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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.
| 351. |
If `(dx)/(dy)=(e^(y)-x)`, where y(0)=0, then y is expressed explicitly asA. `(1)/(2)ln(1+x^(2))`B. `ln(1+x^(2))`C. `ln(x-sqrt(1+x^(2)))`D. `ln(x+sqrt(1-x^(2)))` |
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Answer» Correct Answer - C We have, `(dx)/(dy)=e^(y)-x rArr(dx)/(dy)+x=e^(y)" ...(i)"` This is a linear differential equation with I.F. `=e^(y)`. Multiplying both sides of (i) by `"I.F."=e^(y)` and integrating with respect to y, we obtain `xe^(y)=(1)/(2)e^(2y)+C` It is given y = 0 when x = 0. Putting these values in (ii), we obtain `C=-(1)/(2)`. Substituting `C=-(1)/(2)` in (ii), we obtain `xe^(y)=(1)/(2)e^(2y)-(1)/(2)` `rArr" "e^(2y)-2xe^(y)-1=0` `rArr" "e^(y)=x+sqrt(x^(2)+1)rArry=ln(x+sqrt(x^(2)+1))` |
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| 352. |
Solve the differential equation `y e^(x/y)dx=(x e^(x/y)+y^2)dy(y!=0)`A. `e^(x//y)=cosy+C`B. `e^(x//y)=-siny+C`C. `e^(y//x)=-cosy+C`D. `e^(x//y)=-cosy+C` |
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Answer» Correct Answer - C We have, `ye^(x//y)dx=(xe^(x//y)+y^(2)siny)dy` `rArr" "(ydx-xdy)e^(x//y)=y^(2)sinydy` `rArr" "e^(x//y){(ydx-xdy)/(y^(2))}=sin ydy` `rArr" "e^(x//y)d((x)/(y))=-d(cosy)` On integrating, we obtain `e^(x//y)=-cosy+C`, as the required solution. |
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| 353. |
The family of curves represented by `(dy)/(dx)=(x^(2)+x+1)/(y^(2)+y+1)` and the family represented by `(dy)/(dx)+(y^(2)+y+1)/(x^(2)+x+1)=0`A. `2(x^(3)-y^(3))+3(x^(2)-y^(2))+6(x-y)=c`B. `2(x^(3)-y^(3))+3(x^(2)+y^(2))+6(x-y)=c`C. `2(x^(3)-y^(3))+3(y^(2)-x^(2))+6(y-x)=c`D. `2(x^(3)-y^(3))-3(x^(2)-y^(2))+6(x-y)=c` |
| Answer» Correct Answer - A | |
| 354. |
Solution of the differential equation `(xdy)/(x^(2)+y^(2))=((y)/(x^(2)+y^(2))-1)dx`, isA. `tan^(-1)((y)/(x))=-x+C`B. `tan^(-1)((y)/(x))=x+C`C. `tan^(-1)((x)/(y))=-x+C`D. `tan^(-1)((y)/(x))=-y+C` |
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Answer» Correct Answer - A The given differential equation can be written as `(xdy-ydx)/(x^(2)+y^(2))=-dxrArrd{tan^(-1)((y)/(x))}=-dx` On integrating, we obtain `tan^(-1)((y)/(x))=-x+fC`, which is the required solution. |
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| 355. |
The solution of differential equation `xdx+ydy=a(x^(2)+y^(2))dy`,isA. `x^(2)+y^(2)=Ce^(ay)`B. `x^(2)+y^(2)=Ce^(2ay)`C. `x^(@)+y^(2)=e^(2Cay)`D. none of these |
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Answer» Correct Answer - B We have, `xdx+ydy=a(x^(2)+y^(2))dy` `rArr" "(2xdx+2ydy)/(x^(2)+y^(2))=2adyrArr(d(x^(2)+y^(2)))/(x^(2)+y^(2))=2ady` On integrating, we obtain `log(x^(2)+y^(2))=2ay+logC` `rArr" "x^(2)+y^(2)=Ce^(2ay)` is the required solution. |
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| 356. |
