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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.
| 651. |
The integrating factor of the differentialequation `(dy)/(dx)(x(log)_e x)+y=2(log)_e x`is given by(a)`( b ) x (c)`(d)(b) `( e ) (f) (g) e^(( h ) x (i))( j ) (k)`(l)(c) `( m ) (n) (o)(( p )log)_( q ) e (r) (s) x (t)`(u)(d) `( v ) (w) (x)(( y )log)_( z ) e (aa) (bb)(( c c )(( d d )log)_( e e ) e (ff) (gg) x)( h h )`(ii)A. xB. `e^(x)`C. `log_(e)x`D. `log_(e)(log_(e)x)` |
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Answer» Correct Answer - C `therefore (dy)(dx)+y/(xlog_(e)x)=2/x` `therefore` I.F. `=e^(int1/(xlog_(e)x)dx)` `=e^(log_(e )(log_(e)x)` `=log_(e)x` |
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| 652. |
A homogeneous differential equation of the from `(dx)/(dy)=h(x/y)`can be solved by making the substitutionA. `y=vx`B. `v=yx`C. `x=vy`D. `x=v` |
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Answer» Correct Answer - C Since the given differential equation `(dx)/(dy)=h((x)/(y))` is homogenous, therefore put `x=vy`, `(dx)/(dy)=v+y(dv)/(dy)` `:. V+y(dv)/(dy)=hvimpliesy(dv)/(dy)=v(h-1)` `implies (1)/((h-1)v)dv=(1)/(y)dy` On integration, `(1)/((h-1))int(1)/(v)dv=int(dy)/(y)` `implies (1)/((h-1))log|v|=log|y|+C` |
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| 653. |
If the solution of the differential equation`(dy)/(dx)=1/(xcosy+sin2y)`is `x=c e^(siny)-k(1+siny),`then thevalue of `k`is_______ |
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Answer» Correct Answer - 2 `(dy)/(dx) = 1/(xcosy+2sinycosy)` `therefore (dx)/(dy) = xcosy+2sinycosy` `therefore (dx)/(dy) +(-cosy)x=2sinycosy` `therefore I.F. =e^(-intcosydx)=e^(-siny)` Thus, the solution is `x.e^(-siny)=2inte^(-siny).sinycosydy = -2sinye^(-siny)+2inte^(-siny)cosydy` `=-2sinye^(-siny)+2inte^(-siny)cosydy` `=-2sinye^(-siny)-2e^(-siny)+c` i.e., `x=-2siny-2+ce^(siny)=ce^(siny)-2(1+siny)` `therefore k=2` |
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| 654. |
The differential equations , find the particular solution satisfying the given condition:`2x y+y^2-2x^2(dy)/(dx)=0; y=2`when x = 1 |
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Answer» `2xy+y^(2)-2x^(2)(dy)/(dx)=0` `implies (dy)/(dx)=(2xy+y^(2))/(2x^(2))`……….`(1)` It is a homogenous differential equation. Let, `y=vx` `implies (dy)/(dx)=v+x(dv)/(dx)` From equation `(1)` , `v+x(dv)/(dx)=(2x^(2)v+v^(2)x^(2))/(2x^(2))=v+(1)/(2)v^(2)` `implies x(dv)/(dx)=(1)/(2)v^(2)` `implies (2dv)/(v^(2))=(dx)/(x)` `implies 2int(1)/(v^(2))dv=int(dx)/(x)` `implies -(2)/(v)+c=logx` `implies -(2x)/(y)+c=logx` `implies logx+(2x)/(y)=c` .........`(2)` Given, `y=2` at `x=1` `:. log1+(2xx1)/(2)=c` `implies c=1` Therefore, the particular solution of the given equation is `logx+(2x)/(y)=1` |
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| 655. |
Solve: `(xcosy-ysiny)dy+(x siny+ycosy)dx=0` |
