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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.
| 251. |
`secx dy+secydx=0`A. `cosx+cosy=c`B. `sinx+cosy=c`C. `tanx+tany=c`D. `sinx+siny=c` |
| Answer» Correct Answer - D | |
| 252. |
`cos^(2)ydx-cosxdy=0`A. `sinx+coty=c`B. `cosx+tany=c`C. `cosx+coty=c`D. `secx-coty=c` |
| Answer» Correct Answer - B | |
| 253. |
`sin x cos y+(dy)/(dx)cos x sin y=0`A. `sin(x+y)=c`B. `sec x sec y =c`C. `sin x cos y=c`D. `cos (x+y)=c` |
| Answer» Correct Answer - B | |
| 254. |
`y+(dy)/(dx)sqrt(x^(2)-25)=0`A. `logy=sin^(-1)((x)/(5))+c`B. `y(x+sqrt(x^(2)-25))=c`C. `sin^(-1)((x)/(5))=log(cy)`D. `3y+(25-x^(2))^(3//2)` |
| Answer» Correct Answer - B | |
| 255. |
` (dy)/(dx) = x sqrt(25 -x^(2))`A. `3y+2sqrt(25-x^(2))=c`B. `3y+(25-x^(2))^(3//2)=c`C. `2y+3(25-x^(2))^(3//2)=c`D. `sin^(-1)((x)/(5))=y+c` |
| Answer» Correct Answer - B | |
| 256. |
`tanx sec y dx+dy=0`A. `log(sinx)+cosy=c`B. `log(cosx)+cosy=c`C. `log(cosx)+cosy=c`D. `log(secx)+siny=c` |
| Answer» Correct Answer - D | |
| 257. |
`x^(2) + y^(2) =r^(2) ` is a solution of the D.E. `y =x(dy)/(dx) + r sqrt(1+((dy)/(dx))^(2))`A. `y=xy_(1)-r sqrt(1+(y_(1))^(2))`B. `y_(1)=xy+rsqrt(1+y^(2))`C. `y_(1)=xy+r sqrt(1+x^(2))`D. `y=xy_(1)+r sqrt(1+(y_(1))^(2))` |
| Answer» Correct Answer - D | |
| 258. |
`y sec x = tanx+c` is a solution of the D.E.A. `y_(1)+y tan x= sec x`B. `y+y_(1)tanx=secx`C. `y_(1)=secx tanx`D. `y_(1)+y sec x= tanx` |
| Answer» Correct Answer - A | |
| 259. |
`y=log x` is a solution of the D.E.A. `xy_(2)=y_(1)`B. `xy_(1)+y_(2)=0`C. `xy_(2)+y_(1)=0`D. `xy_(1)=y_(2)` |
| Answer» Correct Answer - C | |
| 260. |
Solution of the differential equation `x(dy)/(dx)=y+sqrt(x^(2)+y^(2))`, isA. `x+sqrt(x^(2)+y^(2))=Cy^(2)`B. `y+sqrt(x^(2)+y^(2))=Cy^(2)`C. `x+sqrt(x^(2)+y^(2))=Cx^(2)`D. `y+sqrt(x^(2)+y^(2))=Cx^(2)` |
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Answer» Correct Answer - D Substituting `y=vx and (dy)/(dx)=v+x(dv)/(dx),` we get `v+x(dv)/(dx)=v+sqrt(1+v^(2))rArr (1)/(sqrt(1+v^(2)))dv=(1)/(x)dx` On integrating, we get `log(v+sqrt(v^(2)+1))=logx+logC` `rArr" "v+sqrt(v^(2)+1)=CxrArr y+sqrt(x^(2)+y^(2))=Cx^(2)` |
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| 261. |
`y=A sin 4x+B cos 4x` is a solution of the D.E.A. `y_(2)=16y`B. `y_(2)+16y=0`C. `y_(2)=16x`D. `y_(2)+16x=0` |
| Answer» Correct Answer - B | |
| 262. |
Obtion the differential equation by elininating arbitrary constants A, B from the equation - `y=A cos (logx)+B sin (logx)`A. `x^(2)y_(2)-xy_(1)+y=0`B. `x^(2)y_(2)+xy_(1)-y=0`C. `x^(2)y_(2)-xy_(1)-y=0`D. `x^(2)y_(2)+xy_(1)+y=0` |
| Answer» Correct Answer - D | |
| 263. |