The general solution of the differential equation `(dy)/(dx)+(2)/(x)y=x^(2)`, isA. `y=cx^(2)+(x^(3))/(5)`B. `y=cx^(-2)+(x^(3))/(5)`C. `y=cx^(3)-(x^(3))/(4)`D. `y=cx^(-3)(x^(2))/(4)` |
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Answer» Correct Answer - B The general solution of the differential equation `(dy)/(dx)+Py=Q`, where `P=(2)/(x) and Q=x^(2)`. I.E. `=e^(intPdx)=e^(int(2)/(x)dx)=e^(2logx)+x^(2)` Multiplying both sides of the differential equation by I.E. `=x^(2)` and integrating with respect to x, we get `yx^(2)=(x^(5))/(5)+c or, y=(x^(3))/(5)+cx^(-2)` |
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| 357. |
The sum of the squares of the perpendicular drawn from the points (0,1) and `(0,-1)` to any tangent to a curve is 2. The equation of the curve, isA. `2y=C(x+2)`B. `y=C(x+1)`C. `y=C(x+2)`D. `y=C(x+2)` |
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Answer» Correct Answer - B The equation of any tangent to a curve `y=f(x)` is `y-y=(dy)/(dx)(X-x)or, X(dy)/(dx)-Y+(y-x(dy)/(dx))=0" ...(i)"` It is given that `|(-1+y-x(dy)/(dx))/(sqrt(((dy)/(dx))^(2)+1))|^(2)+|(1+y-x(dy)/(dx))/(sqrt(((dy)/(dx))^(2)+1))|^(2)=2` `rArr" "2{(y-x(dy)/(dx))^(2)+1}=2{1+((dy)/(dx))^(2)}` `rArr" "(y-x(dy)/(dx))^(2)=((dy)/(dx))^(2)` `rArr" "y-x(dy)/(dx)=pm(dy)/(dx)` `rArr" "(1)/(xpm1)dx=(1)/(dy)dy` `rArr" "log(xpm1)=logy+logC` `rArr" "log(xpm1)=logy+logC` `rArr" "Cy=x pm1 or,y=k(x pm1)` |
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| 358. |
The differential equation `(dy)/(dx)+P y=Q y^n , n >2`can be reduced to linear form y substituting`z=y^(n-1)`b. `z=y^n`c. `z=y^(n+1)`d. `z=y^(1-n)`A. `z=y^(n-1)`B. `z=y^(n)`C. `z=y^(n+1)`D. `z=y^(1-n)` |
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Answer» Correct Answer - D We have, `(1)/(y^(n))(dy)/(dx)+P(1)/(y^(n-1))=Q` It reduces to linear form by substituting `(1)/(y^(n-1))=z` |
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| 359. |
The differential equation `x(dy)/(dx) -y=x^3,` has the general solutionA. `y=x^(3)=2Cx`B. `2y-x^(3)=Cx`C. `2y+x^(2)=2Cx`D. `y+x^(2)=2Cx` |
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Answer» Correct Answer - B We have, `(dy)/(dx)+(-(1)/(x))y=x^(2)` It is a linear differential equation with integrating factor `"I.F. "=e^(int-(1)/(x)dx)=e^(-logx)=(1)/(x)` Multiplying (i) by `(1)/(x)` and integrating, we get `(y)/(x)=(x^(2))/(2)+C or, 2y-x^(3)=2Cx` |
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| 360. |
If integrating factor of `x(1-x^2)dy+(2x^2y-y-a x^3)dx=0`is`e^(intp dx ,)`then `P`is equal to(a)`( b ) (c) (d)(( e )2( f ) x^(( g )2( h ))( i )-a (j) x^(( k )3( l ))( m ))/( n )(( o ) x(( p ) (q)1-( r ) x^(( s )2( t ))( u ) (v)))( w ) (x) (y)`(z)(b) `( a a ) (bb)2( c c ) x^(( d d )3( e e ))( f f )-1( g g )`(hh)(c)`( d ) (e) (f)(( g )2( h ) x^(( i )2( j ))( k )-a)/( l )(( m ) a (n) x^(( o )3( p ))( q ))( r ) (s) (t)`(u)(d) `( v ) (w) (x)(( y )2( z ) x^(( a a )2( b b ))( c c )-1)/( d d )(( e e ) x(( f f ) (gg)1-( h h ) x^(( i i )2( j j ))( k k ) (ll)))( m m ) (nn) (oo)`(pp)A. `(2x^(2)-ax^(3))/(x(1-x^(2))`B. `2x^(3)-1)`C. `(2x^(2)-1)/(ax^(3))`D. `(2x^(2)-1)/(x(1-x^(2))` |