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Answer» We have `(xcosy-ysiny)(dy)/(dx)+(xsiny+ycosy)=0`.............(1) Let `xsiny+ycosy=t` Differentiating w.r.t. `x`, we get `xcosy(dy)/(dx)+siny+cosy(dy)/(dx)-ysiny(dy)/(dx)=(dt)/(dx)` `therefore (xcosy-ysiny)(dy)/(dx)+siny+cosy(dy)/(dx)=(dt)/(dx)` So, from (1), we have `(dt)/(dx)-(siny+(d(siny))/(dx))+t=0` `rArr (d(t-siny))/(dx) = -(tsiny)` `rArr (d(t-siny))/(t-siny)=-dx` `rArr int(d(t-siny))/(t-siny) =-intdx` `rArr log_(e)(t-siny)=-x+log_(e)C` `rArr t-siny=Ce^(-x)` `rArr xsiny+ycosy-siny=Ce^(-x)` |
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| 656. |
The differential equations, find a particular solution satisfying the given condition: `(1+x^2)(dy)/(dx)+2x y=1/(1+x^2); y=0`when `x = 1` |
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Answer» `(1+x^(2))(dy)/(dx)+2xy=(1)/(1+x^(2))` `implies (dy)/(dx)+(2x)/(1+x^(2))y=(1)/((1+x^(2))^(2))` Here, `P=(2x)/(1+x^(2))` and `Q=(1)/((1+x^(2))^(2))` `I.F.=e^(intPdx)=e^(int(2x)/(1+x^(2))dx)` `:. =e^(log(1+x^(2))=(1+x^(2))` and general solution : `y*(1+x^(2))=int(1)/((1+x^(2))^(2))(1+x^(2))dx+c` `=int(1)/(1+x^(2))dx+c` `y(1+x^(2))=tan^(-1)x+c`.........`(1)` Given, `y=0` at `x=1` `0=tan^(-1)1+c` `impliesc=-(pi)/(4)` Therefore, from equation `(1)`, particular solution is `y(1+x^(2))=tan^(-1)x-(pi)/(4)` |
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| 657. |
The differential equations , find the particular solution satisfying the given condition:`[xsin^2(y/x)-y]dx+x dy=0; y=pi/4`when x = 1 |
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Answer» Given that , `sin^(2)((y)/(x))-(y)/(x)+(dy)/(dx)=0` `implies (dy)/(dx)=(y)/(x)-sin^(2)((y)/(x))`……….`(1)` Given differential equation is homogenous. Let `y=vximplies(dy)/(dx)=v+x(dv)/(dx)` From equation `(1)`, `v+x(dv)/(dx)=v-sin^(2)v` `x(dv)/(dx)=-sin^(2)v` `implies cosec^(2)vdv=-(1)/(x)dx` On integration `intcosec^(2)vdv=-int(dx)/(x)implies-cotv=-log|x|+C` `implies log|x|-cotv=Cimplies log|x|-cot((y)/(x))=C`.........`(2)` when `x=1`, then `y=(pi)/(4)implieslog|1|-cot(pi)/(4)=C` `implies C=0-1=-1` put the value of `C` in equation `(2)`, `log|x|-cot((y)/(x))=-1` `implies log|x|-cot((y)/(x)=-loge` `implies cot((y)/(x))=log|ex|` which is the required solution of the given equation. |
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| 658. |
Find the general solution of the differential equations:`(x+3y^2)(dx)/(dy)=y(y >0)` |
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Answer» Here, given equation is, `(x+3y^2)dy/dx = y` `=>dx/dy = (x+3y^2)/y` `=>dx/dy -x/y = 3y` Comparing it with `dx/dy +Px = Q` `P = -1/y and Q = 3y` Integrating factor, `I.F. = e^(intPdy)` `I.F. = e^(int-1/ydy) = e^-lny = y^-lne = y-1 = 1/y` Now, general solution will be, `x(I.F.) = int(I.F.)Qdy` `=>x/y = int(1/y)(3y)dy` `=>x/y = 3y+c` `=>x = 3y^2+cy` |
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| 659. |
General solution of the differential equation `(dy)/(dx)=1+x + y` isA. `y=c.e^(-x^(2)//2)`B. `y=c.e^(x^(2)//2)`C. `u=(x+c).e^(-x^(2)//2)`D. none of these |
| Answer» Correct Answer - D | |
| 660. |
Find the general solution of the differential equation `(x cos y) dy = e^(x) (x log x +1) dx.`A. `siny=(1)/(x)e^(x)+c`B. `siny+e^(x).logx+c=0`C. `siny=e^(x).logx+c`D. none of these |
