x^(3)-logy=c`B. `c=ye^(x^(3))`C. `y^(3)+logx=c`D. `c=xe^(y^(3))` |
| Answer» Correct Answer - B | |
| 611. |
`(3x+1)dy+(2y-3)dx=0`A. `(3x+1)^(2)(2y-3)^(3)=c`B. `(3x+1)^(3)(2y-3)^(2)=c`C. `6(3x+2y)^(5)=c`D. `(3x+1)(2y-3)=c` |
| Answer» Correct Answer - A | |
| 612. |
The solution of the differential equation `xdy/dx+2y=x^(2),(xne0)` with y(1)=1, is(a) `y=x^(2)/4+3/(4x^(2))`(b) `y=x^(3)/5+1/(5x^(2))`(c) `y=3/4x^(2)+3/(4x^(2))`(d) `y=4/5x^(3)+1/(5x^(2))`A. (a) `y=x^(2)/4+3/(4x^(2))`B. (b) `y=x^(3)/5+1/(5x^(2))`C. (c) `y=3/4x^(2)+3/(4x^(2))`D. (d) `y=4/5x^(3)+1/(5x^(2))` |
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Answer» Correct Answer - (a) Given differential equation is `xdy/dx+2y=x^(2),(xne0)` `rArr dy/dx+(2/x)y=x,` shich is a linear differential equation of the form `dy/dx+Py=Q` Here, `P=2/x` and Q=x `therefore IF=e^(int2/x dx) =e^(2log x) =x^(2) ` Since, solution of the given differential equation is `yxxIF=int(QxxIF)dx+C` ` therefore y(x^(2)) = int (x xxx^(2)) dx + C rArr yx^(2) =x^(4)/4=C` ` therefore y(1) =1,`so `1=1/4+C rArr C=3/4` ` therefore yx^(2) =x^(4) /4+3/4 rArr y = x^(2) /4+3/(4x^(2)` |
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| 613. |
`(x^(2)-yx^(2))dy+(y^(2)+xy^(2))dx=0`A. `x+y=log(cxy)`B. `log((x)/(y))=(1)/(x)-(1)/(y)+c`C. `log((y)/(x))=(1)/(x)-(1)/(y)+c`D. `log((x)/(y))=(1)/(x)+(1)/(y)+c` |
| Answer» Correct Answer - D | |
| 614. |
A differential equation associated to theprimitive`y=a+b e^(5x)+c e^(-7x)`is (where `y_n`is `n t h`derivativew.r.t. `x)`(a) `( b ) (c) (d) y_( e )3( f ) (g)+2( h ) y_( i )2( j ) (k)-( l ) y_( m )1( n ) (o)=0( p )`(q)(r) `( s ) (t)4( u ) y_( v )3( w ) (x)+5( y ) y_( z )2( a a ) (bb)-20 (cc) y_( d d )1( e e ) (ff)=0( g g )`(hh)(ii)`( j j ) (kk) (ll) y_( m m )3( n n ) (oo)+2( p p ) y_( q q )2( r r ) (ss)-35 (tt) y_( u u )1( v v ) (ww)=0( x x )`(yy)(d) none of these(zz)`( a a a ) (bbb) w h e r e( c c c )`(ddd)y_n reprersents `( e e e ) (fff) n t h (ggg)`(hhh)order derivative.A. `y_(3)+2y_(2)-y_(1)=0`B. `4y_(3)+5y_(2)-20y_(1)=0`C. `y_(3)+2y_(2)-35y_(1)=0`D. none of these |
| Answer» Correct Answer - C | |
| 615. |
Solve the following differential equation:`3e^xtany dx+(2-e^x)sec^2y dy=0,`given that when `x=0, y=pi/4dot`A. `8(1+e^(x))^(3)tany=1`B. `(1+e^(x))^(3)tany=8`C. `(1+e^(x))^(3)tany=27`D. `2(1+e^(x))=tan^(3)y` |
| Answer» Correct Answer - B | |
| 616. |
`3x^(2)(dy)/(dx)+2y=0` Solve.A. `2logy-(3)/(x)=c`B. `2logy+(3)/(x)=c`C. `3logy-(2)/(x)=c`D. `2logx-(3)/(y)=c` |
| Answer» Correct Answer - C | |
| 617. |
`(x-y^(2)x)dx+(y-x^(2)y)dy=0` Solve the differential equation.A. `x^(2)+y^(2)=x^(2)y^(2)+c`B. `x^(2)+y^(2)=x^(2)y^(2)+c`C. `x^(2)-y^(2)=x^(2)y^(2)+c`D. `x+y=xy` |
| Answer» Correct Answer - B | |
| 618. |