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Answer» Correct Answer - D We have, `(dy)/(dx)+((2x^(2)-1))/(x(1-x^(2)))y=(ax^(2))/(1-x^(2))` This is a linear differential equation of the form `(dy)/(dx)+Py=Q`, where `P=(2x^(2)-1)/(x(1-x^(2))` |
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| 361. |
The solution of differential equation `(dy)/(dx)=(1+y^(2))/(1+x^(2))"is"`A. `y=tan^(-1)x`B. `y=x=k(1+xy)`C. `x=tan^(-1)y`D. `tan(xy)=k` |
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Answer» Given that, `" "(dy)/(dx)=(1+y^(2))/(1+x^(2))` `rArr" "(dy)/(1+y^(2))=(dx)/(1+x^(2))` On integrating both sides, we get `" "tan^(-1)y=tan^(-1)x+C` `rArr" "tan^(-1)y-tan^(-1)x=C` `rArr" "tan^(-1)((y-x)/(1+xy))=C` `rArr" "(y-x)/(1+xy)=tanC` `rArr" "y-x=tanc(1+xy)` `rArr" "y-x=K(1+xy)` where, `" "k=tanC` |
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| 362. |
The differential equation of all circles in the first quadrant which touch the coordiante axes is of orderA. 1B. 2C. 3D. none of these |
| Answer» Correct Answer - A | |
| 363. |
The solution of `x(dy)/(dx)+y=e^(x)"is"`A. `y=(e^(x))/(x)+(k)/(x)`B. `y=(e^(x))/(x)+(k)/(x)`C. `y=xe^(x)+k`D. `x=(e^(y))/(y)+(k)/(y)` |
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Answer» Given, that `x(dy)/(dx)+y=e^(x)` `Rightarrow (dy)/(dx)+(y)/(x)=(e^(x))/(x)` Which is linear differential equation. `IF=e^(int(1)/(2)dx)=e^(logx)=x` The general solution is `y=int((dx)/(x).x)dx` `Rightarrow y.x.=inte^(x)dx` `Rightarrow y.x=e^(x)+k` `Rightarrow y=(e^(x))/(x)+(k)/(x) |
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| 364. |
The integrating factor of differential equation `(dy)/(dx)+y=(1+y)/(x)"is"`A. `(x)/(x^(x)`B. `(e^(x))/(x)`C. `xe^(x)`D. `e^(x)` |
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Answer» Given that, `" "(dy)/(dx)+y=(1+y)/(x)` `rArr" "(dy)/(dx)=(1+y)/(x)-y` `rArr" "(dy)/(dx)=(1+y-xy)/(x)` `rArr" "(dy)/(dx)=(1)/(x)+(y(1-x))/(x)` `rArr" "(dy)/(dx)-((1-x)/(x))y=(1)/(x)` Here, `" "P=(-(1-x))/(x),Q=(1)/(x)` `" "IF=e^(intPdx)=e^(-int(1-x)/(x)dx)=e^(int(x-1)/(x)dx)` `" "=e^(int(1-(1)/(x))dx` `" "e^(intx-logx)` `" "=e^(x)*e^(log((1)/(x)))` `" "=e^(x)*(1)/(x)` |
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| 365. |
The order & the degree of the differential equation whose general solution is, `y =c(x-c)^2`, are respectivelyA. 1,1B. 1,2C. 1,3D. 2,1 |
| Answer» Correct Answer - C | |
| 366. |
The solution of the differential equation `dy/dx=cos(x-y)` isA. `y+cot ((x-y)/(2)) = c`B. `x+cot ((x-y)/(2)) = c`C. `x + tan((x-y)/(2)) = c`D. `x + tan ((x+y)/(2)) = c` |
