Eamcet - Maths - Differential Equations Test

Test Instructions :

1. The Test is 1hr duration.
2. The Test Paper consists of 30 questions. The maximum marks are 30.
3. All the questions are multiple choice question type with three options for each question.
4. Out of the three options given for each question, only one option is the correct answer.
5. Each question is allotted 1 mark for each correct response.
6. 0.25 will be deducted for incorrect response of each question.
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If [x3/(2x-1)(x+2)(x-3)]=A+(B/[2x-1])+(C/[x+2])+(D/[x-3]) then A is equal to

  

  

  

  

The solution of x log x (dy/dx)+y=2 log x is

  

  

  

  

The solution of excot y dx+(1-ex)cosec2ydy=0 is

  

  

  

  

Solution of differential equation dy/dx = (1+y2) (1+x2)-1is

  

  

  

  

The solution of (x+y+1) dy/dx=1 is

  

  

  

  

If a,b,c ≠ 0 and belong to the set { 0,1,2,3,................9 then log10[(a+10b+102c)/(10-4a+10-3b+10-2c)] is equal to

  

  

  

  

The radius of the circle r2-2√2r(cos θ + sin θ)-5=0 is

  

  

  

  

A family of curves has the differential equation (xy)dy/dx = 2y2 - x2. Then the family of curves is

  

  

  

  

The solution of y2 dx+(3xy-1)dy=0 is

  

  

  

  

The solution of extan y dx+(1-ex)sec2ydy=0 is

  

  

  

  

The solution of (x2y3+x2)+(y2x3+y2)dy=0 is

  

  

  

  

If y=(x2-1)n, then (x2-1)yn+2+2xyn+1=

  

  

  

  

The solution ofn(ex+1)y dy+(y+1)dx=0 is

  

  

  

  

The line 4x + 6y + 9=0, touches the parabola y2=4x at the point

  

  

  

  

The solution of (12x+5y-9)dx+(5x+2y-4)dy=0 is

  

  

  

  

The solution of (1+ex/y)dx+ex/y(1-x/y)dy=0 is

  

  

  

  

The solution of 3excos2ydx+(1-ex)cot y dy=0 is

  

  

  

  

If A and B are two square matrices such that B=A-1 BA, then (A+B)2 is equal to

  

  

  

  

The solution of (x+y+1)dx+(3x+4y+4)dy=0 is

  

  

  

  

The solution of x cos(y/x)(y dx+x dy)=y sin y/x(x dy – y dx) is

  

  

  

  

A= sin 780- sin 180+ cos 1320, B= cos 120+ cos 840+ cos 1320+ cos 1560 and C= (sin 750+sin 150)/ (sin 750+cos 150) then by arranging in the ascending order

  

  

  

  

x2+y2=t+(1/t),x4+y4=t2+(1/t2)⇒x3y(dy/dx)=

  

  

  

  

The solution of ex-y dx+ey-x dy=0 is

  

  

  

  

If the equation λx2 – 5xy + 6y2 + x -3y=0, represents a pair of straight lines, then their point of intersection is

  

  

  

  

If x is real, then the minimum value of [(x2-x+1)/(x2+x+1)], is

  

  

  

  

The solution of y dx-x dy+log x dx=0 is

  

  

  

  

x = cos θ, y = sin 5θ  ==>(1-x2) (d2y/dx2)  - x(dy/dx) =

  

  

  

  

The curve represented by X= 2( cos t + sin t ), y=( cos t - sin t ) is

  

  

  

  

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