Solve the differential equation `"dy=cos x(2-y cosec x)dx"` given that `y=2, "when x" d=(pi)/(2)` |
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Answer» Given differential equation, `" "dy=cosx(2-y" cosec"x)dx` `rArr" "(dy)/(dx)=cosx(2-y" cosec"x)` `rArr" "(dy)/(dx)=2cosx-y" cosec"x*cosx` `rArr" "(dy)/(dx)=2cosx-ycotx` `rArr" "(dy)/(dx)+ycotx=2cosx` which is linear differential equation. On comparing it with `(dy)/(dx)+Py=Q`, we get `" "P=cotx, Q=2cosx` `" "IF=e^(intPdx)=e^(intcotxdx)=e^(logsinx)=sinx` The general solution is `" "y*sinx=int2cosx*sinxdx +C` `rArr" "y*sinx=intsin2xdx+C" "[because sin2x=2sinxcosx]` `rArr" "y*sinx=-(cos2x)/(2)+C" "...(i)` When `x=(pi)/(2) and y=2`, then `" "2*sin""(pi)/(2)=-(cos(2xx(pi)/(2)))/(2)+C` `rArr" "2*1=+(1)/(2)+C` `rArr" "2-(1)/(2)=CrArr(4-1)/(2)=C` `rArr therefore" "C=(3)/(2)` On subsituting the value of C in Eq. (i), we get `" "ysinx=-(1)/(2)cos2x+(3)/(2)` |
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| 619. |
`xy^(2)(dy)/(dx) + x(dy)/(dx)=1-x^(2)-x`A. `x^(2)-y^(2)+log[x^(2)(y^(2)-1)]=c`B. `x^(2)-y^(2)-log[x^(2)(y^(2)+1)]=c`C. `y-tan^(-1)y=(logx)-(x^(2))/(2)+c`D. `2x^(2)-y^(2)=log(x.tan^(-1)y)=c` |
| Answer» Correct Answer - C | |
| 620. |
The solution of `(1-x^2)(dy)/(dx)+x y=5x`A. `(y-5)^(2)(1-x^(2))=c`B. `(y+5)(1-x^(2))=c`C. `(y+5)(1-x^(2))^(2)=c`D. `(y+5)^(2)(1-x^(2))=c` |
| Answer» Correct Answer - D | |
| 621. |
General solution of `(y+2)dy=(x^(2)+4x-9)dx` isA. `2y=(x+2)^(2)-26log(x+2)+c`B. `2y=(x-2)^(2)+26log(x+2)+c`C. `y=(x+2)^(2)+13log(x+2)+c`D. None of these |
| Answer» Correct Answer - D | |
| 622. |
Form differential equation for `(x-a)^(2)=4(y-b)`A. `y_(2)=2`B. `2y_(2)=1`C. `2y_(2)=y_(1)`D. `y_(2)=2y_(1)` |
| Answer» Correct Answer - B | |
| 623. |
The differential equation of ` y = c^(2) + c/x` isA. `y=x^(4)(y_(1))^(2)-xy_(1)`B. `y_(1)=x^(4)y^(2)-xy`C. `x^()=y(y_(1))^(2)-xy`D. `y=y_(1)^(2)+(y_(1))/(x)` |
| Answer» Correct Answer - A | |
| 624. |
Solve `2(y+3)-x y(dy)/(dx)=0`, given that `y(1)=-2` |
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Answer» Given that, `" "2(y-3)-xy(dy)/(dx)=0` `rArr" "2(y+3)=xy(dy)/(dx)` `rArr" "2(dx)/(x)=((y)/(y+3))dy` `rArr" "2*(dx)/(x)=((y+3-3)/(y+3))dy` `rArr" "2*(dx)/(x)=(1-(3)/(y+3))dy` On integrating both sides, we get `" "2logx=y-3log(y+3)+C" "...(i)` When x= and y=-2, then `" "2log1=-2-3log (-2+3)+C` `rArr" "2*0=-2-3*0+C` `rArr" "C=2` On subsituting the value of C in Eq. (i), we get `" "2logx=y-3log(y-3)+2` `rArr" "2logx+3log(y+3)=y+2` `rArr" "logx^(2)+log(y+3)^(3)=(y+2)` `rArr" "logx^(2)(y+3)^(3)=(y+2)` `rArr" "x^(2)(y+3)^(3)=e^(y+2)` |
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| 625. |
