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201.

The order and degree of the differential equation `(d^(2)y)/(dx^(2))+((dy)/(dx))^(1//4)+x^(1//5)=0` respectively areA. 2 and 4B. 2 and 2C. 2 and 3D. 3 and 3

Answer» Given that, `" "(d^(2)y)/(dx^(2))+((dy)/(dx))^(1//4)=-x^(1//5)`
`rArr" "(d^(2)y)/(dx^(2))+((dy)/(dx))^(1//4)=-x^(1//5)`
`rArr" "((dy)/(dx))^(1//4)=-(x^(1//5)+(d^(2)y)/(dx^(2)))`
On squaring both sides, we get
`" "((dy)/(dx))^(1//2)=(x^(1//5)+(d^(2)y)/(dx^(2)))^(2)`
Again, on sqaring both sides, we have
`" "(dy)/(dx)=(x^(1//5)+(d^(2)y)/(dx^(2)))^(4)`
order = 2, degree = 4
Discussion
202.

The degree of the differential equation `[1+((dy)/(dx))^(2)]^(3//2)=(d^(2)y)/(dx^(2))"is"`A. 4B. `(3)/(2)`C. not definedD. 2

Answer» Given that `[1+((dy)/(dx))^(2)]^(3//2)=(d^(2)y)/(dx^(2))`
On squaring both sides, we get `[1+((dy)/(dx))^(2)]^(3//2)=((d^(2)y)/(dx^(2)))^(2)`
So, the degree of differential equation is 2.
Discussion
203.

`x(dy)/(dx)=y(logy-logx+1)`A. `logx/y=cy`B. `logy/x=cy`C. `logy/x=cx`D. None of these

Answer» Correct Answer - C
`(dy)/(dx)= y/x[logy/x+1]`
Put `y=vx`
`v+x(dv)/(dx) = vlogv+v`
`therefore (dv)/(vlogv) = (dx)/x`
`therefore log(logv)=logx+logc=logcx`
`therefore logy/x=cx`
Discussion
204.

The degree of the potential equation `((d^(2)y)/(dx^(2)))^(2)+((dy)/(dx))^(2)=x sin ((dy)/(dx))"is`A. 1B. 2C. 3D. not defined

Answer» The degree of above differential equation is not defined because when we expand sin `((dy)/(dx))` we get in infinite series in the increasing power of `(dy)/(dx)`. Therefore its degree is not defined.
Discussion
205.

Solve `x(dy)/(dx)=y(logy-logx+1)`

Answer» Given, `" "x(dy)/(dx)=y(logy-logx+1)`
`rArr" "x(dy)/(dx)=ylog((y)/(x)+1)`
`rArr" "(dy)/(dx)=(y)/(x)(log""(y)/(x)+1)" "`...(i)
which is a homogeneous equation.
Put `" "(y)/(x)=v or y=vx`
`therefore" "(dy)/(dx)=v+x(dv)/(dx)`
On substituting these values in Eq. (i), we get
`rArr" "x(dv)/(dx)=v(logv+1-1)`
`rArr" "x(dv)/(dx)=v(logv)`
`rArr" "(dv)/(vlogv)=(dx)/(x)`
On integrating both sides, we get
`" "int(dv)/(vlogv)=int(dx)/(x)`
On putting logv=u in LHS integral, we get `" "(1)/(v)*dv=du`
`" "int(du)/(u)=int(dx)/(x)`
`rArr" "logu=logx+logC`
`rArr" "logu=logCx`
`rArr" "u=Cx`
`rArr" "logv=Cx`
`rArr" "log((y)/(x))=Cx`
Discussion
206.

`x((d^(3)y)/(dx^(3)))^(2)-((d^(2)y)/(dx^(2)))^(4)=0` . Find order and degree of differential equation.A. 3, 2B. 3, 3C. 3, 4D. 2, 4

Answer» Correct Answer - A
Discussion
207.

`(d^(3)y)/(dx^(3))+6(d^(2)y)/(dx^(2))+y=0` . Find order and degree of differential equation.A. 3, 2B. 2, 3C. 3, 1D. 3, 4

Answer» Correct Answer - C
Discussion
208.

`(d^(3)y)/(dx^(3))+((dy)/(dx))^(3)+y=sinx` . Find order and degree of differential equation.A. 1, 3B. 3, 3C. 3, 0D. 3, 1

Answer» Correct Answer - D
Discussion
209.

`((d^(2)y)/(dx^(2)))^(3)+((dy)/(dx))^(4)=x^(5)` . Find order and degree.A. 2, 1B. 2, 3C. 2, 4D. 2, 5

Answer» Correct Answer - B
Discussion
210.

