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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.
| 1. |
Find the general solution of the differential equations:`(1+x^2)dy+2x y dx=cotx dx(x!=0)` |
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Answer» `(1+x^(2))dy+2xydx=cotxdx` `implies (dy)/(dx)+(2x)/(1+x^(2))y=(cotx)/(1+x^(2))` Here, `P=(2x)/(1+x^(2))` , `Q=(cotx)/(1+x^(2))` `:. I.F.=e^(int(2x)/(1+x^(2)))dx=e^(log(1+x^(2)))=(1+x^(2))` and general solution `y(1+x^(2))=int(cotx)/(1+x^(2))(1+x^(2))dx+c` `=intcotxdx+c` `implies y(1+x^(2))=logsinx+c` |
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| 2. |
If `y=x/(In|cx|)` (where c is an arbitrary constant) is the general solution of the differential equation `(dy)/(dx)=y/x+phi(x/y)` then function `phi(x/y)` is:A. `x^(2)//y^(2)`B. `-x^(2)//y^(2)`C. `y^(2)//x^(2)`D. `-y^(2)//x^(2)` |
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Answer» Correct Answer - D `logc+log|x|=x/y` Differentiating w.r.t. `x,1/x=(y-x(dy)/(dx))/(y^(2))` or `y^(2)/x=y-x(dy)/(dx)` or `(dy)/(dx)=y/x-y^(2)/x^(2)` or `phi(x/y) = -y^(2)/x^(2)` |
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| 3. |
`y(1+logx)(dx)/(dy)-xlogx=0, y=e^(2)" if "x=w`A. `y=exlogx`B. `ey=xlogx`C. `x=eylogy`D. `ex=ylogy` |
| Answer» Correct Answer - A | |
| 4. |
Find the general solution of the differential equations:`xlogx(dx)/(dy)+y=2/xlogx` |
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Answer» Given, `x log x(dy)/(dx)+y=(2)/(x)logx` `implies (dy)/(dx)+(y)/(xlogx)=(2)/(x^(2))` Comparing with differential equation `(dy)/(dx)+Py=Q` `P=(1)/(xlogx)` and `Q=(2)/(x^(2))` `I.F.=e^(intPdx)=e^(int(1dx)/(xlogx))` Let `logx=timplies(1)/(x)=(dt)/(dx)implies(dx)/(x)=dt` `:. I.F. =e^(int((1)/(t)dt)=e^(log|t|)=t=logx` `:.` Solution of given differential equation, `y*I.F.=intQ*I.F.dx+C` `impliesy*logx=int(2)/(x^(2))logxx+C` `implies ylogx=2int(logx*(1)/(x^(2)))dx+C` `=2[logxint(1)/(x^(2))dx-int{(d)/(dx)(logx)int(1)/(x^(2))dx}dx]+C` `=2[logx(-(1)/(x))-int{(1)/(x)*(-(1)/(x))}dx]+C` `=2(-(logx)/(x)+int(1)/(x^(2))dx)+C` `=2(-(logx)/(x)-(1)/(x))+C` `implies ylogx=-(2)/(x)(1+kogx)+C` |
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| 5. |
`e^(dy//dx)=1`A)`y=x+c`B)`y=e^(x)+c`C)`y=c`D)`y=cx+d`A. `y=x+c`B. `y=e^(x)+c`C. `y=c`D. `y=cx+d` |
| Answer» Correct Answer - C | |
| 6. |
`(dy)/(dx)=(y)/(x)-x/(y),` where `y=vx` A)`y^(2)=xlog((c)/(x))` B)`y^(2)=4xlog((c)/(x))` C)`y^(2)=2x^(2)log((c)/(x))` D)`v^(2)=2clogv`A. `y^(2)=xlog((c)/(x))`B. `y^(2)=4xlog((c)/(x))`C. `y^(2)=2x^(2)log((c)/(x))`D. `v^(2)=2clogv` |
| Answer» Correct Answer - C | |
| 7. |
`e^(dy//dx)=x`A)`y=xlog((x)/(e))+c`B)`y=xlogx+c`C)`y=e^(x)+c`D)`x=ylog(ex)+c`A. `y=xlog((x)/(e))+c`B. `y=xlogx+c`C. `y=e^(x)+c`D. `x=ylog(ex)+c` |
| Answer» Correct Answer - A | |
| 8. |
The differential equations, find a particular solution satisfying the given condition: `(x^3+x^2+x+1)(dy)/(dx)=2x^2+x ; y=1`when `x = 0` |
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Answer» Given, `(x^(3)+x^(2)+x+1)(dy)/(dx)=2x^(2)+x` `implies dy=(2x^(2)+x)/((x^(3)+x^(2)+x+1))dx` `implies intdy=int(2x^(2)+x)/(x^(2)(x+1)+1(x+1))dx` `implies intdy=int(2x^(2)+x)/((x+1)(x^(2)+1)))dx` `implies y=int(2x^(2)+x)/((x+1)(x^(2)+1))dx`………..`(1)` Let `(ex^(2)+x)/((x+1)(x^(2)+1))=(A)/((x+1))+(Bx+C)/((x^(2)+1))`.........