A particle falls in a medium whose resistance is propotional to the square of the velocity of the particles. If the differential equation of the free fall is `(dv)/(dt) = g-kv^(2)` (k is constant) thenA. `v=2sqrt(g/k)(e^(2tsqrt(g//t))+1)/(e^(2rsqrt(g//k))-1)`B. `v=sqrt(g/k)(e^(2tsqrt(gk))-1)/(e^(2tsqrt(gk))+1`C. `v to 0` as `t to infty`D. `v to sqrt(g/k)` as `t to infty` |
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Answer» Correct Answer - B::D `(dv)/(dt) = g-kv^(2)` `rArr (dv)/(g-kv^(2))` `rArr (1/k int(dv)/((g/k)-v^(2))) = int(dt+C)` `rArr 1/(2sqrt(gk)) log|(sqrt(g/k)+v)/(sqrt(g//k)-v)|=t+C` At t=0, v=0 `rArr C=0` `rArr v=sqrt(g/k) (1-1/(e^(2tsqrt(gk))/(1-1/(e^(2tsqrt(gk))))))` Clearly when `v to sqrt(g/k)` as `t to infty` |
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| 264. |
An object falling from rest in air is subject notonly to the gravitational force but also to air resistance. Assume that theair resistance is proportional to the velocity with constant ofproportionality as `k >0`, and acts in a direction opposite to motion `(g=9. 8 m/(s^2))dot`Thenvelocity cannot exceed.(c)`( d ) (e) 9.8//k""m//s (f)`(g)(b) `( h ) (i) 98//k""m//s (j)`(k)(c)`( d ) (e) (f) k/( g )(( h ) 9.8)( i ) (j)m//s (k)`(l) (d) None of theseA. 9.8/km/sB. 98/k m/sC. `k/9.8 m//x`D. None of these |
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Answer» Correct Answer - A Let V(t) be the velocity of the object at time t. Given `(dV)/(dt) =9.8-kV` or `(dvV)/(9.8-kV)=dt` Integrating, we get `log(9.8-kV) =-kt+logC` or `9.8-kV=Ce^(-kt)` But `V(0)=0` or `C=9.8` Thus, `9.8 - kV=9.8e^(-kt)` or `V(t)=9.8/k(1-e^(-kt)) lt 9.8/k` for all t Hence, V(t) cannot exceed 9.8k m/s. |
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| 265. |
The differential equation for which `y=a cos x+b sin x ` is a solution, isA. `(d^(2)y)/(dx^(2))+y=0`B. `(d^(2)y)/(dx^(2))-y=0`C. `(d^(2)y)/(dx^(2))+(a+b)y=0`D. `(d^(2)y)/(dx^(2))+(a-b)y=0` |
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Answer» Correct Answer - A We have, `y=a cos x+b sin x` `rArr" "(dy)/(dx)=-a sin x+b cosx` `rArr" "(d^(2)y)/(dx^(2))=-a cos x-b sin xrArr (d^(2)y)/(dx^(2))=-y` |
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| 266. |
Which of the following equation(s) is/are linear?A. `(dy)/(dx)+y/x=logx`B. `y(dy)/(dx)+4x=0`C. `(2x+y^(3))(dy)/(dx)=3y`D. None of these |
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Answer» Correct Answer - A::C Obviously (1) is linear differential equation with `P=1/x` and Q`=logx` `y(dy)/(dx)+4x=0` or `(dy)/(dx)+(4x)/y=0` Hence, it is not linear `(2x+y^(3))(dy)/(dx)=3y` or `(dx)/(dy)=(2x)/(3y)+y^(2)/3` or `(dx)/(dy)-(2x)/(3y)=y^(2)/3` which is linear with `P=2/(3y)` and `Q=y^(2)/3` |
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| 267. |
Which one of the following function(s) is/are homogeneous?A. `f(x,y) = (x-y)/(x^(2)+y^(2))`B. `f(x,y) = x^(1/3)y^(-2/3) tan^(-1)x/y`C. `f(x,y) = x(" ln "sqrtx^(2)+y^(2))-" ln "y+ye^(x//y)`D. `f(x,y) = x[" ln "(2x^(2)+y^(2))x -" ln "(x+y)]+y^(2)tan(x+2y)/(3x-y)` |