The equation of the curve satisfying thedifferential equation `y((dy)/(dx))^2+(x-y)(dy)/(dx)-x=0`can be a(a) circle (b) Straight line(c) Parabola (d)EllipseA. CircleB. Straight lineC. ParabolaD. Ellipse

Answer» Correct Answer - A::B
`y((dy)/(dx))^(2)+(x-y)(dy)/(dx) -x=0`
or `(dy)/(dx) = ((y-x)+-sqrt((x-y)^(2)+4xy))/(2y)`
So, `(dy)/(dx)=1` which gives straight line.
or `(dy)/(dx)=1` which gives straight line
or `(dy)/(dx)=-x/y` which gives circle.
Discussion
211.

The equation of the curve satisfying thedifferential equation `y((dy)/(dx))^2+(x-y)(dy)/(dx)-x=0`can be a(a) circle (b) Straight line(c) Parabola (d)EllipseA. x-y+1 = 0B. `x^(2) + y^(2) = 25`C. `x^(2) + y^(2) - 5x - 10 = 0`D. x+y-7 = 0

Answer» Correct Answer - A::B
Discussion
212.

The differential equation of the rectangular hyperbola whose axes are the asymptotes of the hyperbola, isA. `y(dy)/(dx)=x`B. `x(dy)/(dx)=-y`C. `x(dy)/(dx)=y`D. `x dy +ydx=C`

Answer» Correct Answer - B
The equation of the rectangular hyperbola whose axes are asymptotes of the hyperbola `x^(2)-y^(2)=a^(2)` is
`xy=c^(2)`, where `c^(2)=a^(2)//2`
This is a one parameter family of curves.
Differentiating with respect to x, we get
`x(dy)/(dx)+y=0rArr x(dy)/(dx)=-y`
Discussion
213.

The order and degree of the differential equationof all tangent lines to the parabola `y=x^2`is(a)1,2 (b) 2,3(c) 2,1 (d)1,1A. 1,2B. 2,3C. 2,1D. 1,1

Answer» Correct Answer - A
The parametric form of the given equation is `x=t, y=t^(2)`.
Differentiating, we get `2t=y_(1)`.
Putting this value in the equation of tangent, we get
`2xy_(1)//2=y+(y_(1)//2)^(2)`
or `4xy_(1)=4y+y_(1)^(2)`
The order of this equation is one and degree is two.
Discussion
214.

The differential equation of all parabolas whose axis of symmetry is along x-axis is of orderA. 2B. 3C. 1D. none of these

Answer» Correct Answer - A
The general equation of the said parabolas is `x=ay^(2)+b`, where a, b are arbitrary constants. So, the differential equation is of order 2.
Discussion
215.

The integrating factor of the differential equation `(dy)/(dx)+y=(1+y)/(x)`, isA. `(x)/(e^(x))`B. `(e^(x))/(x)`C. `xe^(x)`D. `e^(x)`

Answer» Correct Answer - B
Discussion
216.

Thesolution of the differential equation `(dy)/(dx)=(x+y)/x`satisfying the condition `y""(1)""=""1`is(1) `y""="ln"x""+""x`(2) `y""=""x"ln"x""+""x^2`(3) `y""=""x e(x-1)`(4) `y""=""x"ln"x""+""x`

Answer» `y = vx`
`dy/dx = (x + vx)/x`
`= 1+v`
diff wrt x
`dy/dx = v + x(dv)/dx= 1+ v`
`x (dv)/(dx) = 1`
`int dv = int dx/x`
`v/x = ln x + c`
`y = x ln x + cx`
`1= c*1 = 1`
`c=1`
so,`y= x ln x + x`
option 4 is correct
Discussion
217.

The tangent at any point `(x , y)`of a curve makes an angle `tan^(-1)(2x+3y)`with x-axis. Find the equation of the curve if itpasses through (1,2).A. `6x+9y+2=26e^(3(x-1))`B. `6x-9y+2=26e^(3(x-1))`C. `6x+9y-2=26e^(3(x-1))`D. `6x-9y-2=26e^(3(x-1))`

Answer» Correct Answer - A
Discussion
218.