`(2)` `implies (2x^(2)+x)/((x+1)(x^(2)+1))=(A(x^(2)+1)+(Bx+C)(x+1))/((x+1)(x^(2)+1))` `implies 2x^(2)+x=Ax^(2)+A+Bx^(2)+Bx+Cx+C` `implies 2x^(2)+x=x^(2)(A+B)+x(B+C)+(A+C)` Comparing constant terms, coefficients of `x` and `x^(2)` on both sides. `A+B=2`, `B+C=1` and `A+C=0` Solving , we get, `A=(1)/(2)`, `B=(3)/(2)` and `C=-(1)/(2)` put these values of `A`, `B` and `C` in equation `(2)` `(2x^(2)+x)/((x+1)(x^(2)+1))=(1)/(2(x+1))+((3)/(2)x-(1)/(2))/(x^(2)+1)` `implies int(2x^(2)+x)/((x^(2)+1)(x+1))dx` `=(1)/(2)int(1)/((x+1))dx+int((3)/(2)x-(1)/(2))/((x^(2)+1))dx` Then from equation `(1)`, `y=(1)/(2)log|x+1|+(3)/(2)int(x)/((x^(2)+1))dx-(1)/(2)int(1)/((x^(2)+1))dx` `implies y=(1)log|x+1|+(3)/(2)int(x)/((x^(2)+1))dx-(1)/(2)tan^(-1)x` Let `x^(2)+1=timplies2x=(dt)/(dx)impliesdx=(dt)/(2x)` `:. y=(1)/(2)log|x+1|+(3)/(2)int(x)/(t)(dt)/(2x)-(1)/(2)tan^(-1)x` `implies y=(1)/(2)log|x+1|+(3)/(4)int(1)/(t)dt-(1)/(2)tan^(-1)x` `y=(1)/(2)log|x+1|+(3)/(4)log|t|-(1)/(2)tan^(-1)x|C` `implies y=(1)/(2)log|x+1|+(3)/(4)log|x^(2)+1|-(1)/(2)tan^(-1)x+C` `implies y=(1)/(4)[2log|x+1|+3log|x^(2)+1|]-(1)/(2)tan^(-1)x+C` `implies y=(1)/(4)[log{(x+1)^(2)(x^(2)+1)^(3)}]-(1)/(2)tan^(-1)x+C`...........`(3)` when `x=0`, then `y=1`, put these in equation `(3)`, `1(1)/(4)log(1)-(1)/(2)tan^(-1)(0)+C` `implies 1=(1)/(4)xx0-(1)/(2)xx0+CimpliesC=1` put `C=1`, in equation `(3)` , `y=(1)/(4)[log{(x+1)^(2)(x^(2)+1)^(3)}]-(1)/(2)tan^(-1)x+1` which is the required particular solution. |
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| 9. |
Find the general solution of the differential equations `(dy)/(dx)=sin^(-1)x` |
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Answer» `(dy)/(dx)=sin^(-1)x` `implies dy=sin^(-1)x.dx` `implies intdy=int1*sin^(-1)xdx+x` `impliesy=sin^(-1)x*int1dx-int((d)/(dx)sin^(-1)x)(int1*dx)dx+c` `implies y=sin^(-1)x` `=int(x)/(sqrt(1-x^(2)))dx+c` Let `1-x^(2)=t` `implies -2x=(dt)/(dx)` `implies xdx=(-dt)/(2)` `implies y=xsin^(-1)x-int(-dt)/(2sqrt(t))+c` `impliesy=xsin^(-1)x+(1)/(2)intt^(-1//2)dt+c` `implies y=x sin^(-1)x+sqrt(t)+c` `implies y=x sin^(-1)x+sqrt(1-x^(2))+c` |
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| 10. |
The differential equation of the family of curves `y=e^x(Acosx+Bsinx),`where `A`and `B`arearbitrary constants is(a) `( b ) (c) (d)(( e ) (f) d^(( g )2( h ))( i ) y)/( j )(( k ) d (l) x^(( m )2( n ))( o ))( p ) (q)-2( r )(( s ) dy)/( t )(( u ) dx)( v ) (w)+2y=0( x )`(y)(z) `( a a ) (bb) (cc)(( d d ) (ee) d^(( f f )2( g g ))( h h ) y)/( i i )(( j j ) d (kk) x^(( l l )2( m m ))( n n ))( o o ) (pp)+2( q q )(( r r ) dy)/( s s )(( t t ) dx)( u u ) (vv)+2y=0( w w )`(xx)(yy)`( z z ) (aaa) (bbb)(( c c c ) (ddd) d^(( e e e )2( f f f ))( g g g ) y)/( h h h )(( i i i ) d (jjj) x^(( k k k )2( l l l ))( m m m ))( n n n ) (ooo)+( p p p ) (qqq)(( r r r ) (sss) (ttt)(( u u u ) dy)/( v v v )(( w w w ) dx)( x x x ) (yyy) (zzz))^(( a a a a )2( b b b b ))( c c c c )+y=0( d d d d )`(eeee)(ffff)`( g g g g ) (hhhh) (iiii)(( j j j j ) (kkkk) d^(( l l l l )2( m m m m ))( n n n n ) y)/( o o o o )(( p p p p ) d (qqqq) x^(( r r r r )2( s s s s ))( t t t t ))( u u u u ) (vvvv)-7( w w w w )(( x x x x ) dy)/( y y y y )(( z z z z ) dx)( a a a a a ) (bbbbb)+2y=0( c c c c c )`(ddddd)A. `(d^(2)y)/(dx^(2))-2(dy)/(dx)+2y=0`B. `(d^(2)y)/(dx^(2))+2(dy)/(dx)-2y=0`C. `(d^(2)y)/(dx^(2))+((dy)/(dx))^(2)+y=0`D. `(d^(2)y)/(dx^(2))-7(dy)/(dx)+2y=0` |
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Answer» Correct Answer - A `y=e^(x)(A cosx+Bsinx)` `(dy)/(dx) = e^(x)[-Asinx+Bcosx]+e^(x)[A cosx+Bsinx]` `(dy)/(dx) = e^(x)[-A sinx + Bcosx]+y` ……………(1) Again, differentiating w.r.t x, we ger `(d^(2)y)/(dx^(2))=(dy)/(dx)-y-y+(dy)/(dx)` [using 1] `(d^(2)y)/(dx)-2(dy)/(dx)+2y=0` |
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| 11. |
The solution of `x^2y_1^2+xyy_1-6y^2=0` are |