| Answer» Correct Answer - B | |
| 367. |
The solution of the differential equation `(dy)/(dx)=e^(y+x)+e^(y-x),` isA. `e^(-y)=e^(x)-e^(-x)+C`B. `e^(-y)=e^(-x)-e^(x)+C`C. `e^(-y)=e^(x)+e^(-x)+C`D. `e^(-y)+e^(x)+e^(-x)=C` |
| Answer» Correct Answer - B | |
| 368. |
The solution the differential equation `"cos x sin y dx" + "sin x cos y dy" =0` isA. `(sin x)/(sin y)=C`B. `sin x sin y =C`C. `sin x+sin y =C`D. `cos x cos y =C` |
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Answer» Given differential equation is `cos x sin ydx+ sin x cos ydy=0` `Rightarrow (cos x)/(sinx )dx=(cos y)/(sin y)dy` `Rightarrow cot x dx=-cot ydy` On integrating both sides, we get `log sin x=-log sin y+log C` `Rightarrow log sin x siny=logC` `Rightarrow sinx. siny=C` |
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| 369. |
`(dy)/(dx) = (xy+y)/(xy+x)`, then the solution of differential equation isA. `y = xe^(x) + c`B. `y = e^(x) + c`C. `y = c x e^(x-y)`D. y = x+c |
| Answer» Correct Answer - C | |
| 370. |
`y=ae^(mx)+be^(-mx)` satisfies which of the following differential equations?A. `(dy)/(dx)+my=0`B. `(dy)/(dx)-my=0`C. `(d^(2)y)/(dx^(2))-m^(2)y=0`D. `(d^(2)y)/(dx)+m^(2)=0` |
| Answer» Correct Answer - C | |
| 371. |
`y=ae^(mx)+b^(-mx)` satisfies which of the following differential equation?A. `(dy)/(dx)+my=0`B. `(dy)/(dx)-my=0`C. `(d^(2)y)/(dx^(2))-m^(2)y=0`D. `(d^(2)y)/(dx^(2))+m^(2)y=0` |
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Answer» Given that, `" "y=ae^(mx)+be^(-mx)` On differentiating both sides w.r.t. x, we get `" "(dy)/(dx)=mae^(mx)-bme^(-mx)` Again, differentiating both sides w.r.t. x, we get `" "(d^(2)y)/(dx^(2))=m^(2)ae^(mx)+bm^(2)e^(-mx)` `rArr" "(d^(2)y)/(dx^(2))=m^(2)(ae^(mn)+be^(-mn))` `rArr" "(d^(2)y)/(dx^(2))=m^(2)y` `rArr" "(d^(2)y)/(dx^(2))-m^(2)y=0` |
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| 372. |
The differential equation ydy+xdx = dx representsA. A set of circles with centre on x axisB. A set of concentric circlesC. A set of ellipseD. A set of circles with centre on y axis |
| Answer» Correct Answer - A | |
| 373. |
Find the degree of the differential equation `((d^(3)y)/(dx^(3))) + 2 ((d^(2)y)/(dx^(2))) + (dy)/(dx) + y = 0`. |
| Answer» The highest power of the highest order derivative i.e., `(d^(3)y)/(dx^(3))` is one, therefore its degree is one. | |
| 374. |
`y=ae^(mx)+be^(-mx)` satisfies which of the following differential equations?A. `(dy)/(dx)-my=0`B. `(dy)/(dx)+xy=0`C. `(d^(2)y)/(dx^(2))+m^(2)y=0`D. `(d^(2)y)/(dx^(2))m^(2)y=0` |
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Answer» Correct Answer - D We have, `y=ae^(mx)+be^(-mx)` `rArr" "(dy)/(dx)=mae^(mx)-mbe^(-mx)` `rArr" "(d^(2)y)/(dx^(2))=m^(2)(ae^(mx)+be^(-mx))` `rArr" "(d^(2)y)/(dx^(2))=m^(2)yrArr (d^(2)y)/(dx^(2))-m^(2)y=0` Hence, option (d) is correct |