| Answer» Correct Answer - C | |
| 661. |
Find the equation of the curve passing through origin and satisfying the differential equation `2xdy = (2x^(2) + x) dx`. |
| Answer» Correct Answer - `2y = x^(2) + x` | |
| 662. |
General solution of : `(dy)/(dx)=sqrt(1-x^(2)-y^(2)+x^(2)y^(2))` isA. `2sin^(-1)y=x sqrt(1-x^(2))+sin^(-1)x+c`B. `cos^(-1)y=x.cos^(-1)x+c`C. `sin^(-1)y=(1)/(2).sin^(-1)x+c`D. `2.sin^(-1)y=x.sqrt(1-y^(2))+c` |
| Answer» Correct Answer - A | |
| 663. |
The differential equations, find a particular solution satisfying the given condition: `(dx)/(dy)+2ytanx=sinx ; y=0`when `x=pi/3` |
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Answer» Comparing the given equation with first order differential equation, `dy/dx+Py = Q(x)`, we get,`P = 2tanx and Q(x) = sinx` So, Integrating factor `(I.F) = e^(int2tanxdx)` `I.F.= e^(2ln secx) = e^(ln sec^2x) = sec^2x` We know, solution of differential equation, `y(I.F.) = intQ(I.F.)dx` `:.`Our solution will be, `ysec^2x = int sinx(sec^2x)dx` `=>ysec^2x = int sinx/cos^2xdx` `=>ysec^2x = int tanx secx dx` `=>ysec^2x =secx+c` `=>y = cosx +c cos^2x` At `x = pi/3, y=0` `=>0 = 1/2 +c/4` `=> -2 = c` So, our solution will be, `=>y = cosx - 2cos^2x` `=>y = cosx(1-2cosx)`, which is the required solution. |
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| 664. |
The equation of a curve passing through the origin and satisfying the differential equation `(dy)/(dx)=(x-y)^(2)`, isA. `e^(2x)(1-x+y)=1+x-y`B. `e^(2x)(1+x-y)=1-x+y`C. `e^(2x)(1-x+y)+(1+x-y)=0`D. `e^(2x)(1+x+y)=1-x+y` |
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Answer» Correct Answer - A We have, `(dy)/(dx)=(x-y)^(2)` Let `(x-y)=v`. Then, `1-(dy)/(dx)=(dv)/(dx)rArr (dy)/(dx)=1-(dv)/(dx)` `therefore" "(dy)/(dx)=(x-y)^(2)` `rArr" "1-(dv)/(dx)=v^(2)` `rArr" "1-v^(2)=(dv)/(dx)` `rArr" "dx=(1)/(1-v^(2))dv` `rArr" "2intdx=2int(1)/(1-v^(2))dv` `rArr" "2x=log((1+v)/(1-v))+logC` `rArr" "C((1+v)/(1-v))=e^(2x)` `rArr" "C((x-y+1)/(y-x+1))=e^(2x)rArrC(x-y+1)=e^(2x)(y-x+1)` Taking C = 1, we find that option (a) is correct. |
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| 665. |
General solution of : `x(dy)/(dx)+y=x.e^(x)` isA. `xy=(x+1)e^(x)+c`B. `xy=(x-1)e^(x)+c`C. `xy=(1-x)e^(x)+c`D. none of these |
| Answer» Correct Answer - B | |
| 666. |
The equation of the curve passing through the origin and satisfying the differential equation `((dy)/(dx))^(2)=(x-y)^(2)`, isA. `e^(2x)(1-x+y)=1+x-y`B. `e^(2x)(1+x-y)=1-x+y`C. `e^(2x)(1-x+y)+(1+x-y)`D. `e^(2x)(1+x+y)=1-x+y` |
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Answer» Correct Answer - A We have, `(dy)/(dx)=pm(x-y)` `ul("CASE I")" When "(dy)/(dx)=(x-y)` In this case, we have `1-(dv)/(dx)=v," where "x-y=v` `rArr" "(dv)/(dx)=1-v` `rArr" "(1)/(1-v)dv=dx` `-log(1-v)=x+logC` `rArr" "(1-v)^(-1)=Ce^(x) rArr(1)/(1-x+y)=Ce^(x)` It passes through the origin. `therefore" "C=1` Hence, `(1)/(1-x+y)=e^(x)" ...(i)"` `ul("CASE II")" When "(dy)/(dx)=(-x-y)=y-x.` In this case, we have `(du)/(dx)+1=u," where"u=y-x` `rArr" "(du)/(u-1)=dx` `rArr" "log(u-1)=x+logC` `rArr" "u-1=Ce^(x)rArr y-x-1=Ce^(x)` It passes through the origin. `therefore" "C=-1` Hence, `y-x-1=-e^(x)or, x-y+1=e^(x)" ...(ii)"` From (i) and (ii), we obtain `(x-y+1)/(1-x+y)=e^(2x)rArr (x-y+1)=(1-x+y)e^(2x)` |