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Answer» `y_1=dy/dx` `x^2y_1^2=xyy_1-6y^2=0` `y_1=(-xypmsqrt(x^2y^2+24xy^2))/(2x^2)` `y_1=(-xypm5xy)/(2x^2)` `y_1=(4xy)/(2x^2),(-6xy)/(2x^2)` `y_1=(2y)/x,(-3y)/x` `dy/dx=(2y)/x,(-3y)/x` `intdy/(2y)=intdx/x` `1/2lny=lnx+lnc` `lny=2(lnxc)` `y=cx^2` `1/2lny=lnx+c` `y_1=(-3y)/x` `dy/dx=(-3y)/x` `intdy/(-3y)=intdx/x` `-1/3lny=lnx+lnc` `lny=-3(lnxc)` `y=(xc)^(-3)` `c=x^3y`. |
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| 12. |
Find the general solution of the differential equations `e^xtanydx+(1-e^x)sec^2ydy=0` |
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Answer» `e^(x)tanydx+(1-e^(x))sec^(2)ydy=0` `implies e^(x)tany(dx=-(1-e^(x))sec^(2)ydy` `implies (e^(x))/((e^(x)-1))dx=(sec^(2))/(tany)dy` `implies int(e^(x))/((e^(x)-1))dx=int(sec^(2))/(tany)dy` Let `e^(x)=1-timplies e^(x)dx=dt` Let `tany=vimplies sec^(2)ydy=dv` `:. int(1)/(t)dt=int(1)/(v)dv` `implies log|t|=log|v|-log|C|` `implies log|e^(x)-1|=log|tany|-log|C|` `implies log|C(e^(x)-1)|=log|tany|` `implies C(e^(x)-1)=tany` which is the required general solution. |
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| 13. |
Find out the differential equation of `y=Acosx^2+Bsinx^2` |
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Answer» `y = Acosx^2+Bsinx^2` `=>dy/dx = A(-sinx^2)(2x)+B(cosx^2)(2x)` `=>dy/dx = 2x(Bcosx^2 - Asinx^2)` `=>(d^2y)/dx^2 = 2x(-Bsinx^2(2x) - Acosx^2(2x))+(Bcosx^2 - Asinx^2)(2)` `=>(d^2y)/dx^2 = -4x^2(Acosx^2+Bsinx^2) +2(Bcosx^2 - Asinx^2)` `=>(d^2y)/dx^2 = -4x^2y +1/x(dy/dx)` `=>(d^2y)/dx^2 -1/x(dy/dx) + 4x^2y = 0`, which is the required differential equation. |
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| 14. |
The general solution of the differential equation `(dy)/(dx) = y tan x - y^(2) sec x` isA. `tan x = (c + sec x)y`B. `sec y = (c + tan y)x`C. `sec x = (c + tan x) y`D. None of these |
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Answer» Correct Answer - C We have `(dy)/(dx) = y tan x - y^(2) sec x` `rArr" "(1)/(y^(2))(dy)/(dx)-(1)/(y) tan x = -sec x` Putting `(1)/(y) = v rArr (-1)/(y^(2))(dy)/(dx)=(dv)/(dx)`, we get `(dv)/(dx) + tan x * v = sec x` which is linear diff. equation `I.F. = e^(int tan xdx) = e^(log sec x) = sec x` `therefore" "`The solution is `v sec x = int sec^(2) xdx + c` `rArr" "(1)/(y) sec x = tan x + c` `rArr" "sec x = y (c + tan x)` |
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| 15. |
`ysin2xdx-(1+y^2+cos^2x)dy=0` |
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Answer» `Pdx+Qdy=0` `ysin2xdx+(-1-y^2-cos^2x)dy=0` `(dp)/(dy)=sin2x` `(dQ)/(dx)=-2cosx*(-sinx)=2sinxcosx` `intPdx=-1/2ycos2x+g(y)` `intQdy=-y-y^33/3-ycos^2x+h(x)` `cos^2x=1/2(1+cos2x)` `g(y):-y-y^3/3-y/2` and `h(x)=0` `-1/2ycos2x-3/2y-y^3/3=0`. |
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| 16. |
The solution of differential equation `x^(2)(x dy + y dx) = (xy - 1)^(2) dx` is (where c is an arbitrary constant)A. `xy - 1 = cx`B. `xy - 1 = cx^(2)`C. `(1)/(xy-1)=(1)/(x)+c`D. None of these |
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Answer» Correct Answer - C Given differential equation can be written as `(xdy+ydx)/((xy-1)^(2))=(dx)/(x^(2)) rArr (d(xy))/((xy-1)^(2))=(dx)/(x^(2))` Integrating both sides `-(1)/((xy-1))=-(1)/(x)+ c rArr (1)/(xy-1)=(1)/(x) + c` |
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| 17. |
Solve: `y^(4)dx+2xy^(3)dy=(ydx-xdy)/(x^(3)y^(3))` |
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Answer» We can rewrite the given equation as `y^(4)dx+2xy^(3)dy+1/(xy^(3))(xdy-ydx)/(x^(2))=0` `rArr xy^(7)dx+2x^(2)y^(6)dy+d(y/x)=0` `rArr x^(2)y^(6)dx+2x^(3)y^(5)dy+x/yd(y/x)=0` `rArr 1/3(3x^(2)y^(6)+6x^(3)y^(5)dy)+(d(y//x))/(y//x)=0` `rArr 1/3intd(x^(3)y^(6))+int(d(x//y)/(y//x))=0` `rArr (x^(3)y^(6))/(3)+log_(e)y/x=c` |