`(x^(2)-x^(2)y)(dy)/(dx)+y^(2)+x^(2)y^(2)=0` A)`x-(x^(-1)+y^(-1))-logy=c` B)`x+(x^(-1)-y^(-1))=log(cy)` C)`x^(2)+xy+log((y)/(x))=c` D)`(x^(-1)-y^(-1))+logy=x+c`A. `x-(x^(-1)+y^(-1))-logy=c`B. `x+(x^(-1)-y^(-1))=log(cy)`C. `x^(2)+xy+log((y)/(x))=c`D. `(x^(-1)-y^(-1))+logy=x+c` |
| Answer» Correct Answer - A | |
| 626. |
General solution of `y-x(dy)/(dx)=5(y^(2)+(dy)/(dx))` is `cy=`A. `(x-5)(1-5y)`B. `(x^(2)+5)(1-5y)`C. `(x+5)(1-5y)`D. `(5x-1)(y+5)` |
| Answer» Correct Answer - C | |
| 627. |
Form differential equation for `cy+logx=0`A. `yy_(1)=xlogx`B. `y/y_(1)=xlogx`C. `xy=y_(1)logx`D. `y_(1)=xy.logx` |
| Answer» Correct Answer - B | |
| 628. |
Solve the following differential equations:`(x+y)(dx-dy)=dx+dy` |
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Answer» Given differential equation is `" "(x+y)(dx-dy)=dx+dy` `rArr" "(x+y)(1-(dy)/(dx))=1+(dy)/(dx)" "...(i)` Put `" "x+y=z` `rArr" "1+(dy)/(dx)=(dz)/(dx)` On substituting these values in Eq. (i), we get `" "z(1-(dz)/(dx)+1)=(dz)/(dx)` `rArr" "z(2-(dz)/(dx))=(dz)/(dx)` `rArr" "2z-z(dz)/(dx)-(dz)/(dx)=0` `rArr" "2z-(z+1)(dz)/(dx)=0` `rArr" "(dz)/(dx)=(2z)/(z+1)` `rArr" "((z+1)/(z))dz=2dx` On integrating both sides, we get `" "int(1+(1)/(z))dz=2intdx` `rArr" "z+logz=2x-logC` `rArr" "(x+y)+log(x+y)=2x-logC" "[because z=x+y]` `rArr" "2x-x-y=logC+log(x+y)` `rArr" "x-y=log|C(x+y)|` `rArr" "e^(x-y)=C(x+y)` `rArr" "(x+y)=(1)/(C)e^(x-y)` `rArr" "x+y=Ke^(x-y)" "[becauseK=(1)/(2)]` |
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| 629. |
Solve the following differential equations:`x y(dy)/(dx)=x^2-y^2`A. `(y^(2))/(x^(2))=2logx+c`B. `(y)/(x)=2logx+c`C. `y=2xlogx+c`D. `x^(2)+y^(2)=x^(2)y^(2)=c` |
| Answer» Correct Answer - A | |
| 630. |
Solve the following differential equations:`(dy)/(dx)=1+x+y+x y`(ii)`y-x(dy)/(dx)=a(y^2+(dy)/(dx))` |
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Answer» Correct Answer - `y/(1-ay)=c(a+y)` `y-x(dy)/(dx)=a(y^(2)+(dy)/(dx))` or `int(dx)/(a+x)=int(dy)/(y-ay^(2))=int(1/y+a/(1-ay))dy` [By partial fractions] Integrating we get `log(a+x)+logc=logy-log(1-ay)` Which is arbitary positive constant. Thus, the solution can be written as `y/(1-ay)=c(a+x)` |
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| 631. |
Solve the following differential equations `(dy)/(dx)=sinx*siny`A. `e^(cosx)tany/2=c`B. `e^(cosx)tany=c`C. `cosxtany=c`D. `cosxsiny=c` |
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Answer» Correct Answer - A `dy-sinxsinydx=0` or `int"cosec "ydy=intsinxdx` or `logtany/2=-cosx+logc` or `log(tany/2)/(c )=-cosx` or `(tany/2)/( c) = e^(-cosx)` or `e^(cosx)tany/2=c` |
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| 632. |
`y-x(dy)/(dx)=3y^(2)+(dy)/(dx)`A)`(1+x)(1-cy)=3y` B)`(1+x)(1-3y)=cy` C)`(1-x)(1-3y)=cy` D)`(1+3x)(1-y)=cx`A. `(1+x)(1-cy)=3y`B. `(1+x)(1-3y)=cy`C. `(1-x)(1-3y)=cy`D. `(1+3x)(1-y)=cx` |