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| 375. |
Find the degree of the differential equation `((d^(2)y)/(dx^(2)))^((2)/(3))-(dy)/(dx) - y = 0` |
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Answer» `((d^(2)y)/(dx^(2)))^((2//3)) = (dy)/(dx) + y` `rArr ((d^(2)y)/(dx^(2)))^(2) = ((dy)/(dx) + y)^(3)` The degree of the differential equation is two. |
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| 376. |
If f (x) and g (x) are two solutions of the differential equation `(a(d^2y)/(dx^2)+x^2(dy)/(dx)+y=e^x, then `f(x)-g(x)` is the solution ofA. `a^(2)(d^(2)y)/(dx^(2))+(dy)/(dx)+y=e^(x)`B. `a^(2)(d^(2)y)/(dx^(2))+y=e^(x)`C. `a(d^(2)y)/(dx^(2))+y=e^(x)`D. `a(d^(2)y)/(dx^(2))+x^(2)(dy)/(dx)+y=0` |
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Answer» Correct Answer - D It is given that f(x) and g(x) are solutions of the differential equation `a(d^(2)y)/(dx^(@))+x^(2)(dy)/(dx)+y=e^(x)` `therefore" "a(d^(2))/(dx^(2)){f(x)}+x^(2)(d)/(dx){f(x)}+f(x)=e^(x)` and, `(d^(2))/(dx^(2)){g(x)}+x^(2)(d)/(dx){g(x)}+g(x)e^(x)` `rArra(d^(2))/(dx^(2)){f(x)-g(x)}+x^(2)(d)/(dx){f(x)-g(x)}+{f(x)-g(x)}=0` `rArr" "f(x)-g(x)` is a solution of the differential equation `a(d^(2)y)/(dx^(2))+x^(2)(dy)/(dx)+y=0` |
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| 377. |
The solution of the differential equation `(1+x^(2))(dy)/(dx)+1+y^(2)=0`, isA. `tan^(-1)x-tan^(-1)y=tan^(-1)C`B. `tan^(-1)y-tan^(-1)x=tan^(-1)C`C. `tan^(-1)y pm tan^(-1)x=tanC`D. `tan^(-1)y+tan^(-1)x=tan^(-1)C` |
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Answer» Correct Answer - D If `tan^(-1)y+tan^(-1)x=tan^(-1)C`, then `(d)/(dx)(tan^(-1)y)+(d)/(dx)(tan^(-1)x)=0` `rArr" "(1)/(1+y^(2))(dy)/(dx)+(1)/(1+x^(2))=0rArr(1+x^(2))(dy)/(dx)+(1)/(1+y^(2))=0` Hence, `tan^(-1)x+tan^(-1)y=tan^(-1)C` is the solution of the given differential. |
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| 378. |
The solution of the differential equation `(1+x^(2))(dy)/(dx)+1+y^(2)=0`, isA. `tan^(-1)x-tan^(-1)y=tan^(-1)C`B. `tan^(-1)y-tan^(-1)x=tan^(-1)C`C. `tan^(-1)y pm tan^(-1)x=tanC`D. `tan^(-1)y+tan^(-1)x=tan^(-1)C` |
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Answer» Correct Answer - D If `tan^(-1)y+tan^(-1)x=tan^(-1)C`, then `(d)/(dx)(tan^(-1)y)+(d)/(dx)(tan^(-1)x)=0` `rArr" "(1)/(1+y^(2))(dy)/(dx)+(1)/(1+x^(2))=0rArr(1+x^(2))(dy)/(dx)+(1)/(1+y^(2))=0` Hence, `tan^(-1)x+tan^(-1)y=tan^(-1)C` is the solution of the given differential. |
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| 379. |
The solution of the differential equation`((dy)/(dx))^2-x(dy)/(dx)+y=0`is(a)`( b ) (c) y=2( d )`(e)(b) `( f ) (g) y=2x (h)`(i)(c)`( d ) (e) y=2x-4( f )`(g)(d) `( h ) (i) y=2( j ) x^(( k )2( l ))( m )-4( n )`(o)A. y = 2B. y = 2xC. y = `2x-4`D. `y=2x^(2)-4` |
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Answer» Correct Answer - C If, `y=2x-4,` then `(dy)/(dx)=2`. Clearly, these values satisfy the differential equation. |