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| 18. |
Finda particular solution of the differential equation`(x" "" "y)" "(dx" "+" "dy)" "=" "dx" "" "dy`, given that `y" "=" "1`, when `x" "=" "0`. (Hint: put `x" "" "y" "=" "t`). |
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Answer» Given differential equation is : `(x-y)(dx+dy)=dx-dy` `impliesdx+dy=(dx-dy)/(x-y)` On integration , `int(dx+dy)=int(dx-dy)/(x-y)+C` `implies x+y-log|x-y|+C`……….`(1)` Given that, when `x=0` , then `y=-1` `:. 0+(-1)=log(0+1)+CimpliesC=-1` put this value in equation `(1)`, `x+y=log|x-y|-1` `implies log|x-y|=x+y+1` which is the required particular solution. |
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| 19. |
Form the differential equation having `y=(sin^(-1) x)^2 +A cos^(-1) x +B` where A and B are arbitrary constants, as its general solution |
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Answer» `y = (sin^-1x)^2+Acos^-1x+B` `=>dy/dx = 2sin^-1x(1/sqrt(1-x^2))+A(-1/(sqrt(1-x^2)))` `=>dy/dx = (2sin^-1x-A)(1/sqrt(1-x^2))` `=>dy/dx(sqrt(1-x^2)) = (2sin^-1x-A)` `=>A = 2sin^-1x-dy/dx(sqrt(1-x^2)) ` Again differentiating w.r.t. `x`, `=>0 = 2(1/sqrt(1-x^2)) -dy/dx(1/(2sqrt(1-x^2)))(-2x) - sqrt(1-x^2)(d^2y)/dx^2)` `=>2 +xdy/dx -(1-x^2)(d^2y)/dx^2 = 0` `=>(1-x^2)(d^2y)/dx^2 -xdy/dx-2 = 0`, which is the required differential equation. |
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| 20. |
STATEMENT-1 : `y = e^(x)` is a particular solution of `(dy)/(dx) = y`. STATEMENT-2 : The differential equation representing family of curve `y = a cos omega t + b sin omega t`, where a and b are parameters, is `(d^(2)y)/(dt^(2)) - omega^(2) y = 0`. STATEMENT-3 : `y = (1)/(2)x^(3)+c_(1)X+c_(2)` is a general solution of `(d^(2)y)/(dx^(2)) = 3x`.A. T F TB. T T TC. F F FD. F F T |
| Answer» Correct Answer - A | |
| 21. |
Which of the following differential equations has y = x as one of its particular solution?(A) `(d^2y)/(dx^2)-x^2(dy)/(dx)+x y=x` (B) `(d^2y)/(dx^2)+x(dy)/(dx)+x y=x` (C) `(d^2y)/(dx^2)-x^2(dy)/(dx)+x y=0` (D) `(d^2y)/(dx^2)+x(dy)/(dx)+x y=0A. `(d^(2)y)/(dx^(2)) - x^(2)(dy)/(dx) + xy = x`B. `(d^(2)y)/(dx^(2)) - x^(2)(dy)/(dx) + xy = 0`C. `(d^(2)y)/(dx^(2)) + x^(2)(dy)/(dx) + xy = x`D. `(d^(2)y)/(dx^(2)) + x(dy)/(dx) + xy = 0` |
| Answer» Correct Answer - B | |
| 22. |
Solution of the differential `(x+2y^(3))=(dx)/(dy)y` isA. `x = y^(2)(c + y^(2))`B. `x = y (c - y^(2))`C. `x = 2y(c - y^(2))`D. `x = y(c + y^(2))` |
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Answer» Correct Answer - D `x dy + 2y^(3) dy = y dx` `rArr" "2y dy = (y dx - x dy)/(y^(2)) rArr 2 ydy = d((x)/(y))` `rArr" "x = y (c + y^(2))`. |
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| 23. |
For `y gt 0 and x in R, ydx + y^(2)dy = xdy` where y = f(x). If f(1)=1, then the value of f(-3) isA. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - C `(x dy-y dx)/(y^(2))=dy` `rArr" "-int d((x)/(y))=int dy` `rArr" "-(x)/(y)=y+c` At `x = 1, y = 1" "therefore" "c = -2` At `x = -3, (3)/(y)=y-2` `rArr" "y^(2)-2y - 3 = 0` `rArr" "y = 3` |
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| 24. |
Solve the differential equation `y e^(x/y)dx=(x e^(x/y)+y^2)dy(y!=0)` |
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Answer» Given differential equation is `ye((x)/(y))dx=(xe^((x)/(y))+y^(2))dy` `implies e^((x)/(y))(dx)/(dy)=(x)/(y)e((x)/(y))+yimpliese^((x)/(y))(dx)/(dy)-(x)/(y)e^((x)/(y))=y` ………`(1)` Let `(x)/(y)=vimpliesx=vyimplies(dx)/(dy)=v+y(dv)/(dx)` From equation `(1)`, `e^(v)(v+y(dv)/(dy))-ve^(v)=y` `implies e^(v)y(dv)/(dy)=yimpliese^(v)(dv)/(dy)=1impliese^(v)dv=dy` On integration, `inte^(v)dv=int1dyimpliese^(v)=y+Cimpliese^((x)/(y))=y+C` |