Solve `x (dy)/(dx)sin(y/x)+x-ysin(y/x)=0` given `y(1)=pi/2`

Answer» `x(dy/dx)sin(y/x)+x-ysin(y/x) = 0`
Dividing given equation with `xsin(y/x)`
`=>dy/dx+1/sin(y/x)-y/x = 0`
Let `y/x = v`
Then, `y = vx =>dy/dx = v+x(dv)/dx`
So, given equation becomes,
`=>v+x(dv)/dx+1/sinv-v = 0`
`=> x(dv)/dx = -1/sinv`
`=> sinvdv = -dx/x`
Integrating both sides,
`=> int sinvdv = int -dx/x`
`=>-cos v = -ln x +c`
`=>-cos(y/x) = -ln x +c`
Now, we are given, `y(1) = pi/2`
So, putting `x = 1 and y = pi/2`
`=>-cos pi/2 = -ln(1) + c => c = 0`
So, our equation becomes,
`=>-cos(y/x) = -lnx`
`=>cos(y/x) = ln x`, which is the required solution.
Discussion
219.

`xy(dy)/(dx)=y+2`A. `y-2log(y+2)=logx+c`B. `y+2log(y-2)=logx+c`C. `x-2log(x+2)=logy+c`D. `y^(2)+2y=logx+c`

Answer» Correct Answer - A
Discussion
220.

Solution of `x(dy)/(dx)+y=xe^(x)`, isA. `xy=e^(x)(x+1)+C`B. `xy=e^(x)(x-1)+C`C. `xy=e^(x)(1-x)+C`D. `xy=e^(y)(y-1)+C`

Answer» Correct Answer - B
Discussion
221.

The solution of x sin `((y)/(x))dy={ysin((y)/(x))-x}dx,` is given byA. `logx+sinv=c`B. `logv-cosx=c`C. `logx-cosv=c`D. `logv+sinx=c`

Answer» Correct Answer - C
Discussion
222.

The solution of differential equation `(1+y^(2))+(x-e^(tan^(-1)y))(dy)/(dx)=0`, isA. `xe^(2tan^(-1)y)=e^(tan^(-1)y+K)`B. `(X-2)=Ke^(tan^(-1)y)`C. `2xe^(tan^(-1)y)=e^(2 tan^(-1)y)+K`D. `xe^(tan^(-1)y)=tan^(-1)y+K`

Answer» Correct Answer - C
We have,
`(1+y^(2))+(x-e^(tan^(-1)))(dy)/(dx)=0`
`rArr" "(dx)/(dxy)+(x)/(1+y^(2))=(e^(tan^(-1)y))/(1+y^(2))" ...(i)"`
The linear differential equation with
`"I.F."=e^(int(1)/(1+y^(2))dy)=e^(tan^(-1)y)`
Multiplying (i) by I.F. and integrating, we get
`xe^(tan^(-1)y)=int(e^(2tan^(-1)y))/(1+y^(2))dy`
`rArr" "xe^(tan^(-1)y)=(1)/(2)inte^(2tan^(-1)y)d(2tan^(-1)y)`
`rArr" "xe^(tan^(-1)y)=(1)/(2)e^(2tan^(-1)y)+C`
`rArr" "2xe^(tan^(-1)y)=e^(2tan^(-1))+K," where K"=2C`
Discussion
223.

`(dy)/(dx)-y=0" at "(0, 1)`A. `y=logx`B. `y=e^(x)`C. `log_(y)x=1`D. `x=e^(y-1)`

Answer» Correct Answer - B
Discussion
224.

Solution of the differential equation `cosxdy=y(sinx-y)dx, 0ltxlt(pi)/(2)`isA. `tanx=(secx+c)y`B. `secx=(tanx+c)y`C. `ysec x=tanx+c`D. `ytanx =secx+c`

Answer» Correct Answer - B
`cosxdy=y(sinx-y)dx`
`rArr (dy)/(dx)=ytanx-y^(2)secx`
`rArr 1/y^(2)(dy)/(dx)-1/ytanx=-secx`
Let `1/y=t`
`therefore -1/y^(2)(dy)/(dx)=(dt)/(dx)`
`rArr -(dt)/(dx) -t tanx=-secx`
`rArr-(dt)/(dx)+(tanx)t=secx`
I.F. `=e^(int(tanxdx))=secx`
The solution is
t(I.F.) = `int(I.F.)secxdx`
`rArr 1/ysecx=tanx+c`
Discussion
225.

`x(dy)/(dx)=y+sqrt(x^(2)-y^(2)),` where `(y)/(x)=b`A. `y=log(v+sqrt(1-v^(2)))+c`B. `logv=sin^(-1)x+c`C. `log(sin^(-1)v)=x+c`D. `sin^(-1)v=(logx)+c`

Answer» Correct Answer - D
Discussion
226.