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Answer» Correct Answer - A::B::C a) `f(lambdax,lambday)=(lambda(x-y))/(lambda^(2)(x^(2)+y^(2)))=lambda^(-1)f(x,y)` Thus, it is homogenous of degree `-1` b) `f(lambdax,lambday)=(lambdax)^(1//3)(lambday)^(-2//3)tan^(-1)x/y` `=lambda^(-1//3)x^(1//3)tan^(-1)x/y` `=lambda^(-1//3)f(x,y)` c) `f(lambdax,lambday) = lambdax("ln"sqrt(lambda^(2)(x^(2)+y^(2))-"ln "lambday))+lambdaye^(x//y)` `=lambdax["ln"((lambdasqrt(x^(2)+y^(2)))/(lambday))]+lambdaye^(x//y)` `lambda[x("ln "sqrt(x^(2)+y^(2))-"ln"y)+ye^(x//y)]` `=lambdaf(x,y)` Thus, it is homogeneous. d) `f(lambdax,lambday)=lambdax["ln "(2lambda^(2)x^(2)+lambda^(2)y^(2))(lambdaxlambda(x+y))]+lambda^(2)x^(2)tan(x+2y)/(3x-y)` `=lambda x["ln "(2x^(2)+y^(2))/(x(x+y))]+lambda^(2)x^(2)tan(x+2y)/(3x-y)` Thus, it is non-homogeneous. |
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| 268. |
Form the differential equation of family of lines concurrent at theorigin.A. `x(dy)/(dx)-y=0`B. `x+(dy)/(dx)=0`C. `(dy)/(dx)=0`D. `(dy)/(dx)=x` |
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Answer» Correct Answer - A The equation of the family of lines passing through the origin is `y=mx" …(i)"` `rArr" "(dy)/(dx)=m" …(ii)"` From (i) and (ii), we get `y=x(dy)/(dx)`, which is the required differential equation. |
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| 269. |
The differential equation whose solution is `(x-h)^2+ (y-k)^2=a^2` is (a is a constant)A. order is 2B. order is 3C. degree is 2D. degree is 3 |
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Answer» Correct Answer - A::C We have `(x-h)^(2)+(y-k)^(2)=a^(2)`…………..(1) Differentiating w.r.t. x, we get `2(x-h)+2(y-k)(dy)/(dx)=0` or `(x-h)+(y-k)(dy)/(dx)=0`……….(2) Differentiating w.r.t. x, we get `1+((dy)/(dx))^(2)+(y-k)(d^(2)y)/(dx^(2))=0` ………..(3) From equation (3), `y-k=-(1+p^(2))/q`, where `p=(dy)/(dx), q=(d^(2)y)/(dx^(2))` Putting the value of `y-k` in equation (2), we get `x-h=(1+p^(2))=a^(2)` or `[1+((dy)/(dx))^(2)]^(3)=a^(2)(d^(2)y)/(dx^(2))^(2)` Which is the required differential equation. |
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| 270. |
`x^(2)dx-y^(2)dy+xdx=dy-ydy-dx` 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(x^(2)+y^(2))-6(x-y)=c`D)`2(x^(3)-y^(3))+3(x^(2)+y^(2))+6(x-y)=c`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(x^(2)+y^(2))-6(x-y)=c`D. `2(x^(3)-y^(3))+3(x^(2)+y^(2))+6(x-y)=c` |
| Answer» Correct Answer - D | |
| 271. |
General solution of `(x^(2)+x^(2)/y^(2))dy+(x^(2)+1)dx=0` isA. `x+y+x^(-1)+y^(-1)=c`B. `x+y-x^(-1)-y^(-1)=c`C. `x+y=cxy`D. `x-y=cxy` |
| Answer» Correct Answer - B | |
| 272. |
General solution of `y-x(dy)/(dx)=8(y^(2)+(dy)/(dx))` isA. `(x+8)(8y+1)=cy`B. `(x+8)(8y-1)=cy`C. `(8x+1)(y-1)=cy`D. `(y+8)(8x+1)=c` |