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| 25. |
The number of arbitrary constants in the general solution of a differential equationof fourth order are:(A) 0 (B) 2 (C) 3 (D) 4 |
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Answer» General equation of 4th order is `(d^4y)/dx^4=k``(d^3y)/(dx^3)=kx+c1` `(d^2y)/(dx^2)=kx^2/2+c1x+c2` `(dy)/(dx)=kx^3/6+(c1)x^2/2+c2x+c3` `y=kx^4/24+(c1)x^3/6+c2x^2/2+c3x+c4` Number of arbitrary constant=`4` |
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| 26. |
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=(x+1)e^(-x)` |
| Answer» Correct Answer - B | |
| 27. |
General solution of differential equation `x^(2)(x+y(dy)/(dx))+(x(dy)/(dx)-y)sqrt(x^(2)+y^(2))=0` isA. `(1)/(sqrt(x^(2)+y^(2)))+(y)/(x)=c`B. `sqrt(x^(2)+y^(2))-(y)/(x) = c`C. `sqrt(x^(2)+y^(2))+(y)/(x)=c`D. `2 sqrt(x^(2)+y^(2))+(y)/(x)=c` |
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Answer» Correct Answer - C `(x dx + y dy)/(sqrt(x^(2)+y^(2)))+(x dy - y dy)/(x^(2))=0` `rArr" "(1)/(2)(d(x^(2)+y^(2)))/(sqrt(x^(2)+y^(2)))+d((y)/(x))=0` `rArr" "sqrt(x^(2)+y^(2))+(y)/(x) = c` |
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| 28. |
Find the particular solution of the differential equation:`(1+e^(2x))dy+(1+y^2)e^x dx=0,`given that `y=1, `when `x=0.` |
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Answer» Given differential equation is `(1+e^(2x))dy+(1+y^(2))e^(x)dx=0` `implies (dy)/(1+y^(2))+(e^(x)dx)/(1+e^(2x))=0` `impliesint(dy)/(1+y^(2))+int(e^(x)dx)/(1+e^(2x))=C` Let `t=e^(x)impliese^(x)dx=dt` `:.tan^(-1)y+int(dt)/(1+t^(2))=C` `implies tan^(-1)y+tan^(-1)t=C` `implies tan^(-1)y+tan^(-1)e^(x)=C` Now, put `x=0` and `y=1`, `:. tan^(-1)1+tan^(-1)e^(0)=C` `implies(pi)/(4)+(pi)/(4)=CimpliesC=(pi)/(2)` put the value of `C` in equation `(1)`, `tan^(-1)y+tan^(-1)e^(x)=(pi)/(2)` which is the required particular solution. |
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| 29. |
Verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:`y cos y = x` : (y sin y + cos y + x) y = y |
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Answer» `y-cosy=x`……..`(1)` differentiating we get, `(dy)/(dx)+siny*(dy)/(dx)=1` `implies (dy)/(dx)=(1)/(1+siny)` Now, `L.H.S=(y sin y+cosy+x)(dy)/(dx)` `=(y sin y+cosy+y-cosy)*(1)/(1+siny)` [from equation `(1)`] `=(y sin y+y)(1)/(1+siny)` `=y=R.H.S` Therefore, `y-cosy=x` is the solution of given differential equation. |
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| 30. |
Verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:`y=sqrt(a^2-x^2)x in (-x , a)` : `x+y(dy)/(dx)=0(y!=0)` |
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Answer» `y=sqrt(a^(2)-x^(2))`………`(1)` differentiate w.r.t.x `(dy)/(dx)=(1)/(2sqrt(a^(2)-x^(2)))*(-2x)=(-x)/(y)` [from equation `(1)`] `implies y*(dy)/(dx)=-x` `implies x+y*(dy)/(dx)=0` Therefore, `y=sqrt(a^(2)-x^(2))` is the solution of given differential equation. |
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| 31. |
The solution of the differential equation `(2y-1)dx-(2x+3)dy=0`, isA. `(2x-1)/(2y+3)=C`B. `(2x+3)/(2y-1)=C`C. `(2x-1)/(2y-1)=C`D. `(2y+1)/(2x-3)=C` |
| Answer» Correct Answer - B | |
| 32. |
Findthe equation of the curve passing through the point `(0,pi/4)`whose differential equation is `s in" "x" "cos" "y" "dx" "+" "cos" "x" "s in" "y" "dy" "=" "0`. |