If `sinx`is an integrating factor of the differentialequation `(dy)/(dx)+P y=Q`, then write the value of `Pdot`A. `log sin x`B. `cot x`C. `sinx`D. `log cos x`

Answer» Correct Answer - B
We have,
I.F. = sin x
`rArr" "e^(intPdx)=sinx`
`rArr" "intPdx=logsinxrArrP=(d)/(dx)(logsinx)=cotx`
Discussion
227.

The solution of differential equation `(1+y^(2))+(x-e^(tan^(-1)y))(dy)/(dx)=0`, isA. `2xe^(tan^(-1)y)=e^(2tan^(-1)y)+k`B. `2xe^(tan^(-1)y)=e^(tan^(-1)y)+k`C. `xe^(tan^(-1)y)=e^(tan^(-1)y)+k`D. `xe^(tan^(-1)y)`

Answer» Correct Answer - A
Discussion
228.

`(x^(2)-x)(dy)/(dx)=y," at "(2, 1)`A. `2xy=x-1`B. `xy=x-2`C. `xy=2(y-1)`D. `xy=2(x-1)`

Answer» Correct Answer - D
Discussion
229.

`1-(dy)/(dx)=sec(x-y)," where "x-y=u`A. `x+cosu=c`B. `x-sinu=c`C. `x+cosu=c`D. `u+cosx=c`

Answer» Correct Answer - B
Discussion
230.

`(x-y+1)(dy)/(dx)=x-y+2`, where `x-y=u`A. `u^(2)+2x-y=c`B. `(u^(2))/(2)+2x-y=c`C. `u^(2)+2(2y-x)=c`D. `u^(2)+x-2y=c`

Answer» Correct Answer - B
Discussion
231.

The solution of the differential equation `(dy)/(dx)+(y)/(x)=x^(2)`, isA. `y=(x^(2))/(4)+Cx^(-2)`B. `y=x^(-1)+Cx^(-3)`C. `y=(x^(3))/(4)+Cx^(-1)`D. `xy=x^(2)+C`

Answer» Correct Answer - C
Discussion
232.

`(dx)/(x)+(dy)/(y)=0`, when `x=2, y=3`A. `xy=6`B. `2y=3`C. `3x=2`D. `6xy=1`

Answer» Correct Answer - A
Discussion
233.

`(x+y)(dx-dy)=dx+dy`, where `x+y=u`A. `x+y=log(x-y)+c`B. `x+y=log(x+y)+c`C. `x-y=logu+c`D. `u=logu+c`

Answer» Correct Answer - C
Discussion
234.

`(x(dy)/(dx)+y)sin(xy)=cosx,` where `xy=u`A. `sinx-cot(xy)=c`B. `cos(xy)-sinx=c`C. `cosx-sin(xy)=c`D. `sinx+cos(xy)=c`

Answer» Correct Answer - D
Discussion
235.

The solution of `(dy)/(dx)+2y tanx=sinx,` isA. `y sec^(3)x=sec^(2)x+C`B. `y sec^(2)x=sec x+C`C. `y sin x=tanx+C`D. none of these

Answer» Correct Answer - B
Discussion
236.

`(dv)/(r^(2))=4pidr,` when `r=0, v=0`A. `v=(1)/(3)pir^(2)h`B. `v=4pir^(3)`C. `v=(4)/(3)pir^(3)`D. `v=4pir^(2)h`

Answer» Correct Answer - C
Discussion
237.

`(x(dy)/(dx)-y)e^(y//x)=x^(2)cosx`, where `v=(y)/(x)`A. `e^(v)=(cosx)+c`B. `e^(sinx)=v+c`C. `e^(cosx)=v+c`D. `e^(v)=(sinx)+c`

Answer» Correct Answer - D
Discussion
238.

`x(dy)/(dx)+(y^(2))/(x)=y`A. `(y)/(x)+logx+c`B. `(y)/(x)=logy+c`C. `-(x)/(y)=logx+c`D. `y=xlogy+c`

Answer» Correct Answer - C
Discussion
239.

`(dy)/(dx)=(y)/(x)+tan((y)/(x))`A. `cx=sin((x)/(y))`B. `cx-sin((y)/(x))`C. `cx=cos((y)/(x))`D. `(y)/(x).cos((y)/(x))=c`

Answer» Correct Answer - B
Discussion
240.

` e^(-x) (dy)/(dx) = y(1+ tanx + tan^(2) x)`A. `logy=e^(x)tanx+c`B. `e^(x)logy=tanx+c`C. `logy=e^(x)sec^(2)x+c`D. `logy+e^(x)tanx=c`

Answer» Correct Answer - A
Discussion
241.