| Answer» Correct Answer - B | |
| 633. |
Let a solution `y=y(x)` of the differential equation `(dy)/(dx)cosx+y sin x-1` satisfy y(0)=1 Statement-1: `y(x)=sin((pi)/(4)+x)` Statement-2: The integrating factor of the given differential equation is sec x.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. |
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Answer» Correct Answer - A We have, `(dy)/(dx)cos x+y sin x=1` `rArr" "(dy)/(dx)+y tan x=sec x` This is a linear differential equation with I.F. given by `"I.F."=e^(inttanxdx)=e^(logsecx)=secx` So, statement-2 is true. Multiplying both sides of by I.F. = sec x and integrating w.r. to x, we get `y sec x=tan x+C" ...(ii)"` It is given that y = 1 when x = 0. `therefore" 1 = C"` Putting C = 1 in (ii), we get `ysec x =tan x+1` `rArr" "y=sin x +cos x=sqrt2 sin ((pi)/(4)+x)` So, statement-1 is true and statement-2 is a correct explanation for statement-1. |
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| 634. |
Solve the following differential equations : ` y -x (dy)/(dx) =3 (1+x^(2) (dy)/(dx))`A. `(1+3x)(3-y)=cy`B. `(3x-1)(y+3)=cx`C. `(1-3x)(3-y)=cy`D. `(3x+1)(y-3)=cx` |
| Answer» Correct Answer - D | |
| 635. |
Let `u(x)`and `v(x)`satisfy thedifferential equation `(d u)/(dx)+p(x)u=f(x)`and `(d v)/(dx)+p(x)v=g(x)`arecontinuous functions. If `u(x_1)`for some `x_1`and `f(x)>g(x)`for all `x > x_1,`prove thatany point `(x , y),`where `x > x_1,`does notsatisfy the equations `y=u(x)`and `y=v(x)dot` |
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Answer» Let `w(x)= u(x) - v(x) ...(i)` and `h(x) = f(x) -g(x)` On differentiating Eq. (i) w.r.t.x `(dw)/dx= (du)/dx-(dv)/dx` `={f(x)-p(x)cdotu(x)}-{g(x)-p(x) v(x)}` [given] `={f(x)-g(x)}-p(x)[u(x)-u(x)]` `rArr (dw)/dx=h(x)-p(x)cdot w(x) …(ii)` `rArr (dw)/dx+p(x) w(x)=h(x)` shich is linear differential equation. The integrating factor is given by `IF=e^(int p(x)dx)=r(x)` [let] on multiplying both sides of Eq. (ii) of r(x), We get `r(x)cdot (dw)/dx+p(x) (r(x)) w(x) = r(x) cdot h (x)` `rArr d/dx[r(x) w(x)]=r(x) cdot h(x) [therefore (dr)/dx=p(x)cdotr(x)]` Now, `r(x) =e^(intp(x)dx)gt0,AAx` and `h(x)=f (x) -g(x)gt0,for xgtx_(1)` Thus, `d/dx[r(x)w(x)]gt0,AAxgt x_(1)` ` r(x)w(x)` increases on the interval ` [x, infty[` Therefore, for all `x gt x_(1)` `r (x) w(x) gt r(x_1)w(x_1)gt0` `[therefore r(x_1)gt0 and u (x_1) gt v(x_1)]` `rArr w(x) gt 0AA x gt x_(1)` `rArr u(x)gtv(x)AAxgtx_(1) [therefore r(x)gt0]` Hence, there cannot exist a point (x,y) such that `xgtx_(1)` and y=u (x) and y=v (x). |
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| 636. |