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Answer» Differential equation of given curve is : `sin xcosydx+cosxsinydy=0` `implies(sinx)/(cosx)dx+(siny)/(cosy)dy=0` `implies tanxdx+tanydy=0` On integration , `inttanxdx+inttanydy=logC` `implieslog(secx)+log(secy)=logC` `implies secx*secy=C`…….`(1)` Curve passes through the point `(0,(pi)/(4))`, so put `x=0`, `y=(pi)/(4)` `sec0sec"(pi)/(4)=CimpliesC=sqrt(2)` put the value of `C` in equation `(1)` , `secx*secy=sqrt(2)` `implies secx*(1)/(cosy)=sqrt(2)implies cosy=(secx)/(sqrt(2))` Therefore, required equation of curve is : `cosy=(secx)/(sqrt(2))`. |
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| 33. |
Find the general solution of the differential equation `(dy)/(dx)+sqrt((1-y^2)/(1-x^2))=0`. |
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Answer» `(dy)/(dx)=-sqrt((1-y^(2))/(1-x^(2)))` `implies (dy)/(sqrt(1-y^(2)))=-(dx)/(sqrt(1-x^(2)))` `implies int(dy)/(sqrt(1-y^(2)))+int(dx)/(sqrt(1-x^(2)))=c` `implies sin^(-1)y+sin^(-1)x=c` |
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| 34. |
For each of the exercises given below, verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.(i) `y=a e^x+b e^(-x)+x^2` : `x(d^2y)/(dx^2)+2y(dy)/(dx)-x y+x^2-2=0`(ii) `y=e^x(acosx+bsin |
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Answer» `(i)` Given, `x(d^(2)y)/(dx^(2))+2(dy)/(dx)-xy+x^(2)-2=0`………`(1)` and `y=ae^(x)+be^(-x)+2x` `implies (dy)/(dx)=ae^(x)-be^(-x)+2x` `implies (d^(2)y)/(dx^(2))-ae^(x)+be^(-x)+2` Now, put the values of `y`, `(dy)/(dx)` and `(d^(2)y)/(dx^(2))` in equation `(1)` `L.H.S =x(d^(2)y)/(dx^(2))+2(dy)/(dx)-xy+x^(2)-2` `=x(ae^(x)+be^(-x)+2)+2(ae^(x)-be^(-x)+2x)-x(ae^(x)+be^(-x)+x^(2))+x^(2)-2` `=(axe^(x)+bxe^(-x)+2x)+(2ae^(x)-2be^(-x)+4x)-(axe^(x)+bxe^(-x)+x^(3))+x^(2)-2` `= axe^(x)+bxe^(-x)+2x+2ae^(x)-2be^(-x)+4x-axe^(x)-bxe^(-x)-x^(3)+x^(2)-2` `=2ae^(x)-2be^(-x)-x^(3)+x^(2)+6x-2ne0` `R.H.S ne L.H.S` Therefore, given function is not the solution of the given differential equation. `(ii)` Given `y=e^(x)(a cosx+bsinx)` `impliese^(-x)y=acosx+bsinx`.........`(1)` `implies e^(-x)(dy)/(dx)-ye^(-x)=-asinx+bcosx` `=e^(-x)(d^(2)y)/(dx^(2))-(dy)/(dx)e^(-x)-(-ye^(-x)+e^(-x)(dy)/(dx))` `=-a cosx-bsinx` `implies e^(-x)(d^(2)y)/(dx^(2))-2e^(-x)(dy)/(dx)+ye^(-x)-e^(-x)(dy)/(dx)` `=-(a cos x-bsinx)` `implies e^(-x)(d^(2)y)/(dx^(2))-2e^(-x)(dy)/(dx)+ye^(-x)=-ye^(-x)` [from equation `(1)` ] `implies e^(-x)(d^(2)y)/(dx^(2))-2e^(-x)(dy)/(dx)+2ye^(-x)=0` `implies e^(-x)((d^(2)y)/(dx)-2(dy)/(dx)+2y)=0` `implies (d^(2)y)/(dx^(2)-2(dy)/(dx)+2y=0` Therefore, given function is the solution of the given differential equation. `(iii)` Given, `y=x sin3x` `implies (dy)/(dX)=x(d)/(dx)(sin3x)+sin3x(d)/(dx)(x)` `(dy)/(dx)=x cos3xxx3+sin3x` `=3xcox3x+sin3x` `implies (d^(2)y)/(dx^(2))=3[x(d)/(dx)cos3x+cos3x(d)/(dx)(x)]+(d)/(dx)(sin3x)` `implies (d^(2)y)/(dx^(2))=3[x(-sin3xx x3)+cos3x]+cos3xx x3` `implies (d^(2)y)/(dx^(2))=-9xsin3x+3cos3x+3cos3x` `implies (d^(2)y)/(dx^(2))=-9xsin3x+6cos3x` `implies (d^(2)y)/(dx^(2))=-9y+6cos3x` [From equation `(1)`, `y=xsin3x`] `implies (d^(2)y)/(dx^(2))+9y-6cos3x=0` Therefore, the given function is the solution of the given differential equation. `(iv)` Given, `x^(2)=2y^(2)logy` `implies2x=2(y^(2)xx(1)/(y)(dy)/(dx)+logyxx2y(dy)/(dx))` `implies 2x=2(y+2ylogy)(dy)/(dx)` `impliesx=(y+2ylogy)(dy)/(dx)` `impliesxy=(y^(2)+2y^(2)logy)(dy)/(dx)` `implies xy=(y^(2)+x^(2))(dy)/(dx)` `implies (x^(2)+y^(2))(dy)/(dx)-xy=0` Therefore, the given function is the solution of given differential equation. |
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| 35. |
The general solution of `e^(x)cos ydx-e^(x)sin ydy=0`, isA. `e^(x)(siny+cosy)=C`B. `e^(x)siny=C`C. `e^(x)=C cos y`D. `e^(x)cosy=C` |
| Answer» Correct Answer - D | |
| 36. |
Show that the general solution of the differentiaequation `(dy)/(dx)+(y^2y+1)/(x^2+x+1)=0`is given by `x+y+1=A(1-x-y-2x y)`where A is a parameter. |