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| 380. |
Solve the differential equation (2 + x) dy = (1 + y) dx` |
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Answer» `(dy)/(1+y) = (dx)/(2 + x)` Integrating both sides, we get, `rArr In|1+y| = In|2+x| + InC` `rArr (1+y) = C(2+x)` |
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| 381. |
Solve the differential equation `(1 + cos 2x) + (1- cos 2y) (dx)/(dy) = 0` |
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Answer» `(1+cos 2x) dy + (1- cos 2y) dx = 0` `rArr (dy)/(1-cos 2y) + (dx)/(1+cos 2x) = 0` Integrating both sides `rArr (1)/(2)int (1)/(sin^(2)y) dx + (1)/(2) int (1)/(cos^(2)x)dx = C_(1)` `rArr int cosec^(2) y dy + int sec^(2) x dx = C_(1)` `rArr -cot y + tan x = C` `rArr tan x - cot y = C` which is the general solution of the given differential equation. |
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| 382. |
Verify that`y=c e^(t a n-1_x)`is a solution of differential equation`(1+x^2)(d^2y)/(dx^2)+x(dy)/(dx)=0.` |
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Answer» `y=c*e^(tan^(-1)x)` …………`(1)` Differentiate with respect to `x` `(dy)/(dx)=c*e^(tan^(-1)x)*(1)/(1+x^(2))` `implies (1+x^(2))(dy)/(dx)=c*e^(tan^(-1)x)` `implies (1+x^(2))(dy)/(dx)=y` [From eq. `(1)`] Again differentiate with respect to `x` `(1+x^(2))(d^(2)y)/(dx^(2))+(dy)/(Dx)*2x=(dy)/(dx)` `implies (1+x^(2))(d^(2)y)/(dx^(2))+(2x-1)(dy)/(dx)=0` `:. y=c*e^(tan^(-1)x)` is a solution of the given differential eqaution. |
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| 383. |
Solve the differential equation `e^(x) tan y dx + (1-e^(x))sec^(2) y dy = 0` |
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Answer» Integrating the equation, we get `int (e^(x))/(1-e^(x))dx + int(sec^(2)y)/(tan y)dy = 0` Put `1-e^(x) = t` for `1^(st)` integral and tan y = u for `2^(nd)` integral `rArr -e^(x)dx = dt` and `sec^(2) y dy = du` `rArr int (-dt)/(t) + int(d u)/(u) = 0` `rArr -In|t| + In|u| = InC` `rArr u = Ct` `rArr tan y = C(1-e^(x))` |
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| 384. |
The equation of the curve whose subnormal is constant isA. `y=ax+b`B. `y^(2)=2ax+b`C. `ay^(2)-x^(2)=a`D. none of these |
| Answer» Correct Answer - B | |
| 385. |
The solution of `(dy)/(dx)+y=e^(-x),y(0)=0`, isA. `y = e^(-x)(x-1)`B. `y = xe^(x)`C. `y = xe^(-x) + 1`D. `y = xe^(-x)` |
| Answer» Correct Answer - D | |
| 386. |
The solution of the differential equation `(x)/(x^(2)_y^(2))dy+((y)/(x^(2)+y^(2))-1)dx`, isA. `y=x cos(C-x)`B. `cos^(-1).(y)/(x)=(-x+C)`C. `y=x tan(C-x)`D. `(y^(2))/(x^(2))=x tan (C-x)` |
| Answer» Correct Answer - C | |
| 387. |
The equation of the curve whose subnormal is twice the abscissa, isA. a circleB. a parabolaC. an ellipseD. a hyperbola |