If `(dy)/(dx)=y+3 and y(0)=2`, then y(ln 2) is equal toA. 7B. 5C. 13D. `-2`

Answer» Correct Answer - A
We have,
`(dy)/(dx)=y+3`
`rArr" "(1)/(y+3)dy=dx`
`rArr" "int(1)/(y+3)dy=int 1.dx`
`rArr" "log(y+3)=x+C" …(1)"`
It is givne that y(0) = 2 i.e. y = 2 when x = 0.
`therefore" "log 5 =C" [Putting y = 2, x = 0 in (i)]"`
Substituting the value of C in (i), we get
`log(y+3)=x+log5`
Putting x = ln 2, we get
`y+3=5e^(log2)rArr y+3=10rArr y = 7`
Discussion
242.

`x+y(dy)/(dx)=sec(x^(2)+y^(2)),` where `x^(2)+y^(2)=u`A. `2x-sinu=c`B. `x-sin2u=c`C. `2x=tanu+c`D. `2u=cosx+c`

Answer» Correct Answer - A
Discussion
243.

`(dy)/(dx)=sin(x+y)+cos(x+y),` where `x+y=u`A. `1+tanu=e^(x)+c`B. `1+tan((u)/(2))=e^(x)+c`C. `1+tan((u)/(2))=ce^(x)`D. `tan(x+y)=ce^(x)`

Answer» Correct Answer - C
Discussion
244.

`1+(dy)/(dx)=cos(x+y),` where `x+y=u`A. `-cos(x+y)=x+c`B. `-cot^(2)(x+y)=x+c`C. `x+sin(x+y)=c`D. `(x+y)+cos(x+y)=c+c`

Answer» Correct Answer - A
Discussion
245.

`(dy)/(dx)=tan^(2)x`A. `x-y=tanx+c`B. `x+y=tanx+c`C. `x+y=sec^(2)x+c`D. `x-y=tany+c`

Answer» Correct Answer - B
Discussion
246.

Solution of the differential equation `cosxdy=y(sinx-y)dx, 0ltxlt(pi)/(2)`isA. `y tan x = sec x+C`B. `tan x= (sec x+C)y`C. `sec x =(tanx+C)y`D. `ysec x=tan x+C`

Answer» Correct Answer - C
We have,
`cosxdy=y(sinx-y)dx`
`rArr" "(dy)/(dx)=y tan x-y^(2)secx`
`rArr" "(1)/(y^(2))(dy)/(dx)+tanx(-(1)/(y))=-secx`
`rArr" "(dv)/(dx)+(tanx)v=secx," where v"=-(1)/(y)`
This is a linear differential equation with integrating factor
`e^(inttanxdx)=secx.` So, its solution is given by
`v sec x=-int sec^(2)xdx+C`
`rArr" "v secx=-tanx-C`
`rArr" "-(1)/(y)secx=-tanx-C`
`rArr" "secx=y(tanx+C)`
`ul("ALITER")` We have,
`cosxdy=y(sinx-y)dx`
`rArr" "cosxdy-y sinxdy=-y^(2)dx`
`rArr" "(secxdy-y tanx sec x dx)/(y^(2))=-sec^(2)xdx`
`rArr" "d((secx)/(y))=sec^(2)xdx`
`rArr" "(secx)/(y)=tanx+C`
`rArr" "secx=y(tanx+C)`
Discussion
247.

`(e^(y)+1)cos x dx+e^(y)sin x dy =0` A)`(e^(y)+1)sin x =c` B)`(e^(y)+1)=c tan x` C)`e^(y)sin x + cos x=c` D)`(e^(x)+1)sin y=c`A. `(e^(y)+1)sin x =c`B. `(e^(y)+1)=c tan x`C. `e^(y)sin x + cos x=c`D. `(e^(x)+1)sin y=c`

Answer» Correct Answer - A
Discussion
248.

`1+(dy)/(dx)=cos(x+y),` where `x+y=u`A. `x +cosu=c`B. `x sin u=c`C. `x+cotu=c`D. `u-sinx=c`

Answer» Correct Answer - A
Discussion
249.

`cos x cos y dy - sin x sin y dx=0`A. `cos x = c sin y`B. `cos x+siny=c`C. `cos x sin y=c`D. `secx.cosy=c`

Answer» Correct Answer - C
Discussion
250.

`(1+x^(2))dy=y^(2)dx`A. `y^(-1)+tan^(-1)x=c`B. `x^(-1)+tan^(-1)y=c`C. `x+(x^(3))/(3)=(y^(3))/(3)+c`D. `tan^(-1)x=y^(3)+c`

Answer» Correct Answer - A
Discussion