| Answer» Correct Answer - B | |
| 273. |
General solution of `2e^(x)cos^(2)ydx+(1-e^(x))cotydy=0` isA. `tany=ce^(2x)`B. `tany=c(e^(x)-1)^(2)`C. `tany=c(e^(x)+1)^(2)`D. `coty=c(e^(x)-1)^(2)` |
| Answer» Correct Answer - B | |
| 274. |
The differential equation whose solution is `(x-h)^2+ (y-k)^2=a^2` is (a is a constant)A. `[1+((dy)/(dx))^(2)]^(3)=a^(2)(d^(2)y)/(dx^(2))`B. `[1+((dy)/(dx))^(2)]^(3)=a^(2)((d^(2)y)/(dx^(2)))^(2)`C. `[1+((dy)/(dx))]^(3)=a^(2)((d^(2)y)/(dx^(2)))^(2)`D. none of these |
| Answer» Correct Answer - B | |
| 275. |
The general solution of `e^(x) cos ydx-e^(x) sin dy=0` isA. `e^(x)cosy=k`B. `e^(x)siny=k`C. `e^(x)=k cos y`D. `e^(x)=k sin y` |
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Answer» Given that, `" "e^(x)cosydx-e^(x)sinydy=0` `rArr" "e^(x)cosydx=e^(x)sinydy` `rArr" "(dx)/(dy)=tany` `rArr" "dx=tanydy` On integrating both sides, we get `" "x=logsecy+C` `rArr" "x-C=logsecy` `rArr" "secy=e^(x-C)` `rArr" "secy=e^(x)e^(-C)` `rArr" "(1)/(cosy)=(e^(x))/(e^(C))` `rArr" "e^(x)cosy=e^(C)` `rArr" "e^(x)cosy=K" "` [where, `K=e^(C)`] |
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| 276. |
The number of solution of `dy / dx = (y +1) / (x-1)` when `y(1) = 2` isA. NoneB. OneC. TwoD. Infinite |
| Answer» Correct Answer - A | |
| 277. |
The differential equationy `y (dy)/(dx)+x=a` (a is any constant) representsA. a set of circles having centre on the y-axisB. a set of circle centre on the x-axisC. a set of ellipsesD. none of these |
| Answer» Correct Answer - B | |
| 278. |
The differential equation `y(dy)/(dx)+x=C` representsA. family of hyperbolasB. family of parabolasC. family of ellipsesD. family of circles |
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Answer» Given, that `y(dy)/(dx)+x=C` `Rightarrow y(dy)/(dx)=C-x` `Rightarrow ud y=(C-x)dx` On integrating both sides, we get `(y^(2))/(2)=Cx-(x^(2))/(2)+K` `Rightarrow (x^(2))/(2)+(y^(2))/(2)=Cx+K` `Rightarrow (x^(2))/(2)+(y^(2))/(2)=Cx+K` `Rightarrow (x^(2))/(2)+(y^(2))/(2)=Cx=K` Which represent family of circles. |
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| 279. |
The differential equation `y(dy)/(dx) + x = c` representsA. Family of hyperbolasB. Family of parabolasC. Family of ellipseD. Family of circles |
| Answer» Correct Answer - D | |
| 280. |
The integrating factor of differential equation `dy/dx+y=(1+y)/x` isA. `(x)/(e^(x))`B. `(e^(x))/(x)`C. `ex^(x)`D. `e^(x)` |
| Answer» Correct Answer - B | |
| 281. |
The general solution of `(dy)/(dx) = 2x e^(x^(2)-y)` isA. `e^(x^(2)-y) = c`B. `e^(-y) + e^(x^(2)) = c`C. `e^(y) = e^(x^(2)) + c`D. `e^(x^(2)+y) = c` |
| Answer» Correct Answer - C | |
| 282. |
The differential equation for the family of curves `x^(2)-y^(2)-2ay=0`, where a is an arbitrary constant, isA. `(x^(2)-y^(2)) (dy)/(dx) = 2xy`B. `2(x^(2) + y^(2)) (dy)/(dx) = xy`C. `(x^(2) - y^(2))(dy)/(x) = xy`D. `(x^(2)+y^(2))(dy)/(dx) = 2xy` |