| Answer» Correct Answer - D | |
| 388. |
Which of the following is general solution of `(d^(2)y)/(dx^(2)) - 2(dy)/(dx) + y = 0` ?A. `y = (Ax + B)e^(x)`B. `y = (Ax + B)e^(-x)`C. `y = A cos x + B sin x`D. `y = Ae^(x) + Be^(-x)` |
| Answer» Correct Answer - A | |
| 389. |
The solution of the differential equation `ydx+ (x +x^2 y) dy =0` isA. `logy=Cx`B. `-(1)/(xy)+logy=C`C. `(1)/(xy)+logy=C`D. `-(1)/(xy)=C` |
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Answer» Correct Answer - B::D We have, `ydx+(x+x^(2)y)dy=0` `rArr" "ydx+xdy+x^(2)ydy=0` `rArr" "(d(xy))/((xy)^(2))+(1)/(y)dy=0" "["Dividing throughout by "(xy)^(2)]` On integrating, we get `-(1)/(xy)+logy=C` |
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| 390. |
Find the general solution of the differential equation `(dy)/(dx)+sqrt((1-y^2)/(1-x^2))=0`.A. `tan^(-1)x+cot^(-1)x=C`B. `sin^(-1)x+sin^(-1)y=C`C. `sec^(-1)x+"cosec"^(-1)x=C`D. none of these |
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Answer» Correct Answer - B We have, `(dy)/(dx)+sqrt((1-y^(2))/(1-x^(2)))=0rArr (1)/(sqrt(1-y^(2)))dy+(1)/(sqrt(1-x^(2)))dx=0` On integration, we have `sin^(-1)y+sin^(-1)x=+C` |
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| 391. |
Solution of the differential equation `xdy-ydx=0` representsA. A rectangular hyperbolaB. Parabola whose vertex is at originC. Straight line passing through originD. A circle whose centre is at origin |
| Answer» Correct Answer - C | |
| 392. |
Solution of the differential equation `xdy-ydx=0` representsA. a parabola whose vertax is at the originB. a circle whose centre is at the originC. a rectangular hyperbolaD. straight lines passing through the origin |
| Answer» Correct Answer - D | |
| 393. |
Solution of the differential equation `xdy-ydx=0` representsA. a rectangular hyperbolaB. a straight line passing through the originC. parabola whose vertex is at the originD. circle whose centre is at the origin |
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Answer» Correct Answer - B We have, `xdy-ydx=0` `rArr" "(1)/(y)dy-(1)/(x)dx=0` `rArr" "logy-logx=logC" [On integrating]"` `rArr" "(y)/(x)=C rArr y=Cx` Clearly, it represents a straight line passing through the origin. |
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| 394. |
The solution of the differential equation `sinxcosy dy + cosx siny dx= 0` isA. `(sinx)/(siny)=C`B. `cosx+cosy=C`C. `sinx+siny=C`D. `sinxsiny=C` |
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Answer» Correct Answer - D We have, `cosx sinydx +sinx cosy dy =0` `rArr" "sinyd(sinx)+sinxd(siny)=0` `rArr" "d(sin x siny)=0` On integrating, we get `sin x sin y =C`. |
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| 395. |
The order of the differential equation of ellipse whose major and minor axes are along x-axis and y-axis respectively, isA. `xy y_(2) - xy_(1)^(2) + y y_(1) = 0`B. `xy y_(2) + xy_(1)^(2) - y y_(1) = 0`C. `xy y_(2) + xy_(1)^(2) + y y_(1) = 0`D. `xy y_(2) + xy_(1)^(2) = 0` |