Let a solution y = y(x) of the differential equation `xsqrt(x^2-1) dy - y sqrt(y^2-1) dx=0`, satisfy `y(2)= 2/sqrt 3`A. (a) Statement I is true, Statement II is also true, Statement II is the correct explanation of Statement I.B. (b) Statement I is true, Statement II is also true, Statement II is not the correct explanation of Statement I.C. (c) Statement I is true, Statement II is false.D. (d) Statement I is false, Statement II is true. |
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Answer» Correct Answer - (c) Given , `dy/dx=(ysqrt(y^(2)-1))/(xsqrt(x^(2)-1))` `int dy/ (ysqrt(y^(2)-1))=intdx/(xsqrt(x^(2)-1))` `rArr sec^(-1) y=sec^(-1) x+c` `At x=2, y = 2/sqrt(3), pi/6=pi/3+c` `rArr c=- pi/6` Now, `y = sec (sec^(-1)x-pi/6)` `=cos [cos^(-1)frac {1}{x}-cos^(-1) frac{sqrt(3)}{2}]` `= cos [ cos ^(-1) (sqrt(3)/(2x)+sqrt(1-1/x^(2))sqrt(1-3/4)]` `y=sqrt(3)/(2x)+1/2 sqrt(1-1/x^(2))` |
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| 637. |
`(tany)(dy)/(dx)=sin(x+y)-sin(x-y)`A)`secy tany = 2 sin x+c` B)`log(cosy-coty)=2sin x+c` C)`log(secy+tany)=2sin x+c` D)`secy+tany=c.e^(sinx)`A. `secy tany = 2 sin x+c`B. `log(cosy-coty)=2sin x+c`C. `log(secy+tany)=2sin x+c`D. `secy+tany=c.e^(sinx)` |
| Answer» Correct Answer - C | |
| 638. |
If the integrating factor of the differential equation `(dy)/(dx) +P` (x) y = Q (x) is x , then P (x) isA. xB. `(1)/(2)x^(2)`C. `(1)/(x)`D. `(1)/(x^(2))` |
| Answer» Correct Answer - C | |
| 639. |
Integrating factor of the differential equation `(x.logx)(dy)/(dx)+y=2logx` isA. `e^(x)`B. `logx`C. `log(logx)`D. `x` |
| Answer» Correct Answer - B | |
| 640. |
An integrating factor of the DE : `(1+x)(dy)/(dx)-xy=1-x` isA. `(1+x).e^(-x)`B. `log(1+x)`C. `1+x`D. `x.e^(x)` |
| Answer» Correct Answer - A | |
| 641. |
The integrating factor of differential euation `(1-x^(2))(dy)/(dx)-xy=1 "is"`A. `-x`B. `(x)/(1+x^(2))`C. `sqrt(1-x^(2))`D. `(1)/(x)log(1-x^(2))` |
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Answer» Given that, `" "(1-x^(2))(dy)/(dx)-xy=1` `rArr" "(dy)/(dx)-(x)/(1-x^(2))y=(1)/(1-x^(2))` which is a linear differential equation. `therefore" "IF=e^(-int(x)/(1-x^(2))dx)` Put `" "1-x^(2)=trArr-2xdx=dtrArrxdx=-(dt)/(2)` Now, `" "IF=e^((1)/(2)int(dt)/(t))=e^((1)/(2)logt)=e^((1)/(2)log(1-x^(2)))=sqrt(1-x^(2))` |
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| 642. |
The integrating solution of differential equation `(dy)/(dx) = e^((x^(2))/(2)) + xy` isA. `y = ce^(-(x^(2))/(2))`B. `y = ce^((x^(2))/(2))`C. `y = (x+c)e^((x^(2))/(2))`D. `y = (c-x)e^((x^(2))/(2))` |
| Answer» Correct Answer - C | |
| 643. |
The integrating factor of the differential equation `(1-x^(2))(dy)/(dx)-xy=1`, isA. `-x`B. `(x)/(1x^(2))`C. `sqrt(1-x^(2))`D. `(1)/(2)log(1-x^(2))` |