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| 667. |
The equation of curve passing through origin and satisfying the differential equation `(1 + x^2)dy/dx + 2xy = 4x^2` , isA. `(1+x^(2))y=x^(3)`B. `2(1+x^(2))y=3x^(3)`C. `3(1+x^(2))y=4x^(3)`D. none of these |
| Answer» Correct Answer - C | |
| 668. |
If `y(x)` is solution of `x(dy)/(dx)+2y=x^(2), y(1)=1` then value of `y(1/2)=` (a) `-(49)/(16)` (b) `(49)/(16)` (c) `(45)/(8)` (d) `-(45)/(8)`A. (a) 13/16B. (b) 1/4C. (c) 49/16D. (d) 7/64 |
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Answer» Correct Answer - (c) Given differential equation can be rewritten as `dy/dx + (2/x)cdoty=x,` which is a linear differential equation of the form `dy/dx + Py =Q.` where`P=2/x and Q=x` Now, integrating factor `(IF) = e^(int 2/x dx) =e^(2log x) =e^(log x^(2)=x^(2)` `[therefore e^(logf(x))=f(x)]` and the solution is given by `y(IF)= int (QxxIF) dx + C` ` rArr yx^(2)=int x^(3) dx +C` ` rArr yx^(2)= x^(4)/4+C …(i)` Since, it is given that Y = 1 when x = 1 `therefore` From Eq. (i). we get `1=1/4+c rArr c =3/4 …(ii)` `therefore 4x^(2)y=x^(4)+3` [using Eqs. (i) and (ii) ] `rArr y=(x^(4)+3)/(4x^(2))` Now,`y(1/2)=(1/16+3)/(4xx1/4)= 49/16` |
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| 669. |
If a curve `y=f(x)`passes through the point `(1,-1)`and satisfies the differentialequation `,y(1+x y)dx""=x""dy`, then `f(-1/2)`is equal to:(1) `-2/5`(2) `-4/5`(3) `2/5`(4) `4/5`A. `-(2)/(5)`B. `-(4)/(5)`C. `(2)/(5)`D. `(4)/(5)` |
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Answer» Correct Answer - D The differential equation is `y(1+xy)dx=xdy` `rArr" "ydx-xdy=-xy^(2)dx` `rArr" "(ydx-xdy)/(y^(2))=-xdx` `rArr" "d((x)/(y))=-xdx` On integrating, we obtain `(x)/(y)=-(x^(2))/(2)+C` It is given that the curve given by (i) passes though the point `(1,-1)` `therefore" "-1=-(1)/(2)+CrArr C=-(1)/(2)` Putting `C=-(1)/(2)` in (i), we obtain `y(x^(2)+1)+2x=0` Putting `x=-(1)/(2)` in (ii), we obtain `y=(4)/(5)`. Hence, `f(-(1)/(2))=(4)/(5)` |
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| 670. |
The foci of the curve which satisfies the equation `(1+y^(2))dx - xy dy = 0` and passes through the point (1, 0) areA. `(sqrt(2), 0)`B. `(0, sqrt(2))`C. `(sqrt(-2), 0)`D. `(0, -sqrt(2))` |
| Answer» Correct Answer - A::C | |
| 671. |
Lety(x) be the solution of the differential equation `(xlogx)(dy)/(dx)+y=2xlogx ,(xgeq1)dot`Then y(e) is equal to :(1)e (2) 0 (3) 2 (4) 2eA. (a) eB. (b) 0C. (c) 2D. (d) 2e |
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Answer» Correct Answer - (c) Given differential equation is `(x log x)dy/dx+y=2xlogx` `rArr dy/dx+y/(xlogx)=2` This is a linear differential equation. `therefore IF= e^(int1/(x logx)dx)=e^(log(logx))=log x` Now, the solution of given differential equation is given by `y cdot log x = int log x cdot 2dx` `rArr y cdot log x = 2 int log x dx` `rArr y cdot log x = 2 [x log x-x]+c` At `x = 1 rArr c = 2` `rArr y cdot log x = 2 [x log x-x]+2` At `x=e, y=2(e-e)+2` `rArr y = 2` |