| Answer» Correct Answer - B | |
| 396. |
Let F be the family of ellipse whose centre is the origin and major axis is the y-axis. Then the differential equation of family F isA. `(d^(2)y)/(dx^(2))+(dy)/(dx)(x(dy)/(dx)-y)=0`B. `xy(d^(2)y)/(dx^(2))-(dy)/(dx)(x(dy)/(dx)-y)=0`C. `xy(d^(2)y)/(dx^(2))+(dy)/(dx)(x(dy)/(dx)-y)=0`D. `(d^(2)y)/(dx^(2))-(dy)/(dx)(x(dy)/(dx)-y)=0` |
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Answer» Correct Answer - A Let the equation of the family F of required ellipses be `Ax^(2)+By^(2)=1" …(i)"` It is a two parameter family of curves. Differentiating (i) with respect to x, we get `ax+Byy_(1)=0" …(ii)"` Differentiating (ii) with respect to x, we get `A+By_(1)^(2)+Byy_(2)=0" ...(iii)"` Multiplying (iii) by x and subtracting from (ii), we get `B[yy_(1)-xy_(1)^(2)-xyy_(2)]=0` `rArr" "xyy_(2)+y_(1)(xy_(1)-y)=0rArrxy(d^(2)y)/(dx^(2))+(dy)/(dx)(x(dy)/(dx)-y)=0` This is the required differential equation. |
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| 397. |
The family of curves represented by `(dy)/(dx)=(x^(2)+x+1)/(y^(2)+y+1)` and the family represented by `(dy)/(dx)+(y^(2)+y+1)/(x^(2)+x+1)=0`A. touch each otherB. are orthogonalC. are one and the differentialD. none of these |
| Answer» Correct Answer - B | |
| 398. |
The differential equation of the family of parabola with focus at the origin and the x-axis an axis, isA. `y((dy)/(dx))^(2)+4x(dy)/(dx)=4y`B. `y((dy)/(dx))^(2)=2x(dy)/(dx)-y`C. `y((dy)/(dx))^(2)+y=2xy(dy)/(dx)`D. `y((dy)/(dx))^(2)+2xy(dy)/(dx)+y-0` |
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Answer» Correct Answer - B The equation of the family of parabolas with focus at the origin and the x-axis as axis is `y^(2)=2a(x-1)`, where a is parameter.`" …(i)"` Differentiating with respect to x, we get `2y(dy)/(dx)=4arArr a=(y)/(2)(dy)/(dx)` Sunstituting the value of a in (i), we get `y^(2)=2y(dy)/(dx)(x-(y)/(2)(dy)/(dx))` `rArr" "y^(2)=y(dy)/(dx)(2x-y(dy)/(dx))rArry((dy)/(dx))^(2)-2x(dy)/(dx)+y=0` This is the required differential equation. |
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| 399. |
The differential equation of all conics whose centre lie at the origin is of orderA. 2B. 3C. 4D. none of these |
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Answer» Correct Answer - B The general equation of all conics whose centre lie at the origin is `ax^(2)+2hxy+by^(2)=1` Clearly, it has three arbitrary constants. So, the differential equation will be of order 3. |
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| 400. |
The form of the differential equation of the central conics, isA. `x=y(dy)/(dx)`B. `x+y(dy)/(dx)=0`C. `x((dy)/(dx))^(2)+xy(d^(2)y)/(dx^(2))=y(dy)/(dx)`D. none of these |
| Answer» Correct Answer - C | |