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Answer» Correct Answer - C We have, `(1-x^(2))(dy)/(dx)=xy-1rArr(dy)/(dx)-((x)/(1-x^(2)))y=(1)/(1-x^(2))` It is a linear differential equation with I.F. given by `"I.F. "=e^(-int(x)/(1-x^(2))dx)=e^((1)/(2)log(1-x^(2)))=sqrt(1-x^(2))` |
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| 644. |
Find the real value of `m`for which the substitution `y=u^m`will transform the differentialequation `2x^4y(dy)/(dx)+y^4=4x^6`in to a homogeneous equation. |
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Answer» `y=u^(m)` or `(dy//dx) = mu^(m-1)(du)/(dx)` The given differential equation becomes `2x^(4).u^(m). Mu^(m-1)(du)/(dx) +u^(4m)=4x^(6)` for homogenous equation, degree of each term should be same in the numerator and the denominator. Hence, `6=4m=4+2m-1` or `m=3//2` |
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| 645. |
The differential equation `(dy)/(dx)=(x+y-1)/(x+y+1)` reduces to variable separable form by making the substitutionA. `x+y=v`B. `x-y=v`C. `y=vx`D. `x=vy` |
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Answer» Correct Answer - A Let `x+y=v`. Then, `1+(dy)/(dx)=(dv)/(dx)` Substituting these values in the given differential equaiton, we get `(dv)/(dx)-1=(v-1)/(v+1)rArr (dv)/(dx)=(2v)/(v+1)rArr(v+1)/(2v)dv=dx` Clearly, it is in variable separable form. |
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| 646. |
If `x(dy)/(dx)=y(log y -logx+1),` then the solution of the equation isA. `log((x)/(y))=Cy`B. `log((y)/(x))=Cx`C. `x log((y)/(x))=Cy`D. `ylog((x)/(y))=Cx` |
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Answer» Correct Answer - B We have, `x(dy)/(dx)=ylog((y)/(x))+y` Putting `y=vx` and `(dy)/(dx)=v+x(dv)/(dx)`, we get `v+x(dv)/(dx)=vlog v +v` `rArr" "(1)/(v log v) dv=(1)/(x)dx` `rArr" "log(logv)=logx+logC" [On integrating]"` `rArr" "logv=Cx` `rArr" "log((y)/(x))=Cx`, which is the required solution. |
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| 647. |
The integrating factor of the differential equation `(1-x^(2))(dy)/(dx)-xy=1`, isA. `-x`B. `(-x)/(1+x^(2))`C. `sqrt(1-x^(2))`D. `(1)/(2).log(1-x^(2))` |
| Answer» Correct Answer - C | |
| 648. |
Which of the following is a homogenous differential equation ?A. `(4x+6y+5)dy-(3y+2x+4)dx=0`B. `(xy)dx-(x^(3)+y^(3))dy=0`C. `(x^(3)+2y^(2))dx+2xydy=0`D. `y^(2)dx+(x^(2)-xy-y^(2))dy=0` |
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Answer» `(a)` Given equation `(4x+6y+5)dy-(3y+2x+4)dx=0` can be written as `(dy)/(dx)=(3y+2x+4)/(4x+6y+5)implies(dy)/(dx)=(x(3*(y)/(x)+2+(4)/(x))/(x(4+6(y)/(x)+(5)/(x))` whichis non-homogenous differential equation. `(b)` Given differential equation `xydx-(x^(3)+y^(3))dy=0` from the separating of variables, `(x^(3)+y^(3))dy=xydximplies(dy)/(dx)=(xy)/(x^(3)+y^(3))` `implies (dy)/(dx)=((y)/(x^(2)))/(1+((y)/(x))^(3))` which is non-homogenous differential equation `(c )` Given differential eqaution `(x^(3)+2y^(2))dx+2 xy dy=0` `2xy dy=-(x^(3)+2y^(2))dx` From the separating of variables, `(dy)/(dx)=-(x^(3)+2y^(2))/(2xy)implies(dy)/(dx)=(1+(2y^(2))/(x^(3)))/((2y)/(x^(2)))` which is non-homogenous differential equation. `(d)` Given differential equation `y^(2)dx+(x^(2)-xy-y^(2))dy=0` From the separating of variables, `(x^(2)-xy-y^(2))dy=-y^(2)dx` `implies (dy)/(dx)=-(y^(2))/(x^(2)-xy-y^(2))` `implies (dy)/(dx)=-(((y)/(x))^(2))/([1-(y)/(x)-((y)/(x))^(2)])` which is a homogenous differential equation. |
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| 649. |
The differential equation `(dy)/(dx)=(7x-3y-7)/(-3x+7y+3)` reduces to homogeneous form by making the substitutionA. `x=X+1, y=Y+0`B. `x=X+1,y=Y+1`C. `x=X-1, y=Y+1`D. `x=X+0, y=Y+1` |
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Answer» Correct Answer - A Let `x=X+h, y=Y+k`. Then, the given equation becomes `(dY)/(dX)=((7X-3Y)+(7h-3k-7))/((-3X+7Y)+(03h+7k+3))` This will reduces to a homogeneous differential equation, if `7h-3k-7=0 and -3h+7k+3=0 rArr h=1, k=0` Hence, `x=X+1, y=Y+0` reduces the given equation to homogeneous form. |
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| 650. |
Thedifferential equations , find the particular solution satisfying the givencondition:(x + y) dy + (x y) dx = 0; y = 1 when x =1 |
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Answer» Given that, `(x+y)dy+(x-y)dx=0` `implies (dy)/(dx)=(y-x)/(x+y)`………`(1)` Given differential equation is homogenous. Let `y=vx` `implies (dy)/(dx)=v+x(dv)/(dx)` `implies` From equation `(1)`, `v+x(dv)/(dx)=(vx-x)/(x+vx)` `implies v+x(dv)/(dx)=(x(v-1)/(x(1+v)impliesv+x(dv)/(dx)=(v-1)/(1+v)` `impliesx(dv)/(dx)=(v-1)/(1+v)-v` `impliesx(dv)/(dx)=(v-1-v-v^(2))/(1+v)` `implies -x(dv)/(dx)=(1+v^(2))/(1+v)=((v+1))/(1+v^(2))dv=-(1)/(x)dx` On integration, `int((v+1))/(1+v^(2))dv=-int(dx)/(x)` `impliesint(v)/(1+v^(2))dv+int(1)/(1+v^(2))dv=-int(dx)/(x)` `implies(1)log|v^(2)+1|+tan^(-1)v+log|x|=C` `implies(1)/(2)log((y^(2)+x^(2))/(x^(2)))+log|x|+tan^(-1)((y)/(x))=C` `implies log((y^(2)+x^(2))/(x^(2)))+2log|x|+2tan^(-1)((y)/(x))=2C` `implieslog((y^(2)+x^(2))/(x^(2)))+logx^(2)+2tan^(-1)((y)/(x))=2C` `implies log((y^(2)+x^(2))x^(2))/(x^(2))+2tan^(-1)((y)/(x))=2C` `implies log(x^(2)+y^(2))+2tan^(-1)((y)/(x))=A` (put `2C=A`) .........`(2)` Given, `x=1`, `y=1` `log2+(2xx(pi)/(4))=A` `implies A=(pi)/(2)+log2` put the value of `A` in equation `(2)`, `log(x^(2)+y^(2))+2tan^(-1)((y)/(x))=(pi)/(2)+log2` which is the required solution of the given differential equation |
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