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| 672. |
Let `y=g(x)` be the solution of the differential equation `sin (dy)/(dx)+y cos x=4x, x in (0,pi)` If y(pi/2)=0`, then `y(pi/6)` is equal toA. (a) `4/(9sqrt3)pi^(2)`B. (b) `(-8)/(9sqrt3)pi^(2)`C. (c) `-8/(9)pi^(2)`D. (d) `-4/(9)pi^(2)` |
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Answer» Correct Answer - (c) Wh have, `sin x dy/dx +y cos x = 4x rArr dy/dx+ y cot x = 4 x cosec x` This is a linear differential equation of form `dy/dx+ Py=Q` where `P=cot x,Q=4x cosec x` Now, `IF=e^(int pdx)=e^(int cot xdx)=e^(logsinx)=sinx` Solution of the differential equation is `y cdot sin x = int 4x cosec x sin x dx +C` `rArr ysin x = int 4xdx +C = 2x^(2)+C` Put `x=pi/2,y=0,` we get `C= -pi^(2)/2 rArr y sin x = 2x^(2) - pi^(2)/2` Put `x =pi/6` `therefore y(1/2)=2(pi^(2)/36)-pi^(2)/2` `rArr y=pi^(2)/9-pi^(2) rArr y=-(8pi^(2))/9` Alternate Method We have , `sin x dy/dx+y cos x = 4x,` which can be written as `d/dx(sin x cdoty) =4x` On integrating both sides, we get `int d/dx (sin x cdoty)cdot dx =int 4x cdot dx` `rArr y cdot sin x =(4x^(2))/2 + C rArr y cdot sin x =2x^(2)+C` Now, as y=0 when `x = pi/2` `therefore C=-pi^(2)/2` `rArr y cdot sin x = 2x^(2) -pi^(2)/2` Now, putting `x=pi/6,` we get `y(1/2)=2((pi^(2))/36)-pi^(2)/2 rArr y = pi^(2)/9-pi^(2)=-(8pi^(2))/9` |
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| 673. |
If a curve `y=f(x)`passes through the point `(1,-1)`and satisfies the differentialequation `,y(1+x y)dx""=x""dy`, then `f(-1/2)`is equal to:(1) `-2/5`(2) `-4/5`(3) `2/5`(4) `4/5`A. (a) -2/5B. (b) -4/5C. (c) 2/5D. (d) 4/5 |
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Answer» Correct Answer - (d) Given differential equation is `y(1+xy)dx=x dy` `rArr y dx+xy^(2)dx=x dy` `rArr (x dy- y dx)/y^(2) =x dx` `rArr -((x dy- y dx))/y^(2) =x dx rArr -d (x/y)=x dx` On integrating both sides, we get `-x/y=x^(2)/2+C …(i)` `therefore` It passes through (1,-1). `therefore 1=1/2+C rArr C=1/2` Now, from Eq. (i) `-x/y=x^(2)/2+1/2` ` rArr x^(2)+1=-(2x)/y` ` rArr y=-(2x)/(x^(2)+1)` `therefore f(-1/2)=4/5` |
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| 674. |
STATEMENT-1 : The differential equation `(d^(2)y)/(dx^(2)) + cos x.(dy)/(dx) + (x^(3) + 7)y = e^(x)` is a linear equation and STATEMENT-2 : Every first degree equation is a linear equation.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 FalseD. Statement-1 is False, Statement-2 is True |
| Answer» Correct Answer - C | |
| 675. |
A curve passes through `(1,0)` and satisfies the differential equation `(2x cos y + 3x^2y) dx +(x^3 - x^2 siny - y)dy = 0`A. The equation of curve is `x^(2) cos y + x^(3)y - y^(2) = 1`B. The equation of curve is `x^(2) cos y + x^(3)y - (y^(2))/(2) = 1`C. The equation of normal at (1, 0) is y = 0D. The equation of tangent at (1, 0) is x = 1 |
| Answer» Correct Answer - B::C::D | |
| 676. |
The solution of differential equation `x^(2)y^(2)dy = (1-xy^(3))dx` isA. `x^(3)y^(3) = x^(2) + C`B. `2x^(3)y^(3) = 3x^(2) +C`C. `x^(3)y^(3) = x^(2) + x + C`D. `x^(3)y^(3) = 3x^(2) + C` |