| Answer» Correct Answer - A | |
| 283. |
General solution of `(y+y^(2))dy=(x+x^(2))dx` isA. `2(x^(3)-y^(3))+3(x^(2)-y^(2))=c`B. `2(x^(3)+y^(3))+3(x^(2)+y^(2))=c`C. `2(x^(3)+y^(3))-3(x^(2)+y^(2))=c`D. None of these |
| Answer» Correct Answer - A | |
| 284. |
General solution of `(dy)/(dx)=e^(x-y)` isA. `e^(x)+e^(y)=c`B. `e^(x)=ce^(y)`C. `e^(y)-e^(x)=c`D. `e^(y)=ce^(x)` |
| Answer» Correct Answer - C | |
| 285. |
General solution of `(dy)/(dx)=e^(y-x)` isA. `e^(-x)-e^(-y)=c`B. `e^(-x)+e^(-y)=c`C. `e^(-y)+e^(x)=c`D. `e^(y)-e^(-x)=c` |
| Answer» Correct Answer - A | |
| 286. |
General solution of `(x)/(dx)=(y)/(dy)` isA. `xy=c`B. `x=cy`C. `y=cx`D. `e^(x)=e^(y)+c` |
| Answer» Correct Answer - B | |
| 287. |
General solution of `e^(x-y)dx+e^(y-x)dy=0` isA. `x-y=c`B. `e^(x)+e^(y)=c`C. `e^(2x)+e^(2y)=c`D. `e^(x)-e^(y)=c` |
| Answer» Correct Answer - C | |
| 288. |
If `(dy)/(dx)=3x^(2)+1 and y=30` when `x = 3`, then `y=`A. `x^(3)-x+2`B. `x^(3)+x+2`C. `x^(3)+x-2`D. `x^(3)+x` |
| Answer» Correct Answer - D | |
| 289. |
Solve `((dy)/(dx))=e^(x-y)(e^x-e^y)dot` |
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Answer» Multiplying the given equation by `e^(y)`, we get `e^(y)(dy)/(dx) + e^(x)e^(y)=e^(2x)` Putting `e^(y)=v`, so that `e^(y)(dy)/(dx) = (dv)/(dx)`, and equation (1) transform to `(dv)/(dx) +e^(x)v=e^(2x)` I.F. `=e^(inte^(x)dx) = e^(e^(x))` Hence, solution is `ve^(e^(x)) = int e^(2x)e^(e^(x))dx+c` Let `e^(x)=t` or `e^(x)dx=intte^(t)dt+c` or `e^(y)e^(e^(x))=e^(x)e^(e^(x))-e^(e^(x))+c` |
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| 290. |
General solution of `(x^(3)y^(2)+y^(2))(dy)/(dx)+(x^(2)y^(3)+x^(2))=0` isA. `(x^(3)-1)(y^(3)+1)=c`B. `(x^(3)+1)(y^(3)+1)=c`C. `(x^(3)+1)(y^(3)-1)=c`D. `(x^(3)-1)(y^(3)-1)=c` |
| Answer» Correct Answer - B | |
| 291. |
General solution of `(dy)/(dx)=e^(2x-y)+x^(3)e^(-y)` isA. `4e^(y)=2e^(2x)-x^(4)-x^(2)+c`B. `4e^(y)=2e^(2x)+x^(4)-x^(2)+c`C. `4e^(y)=2e^(2x)+x^(4)+c`D. `4e^(x)=2e^(2y)-y^(4)-y^(2)+c` |
| Answer» Correct Answer - C | |
| 292. |
Solve `(dy)/(dx)=cos(x+y)-sin(x+y)`. |
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Answer» Correct Answer - `-log_(e)|1-tan(x+y)/(2)|=x+c` `(dy)/(dx) = cos(x+y)-sin(x+y)` Putting `x+y=t`, we get `(dy)/(dx)=(dt)/(dx)-1` Therefore, `(dt)/(dx)-1=cost-sint` or `(dt)/(1+cost-sint)=dx` or `(sec^(2)t/2dt)/(2(1-tant/2))=dx` or `-"ln "|1-tan(x+y)/(2)|=x+c` |
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| 293. |
A tangent and a normal to a curve at any point P meet the x and y axes at A, B and C, D respectively. Find the equation of the curve passing through `(1, 0)` if the centre of circle through `O.C, P and B` lies on the line y = x (where O is the origin). |
| Answer» `log(sqrt(x^(2) + y^(2))) + tan^(-1) ((y)/(x)) = 0` | |