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Answer» Given differntial equation is : `(dy)/(dx)+(y^(2)+y+1)/(x^(2)+x+1)=0` `implies (dy)/(y^(2)+y+1)+(dx)/(x^(2)+x+1)=0` On integration `int(dy)/(y^(2)+y+1)+int(dx)/(x^(2)+x+1)=C` `impliesint(dy)/(y^(2)+y+1+((1)/(2))^(2)-((1)/(2))^(2))+int(dx)/(x^(2)+x+1+((1)/(2))^(2)-((1)/(2))^(2))=C` `impliesint(dy)/((y+(1)/(2))^(2)+(1-(1)/(4)))+int(dx)/((x+(1)/(2))^(2)+(1-(1)/(4)))=C` `impliesint(dy)/((y+(1)/(2))^(2)+((sqrt(3))/(2))^(2))+int(dx)/((x+(1)/(2))^(2)+((sqrt(3))/(2))^(2))=C` `implies(2)/(sqrt(3))tan^(-1)((y+(1)/(2))/((sqrt(3))/(2)))+(2)/(sqrt(3))tan^(-1)((x+(1)/(2))/((sqrt(3))/(2)))=C` `implies tan^(-1)(2y+1)/(sqrt(3))+tan^(-1)(2x+1)/(Sqrt(3))=(sqrt(3)C)/(2)=k` (say) `impliestan^(-1)[((2y+1)/(sqrt(3))+(2x+1)/(sqrt(3)))/(1-((2y+1)/(sqrt(3)))((2x+1)/(sqrt(3))))]=k` `impliestan^(-1)[((2y+1+2x+1)/(sqrt(3)))/(1-((4xy+2x+2y+1)/(3)))]=k` `implies(2sqrt(3)(x+y+1))/(3-(4xy+2x+2y+1))=tank` `implies(2sqrt(3)(x+y+1))/(2(1-x-y-2xy))=tank` `implies x+y+1=(1)/(sqrt(3))tank(1-x-y-2xy)` Let `A=(1)/(sqrt(3))tank` which is an arbitrary constant. `implies x+y+1=A(1-x-y-2xy)` |
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| 37. |
General solution of `(dy)/(dx)+(2xy)/(1+x^(2))=0` isA. `y(1+x^(2))=c`B. `y(1+x)=c`C. `1+x^(2)=cy`D. None of these |
| Answer» Correct Answer - A | |
| 38. |
Form the differential equation representing the family of curves given by `(x-a)^2+2y^2=a^2`, where a is an arbitrary constant. |
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Answer» Given, family of curve, `(x-a)^(2)+2y^(2)=a^(2)`, where `a` is constant. `implies x^(2)-2ax+2y^(2)=0`……`(1)` Differentiate w.r.t.x, `2x-2a+4y(dy)/(dx)=0`…….`(2)` Multiply equation `(2)` by `x` and subtract equation `(1)` from it, `x(2x-2a+4y(dy)/(dx))-(x^(2)-2ax+2y^(2))=0` `2x^(2)-2ax+4xy(dy)/(dx)-x^(2)+2ax-2y^(2)=0` `implies 4xy(dy)/(dx)+x^(2)-2y^(2)=0` `implies (dy)/(dx)=(2y^(2)-x^(2))/(4xy)` which is the required differential equation. |
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| 39. |
The solution of the differential equation `(d^(2)y)/(dx^(2))=e^(-2x)`, isA. `(1)/(4)e^(-2x)`B. `(1)/(4)e^(-2x)+cx+d`C. `(1)/(4)e^(-2x)+cx^(2)+d`D. `(1)/(4)e^(-2x)+c+d` |
| Answer» Correct Answer - B | |
| 40. |
The general solution of the differential equaiton `(1+y^(2))dx+(1+x^(2))dy=0`, isA. (x-y) = c(1-xy)B. (x-y) = c(1+xy)C. (x+y) = c(1-xy)D. (x+y) = c(1+xy) |
| Answer» Correct Answer - C | |
| 41. |
The solution of the differential equaitonA. `(dy)/(dx)+y tan x =sec x,` isB. `y sec x=tan x+C`C. `y tan x=sec x+C`D. `tan x=y tan x +C` |
| Answer» Correct Answer - A | |
| 42. |
The general solution of the equation `(dy)/(dx)=1+x y`is(a)`( b ) (c) y=c (d) e^( e ) (f)-( g )(( h ) (i) x^((( j )2( k ))( l ))/( m )2( n ) (o) (p))( q ) (r)`(s)(b) `( t ) (u) y=c (v) e^( w ) (x) (y)(( z ) (aa) x^((( b b )2( c c ))( d d ))/( e e )2( f f ) (gg) (hh))( i i ) (jj)`(kk)(c)`( d ) (e) y=(( f ) (g) x+c (h)),( i ) e^( j ) (k)-( l )(( m ) (n) x^((( o )2( p ))( q ))/( r )2( s ) (t) (u))( v ) (w)`(x) (d) None of theseA. `y=ce^(-x^(2)//2)`B. `y=ce^(x^(2)//2)`C. `y=(x+c),e^(-x^(2)//2)`D. None of these |
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Answer» Correct Answer - D `(dy)/(dx)=1+xy` or `(dy)/(dx)-xy=1` I.F. `=e^(int-xdx)=e^(-x^(2)//2)` Hence solution is `y.e^(-x^(2)//2) = inte^(-x^(2)+2)dx+c`. `inte^(-x^(2)//2)dx` is not further integrable. |
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| 43. |
Prove that `x^2-y^2=c(x^2+y^2)^2`is the general solution of differential equation `(x^3-2xy^2)dx=(y^3-3x^2y)dy`, where c is a parameter. |