| Answer» Correct Answer - B | |
| 677. |
Let a curve passes through (3, 2) and satisfied the differential equation (x-1) dx + 4(y -2) dy = 0A. It represents equation of a circleB. It represents equation of an ellipseC. Area euclosed by the curve is `2pi`.D. Line y = 1 is a tengent to the curve |
| Answer» Correct Answer - B::C::D | |
| 678. |
Find the order of the following differential equations: `((d^(2)y)/(dx^(2)))^(2) + 3((dy)/(dx))^(3) = 5 x^(2)` |
| Answer» Correct Answer - 2 | |
| 679. |
The curve in which the slope of the tangent at any point equal the ratio of the abscissa to the ordinate of the point isA. an ellipseB. a parabolaC. a rectangular hyperbolaD. a circle |
| Answer» Correct Answer - C | |
| 680. |
`a^(x)(y^(2)+1)dx=ydy`A)`a^(x)=tan^(-1)y+c` B)`tan^(-1)y=a^(x)log_(a)e+c` C)`a^(x)log_(a)e=log(y^(2)+1)+c` D)`(a^(x))/(loga)=logsqrt(y^(2)+1)+c`A. `a^(x)=tan^(-1)y+c`B. `tan^(-1)y=a^(x)log_(a)e+c`C. `a^(x)log_(a)e=log(y^(2)+1)+c`D. `(a^(x))/(loga)=logsqrt(y^(2)+1)+c` |
| Answer» Correct Answer - D | |
| 681. |
Find the general solution of the following differential equation :`(1y2)+((x-e^"tan"^((-1_"y")))dy)/(dx)=0` |
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Answer» Given, differential equation is `" "(1+y^(2))=(x-e^(tan^(-1)y)(dy)/(dx)` `rArr" "(1+y^(2))=-(x-e^(tan^(-1)y))(dy)/(dx)` `" "(1+y^(2))(dx)/(dy)=-x+e^(tan^(-1))y` `rArr" "(1+y^(2))(dx)/(dy)+x=e^(tan^(-1))y` `rArr" "(dx)/(dy)+(x)/(1+y^(2))=(e^(tan^(-1))y)/(1+y^(2))" "` [dividing throughout by `(1+y^(2))`] which is a linear differential equation. On comparing it with `(dx)/(dy)+Px=Q`, we get `" "P=(1)/(1+y^(2)),Q=(e^(tan^(-1)y))/(1+y^(2))` `" "IF=e^(intPdy)=e^(int(1)/(1+y^(2)))=e^(tan^(-1)y)` The general solution is `x*e^(tan^(-1)y)=int(e^(tan^(-1)y))/(1+y^(2))*e^(tan^(-1)y)dy+C` `rArr" "x*e^(tan^(-1)y)=int((e^(tan^(-1)y)))/(1+y^(2))*dy+C` Put `tan^(-1)y=trArr " "(1)/(1+y^(2))dy=dt` `therefore" "x*e^(tan^(-1)y)=inte^(2t)dt+C` `rArr" "x*e^(tan^(-1)y)=(1)/(2)e^(2tan^(-1)y)+C` `rArr" "2xe^(tan^(-1)y)=e^(2tan^(-1)y)+2C` `rArr" "2xe^(tan^(-1)y)=e^(2tan^(-1)y)+K" "[becauseK=2C]` |
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| 682. |
Solve the following differential equation: `(x^2dy)/(dx)=x^2+x y+y^2` |
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Answer» Given that, `" "x^(2)(dy)/(dx)=x^(2)+xy+y^(2)` `rArr" "(dy)/(dx)=1+(y)/(x)+(y^(2))/(x^(2))" "...(i)` Let `" "f(x,y)=1+(y)/(x)+(y^(2))/(x^(2))` `" "f(lamdax,lamday)=1+(lamday)/(lamdax)+(lamda^(2)y^(2))/(lamda^(2)x^(2))` `" "f(lamdax, lamday)=lamda^(0)(1+(y)/(x)+(y^(2))/(x^(2)))` `" "=lamda^(0)f(x,y)` which is homogeneous expression of degree 0. Put `" "y=vxrArr=(dy)/(dx)=v+x(dv)/(dx)` On substituting these values in Eq. (i), we get `" "(v+x(dv)/(dx))=1+V+V^(2)` `rArr" "x(dv)/(dx)=1+v+v^(2)-v` `rArr" "x(dv)/(dx)=1+v^(2)` `rArr" "(dv)/(1+v^(2))=(dx)/(x)` On integrating both sides, we get `" "tan^(-1)v=log|x|+C` `rArr" "tan^(-1)((y)/(x))=log|x|+C` |
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