| 294. |
Solve the following differential equation: ` x(dy)/(dx)-y=2 sqrt(y^2-x^2)` |
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Answer» Correct Answer - `1/2log_(e)y+sqrt(y^(2)-x^(2))/(x)=log_(e)cx` Putting `y=vx` and `(dy)/(dx)=v+x(dv)/(dx)`, we get `xv+x^(2)(dv)/(dx)=vx+2xsqrt(v^(2)-1)` or `int(dv)/(2sqrt(v^(2)-1))=int(dx)/x`, Integrating, we get `1/2" ln "(v+sqrt(v^(2)-1))="ln "cx` or `1/2"ln "(y+sqrt(y^(2)-x^(2)))/x="ln "cx` |
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| 295. |
`x(dy)/(dx)=y(logy-logx+1)` |
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Answer» Correct Answer - `y=xe^(cx), c gt 0` `x(dy//dx)=y(logy-logx+1)` or `(dy)/(dx)=y/x[logy/x+1]` Putting `y=vx`, we get `(dy)/(dx)=v+x(dv)/(dx)` and the given equation transforms to `v+x(dv)/(dx)=v[logv+1]` or `x(dv)/(dx)=vlogv` or `int(dv)/(dx)=vlogv` or `int(dv)/(vlogv)=int(dx)/x` or `loglogv=logx+logc, c gt 0` or `cx=log(y//x)` or `y=xe^(cx), c gt 0` |
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| 296. |
Solve `[2sqrt(x y)-x]dy+ydx=0` |
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Answer» Correct Answer - `log_(e)y+sqrt(x/y)=C` We have `(dy)/(dx)=(y/x)/(1-2sqrt(y/x))` which is homogeneous. Put `y=vx` so that `(dy)/(dx) = x(dv)/(dx)+v` `therefore x(dv)/(dx) = v/(1-2sqrt(v))-v=(2v^(3//2))/(1-2sqrt(v))` or `(dx)/x=(1-2sqrt(v))/(2v^(3//2))dv=1/(2v^(3//2)-1/v)dv` Integrating, we get `-C+logx=-v^(-1//2)-logv=-sqrt(x/y)-logy+logx` or `logy + sqrt(x/y)=C` |
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| 297. |
Solve `(x-1)dy+ydx=x(x-1)y^(1/3)dxdot` |
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Answer» Dividing by `dxy^(1//3)(x-1)`, the given equation reduces to `y^(-1//3)(dy)/(dx)+1/(x-1)y^(2//3)=x` put `y^(2//3)`=z, so that, `2/3y^(-1//3)(dy)/(dx)=(dz)/(dx)` Then given equation reduces to `(dz)/(dx) + 2/(3(x-1))z = 2/3x` (linear form) Hence, solution is given by `z(x-1)^(2/3)=2/3intx(x-1)^(2//3) dx+c` Putting `(x-1)=t^(3)` in the R.H.S, we get `intx(x-1)^(2//3)dx = intt^(3)+1(t^(2)3t^(2))dt` `=3int(t^(7)+t^(4))dt` `=3[(1//8)t^(8)+(1//5)t^(5)]` `=(3//8)(x-1)^(8//3)+(3//5)(x-1)^(5//3)` Hence, the solution is `y^(2//3) = 1/4(x-1)^(2)+2/5(x-1)+c(x-1)^(-2//3)`. |
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| 298. |
General solution of `(dy)/(dx)+(y)/(3)=1` isA. `y=3+ce^(x//3)`B. `3y=c+e^(x//3)`C. `y=3+ce^(-x//3)`D. `3y=c+e^(-x//3)` |
| Answer» Correct Answer - C | |
| 299. |
General solution of `y(1+x)dx+x(1+y)dy=0` isA. `x-y+log(xy)=c`B. `x-y-log(xy)=c`C. `x+y+log(xy)=c`D. `x+y-log(xy)=c` |
| Answer» Correct Answer - C | |
| 300. |
`(dy)/(dx)+sqrt((b^(2)-y^(2))/(a^(2)-x^(2)))=0`A. `ax+by=c(1-abxy)`B. `sin^(-1)((x)/(a))+cos^(-1)((y)/(b))=c`C. `cos^(-1)((x)/(a))+sin^(-1)((y)/(b))=c`D. `cos^(-1)((x)/(a))+cos^(-1)((y)/(b))=c` |
| Answer» Correct Answer - D | |