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Answer» Given, `(x^(3)3xy^(2))dx=(y^(3)-3x^(2)y)dy` `(dy)/(dx)=(x^(3)-3xy^(2))/(y^(3)-3x^(2)y)` `implies v+x(dv)/(dx)=(x^(3)-3x^(3)v^(2))/(v^(3)x^(3)-3x^(3)v)` Let `y=vx` `implies (dy)/(dx)=v+x(dv)/(dx)` `=(1-3v^(2))/(v^(3)-3v)` `implies x(dv)/(dx)=(1-3v^(2))/(v^(3)-3v)-v` `=(1-3v^(2)-v^(4)+3v^(2))/(v^(3)-3v)` `=(1-v^(4))/(v^(3)-3v)` `implies (v^(3)-3v)/(1-v^(4))dv=(dx)/(x)` `implies int(2v(v(2)-3))/(1-v^(4))dv=2int(dx)/(x)+logc` `implies int(t-3)/(1-t^(2))dt=2logx+logc` (Let `v^(2)=timplies2vdv=dt`) `=int{(-1)/(1-t)-(2)/(1+t)}dt=2logx+logc` (Using partial fractions) `implies (log(1-t)-2log(1+t)=log(cx^(2))` `implieslog(1-t)/((1+t)^(2))=logcx^(2))` `implies (1-t)/((1+t)^(2))=cx^(2)` `implies1-v^(2)=cx^(2)(1+v^(2))^(2)` `implies 1-(y^(2))/(x^(2))=cx^(2)(1+(y^(2))/(x^(2)))^(2)` `implies x^(2)-y^(2)=c(x^(2)+y^(2))^(2)` |
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| 44. |
`y=a e^(-1/x)+b`is asolution of `(dy)/(dx)=y/(x^2),`then(a)`( b ) (c) a in R (d)`(e) (b) 0(c)`( d ) (e) b=1( f )`(g)(d) `( h ) a (i)`(j) takesfinite number of valuesA. `x in R-{0}`B. `b=0`C. `b=1`D. a takes finite number of values |
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Answer» Correct Answer - A::B `(dy)/(dx)=y/x^(2)` or `(dy)/(y) = (dx)/x^(2)` or `"ln "y=-1/x+"ln "c` or `y/c=e^(-1/x)` or `y=ce^(-1//x)` Comparing with `y=ae^(-1//x)+b, a in R-{0},b=0` |
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| 45. |
The graph of the function y = f(x) passing through the point (0, 1) and satisfying the differential equation `(dy)/(dx) + y cos x = cos x` is such thatA. It is constant functionB. It is periodicC. it is neither an even nor an odd function.D. it is continuous and differentiable for all x |
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Answer» Correct Answer - A::B::D `(dy)/(dx)+ycosx=cosx` (linear) I.F. =`e^(intcosxdx)=e^(sinx)` Thus, solution is `ye^(sinx)=inte^(sinx)cosxdx=e^(sinx)+c` When `x=0, y=1`, then c=0 Thus, y=1. Hence, option (1), (2), (4) are true. |
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| 46. |
The solution of the differential equaton `y-x(dy)/(dx)=a(y^(2)+(dy)/(dx))`, isA. `(x+a)(x+ay)=Cy`B. `(x+a)(1-ay)=Cy`C. `(x+a)(1-ay)=C`D. none of these |
| Answer» Correct Answer - B | |
| 47. |
solution of the differential equation `xdy-ydx=sqrt(x^2+y^2 )dx` isA. `x+sqrt(x^(2)+y^(2))=Cx^(2)`B. `y-sqrt(x^(2)+y^(2))=Cx`C. `x-sqrt(x^(2)+y^(2))=Cx`D. `y+sqrt(x^(2)+y^(2))=Cx^(2)` |
| Answer» Correct Answer - D | |
| 48. |
The solution of the differential equation `((x+2y^3)dy)/(dx)=y`is(a)`( b ) (c) (d) x/( e )(( f ) (g) y^(( h )2( i ))( j ))( k ) (l)=y+c (m)`(n)(b) `( o ) (p) (q) x/( r ) y (s) (t)=( u ) y^(( v )2( w ))( x )+c (y)`(z)(c)`( d ) (e) (f)(( g ) (h) x^(( i )2( j ))( k ))/( l ) y (m) (n)=( o ) y^(( p )2( q ))( r )+c (s)`(t)(d) `( u ) (v) (w) y/( x ) x (y) (z)=( a a ) x^(( b b )2( c c ))( d d )+c (ee)`(ff)A. `x/y^(2)=y+c`B. `x/y=y^(2)+c`C. `x^(2)/y=y^(2)+c`D. `y/x=x^(2)+c` |
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Answer» Correct Answer - B `(dy)/(dx) = (x+2y^(3))/(y)` or `(dx)/(dy)-1/yx=2y^(2)` which is linear I.F. `e^(int-1/ydy)=e^(-logy) = 1/y` Thus, solution is `1/yx = int1/y2y^(2)dy+c=y^(2)+c` or `x/y=y^(2)+c` |
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| 49. |
The general solution of the differential equaiton `(1+y^(2))dx+(1+x^(2))dy=0`, isA. `x-y=C(1-xy)`B. `x-y=C(1+xy)`C. `x+y=C(1-xy)`D. `x+y=C(1+xy)` |
| Answer» Correct Answer - C | |
| 50. |
The general solution of the differential equation `(dy)/(dx)=x^2/y^2` isA. `x^(3)-y^(3)=C`B. `x^(3)+y^(3)=C`C. `x^(2)+y^(2)=C`D. `x^(2)-y^(2)=C` |
| Answer» Correct Answer - A | |