Solving Linear Differential Equations of First Order
Trending Questions
Q.
If and denote the set of real numbers and complex numbers, respectively. Then, the function defined by is
one-one
onto
bijective
neither one-one nor onto
Q.
The value of for which the equation represents a pair of straight lines, is
Q. Let y=y(x) be the solution of the differential equation dydx+2y=f(x), where f(x)={1 , x∈[0, 1]0 , otherwise
If y(0)=0, then y(32) is
If y(0)=0, then y(32) is
- e2+12e4
- 12e
- e2−1e3
- e2−12e3
Q. Let y=y(x) be the solution of the differential equation, (x2+1)2dydx+2x(x2+1)y=1 such that y(0)=0. If √a y(1)=π32, then the value of a is:
- 116
- 12
- 1
- 14
Q. If a curve passes through the point (1, −2) and has slope of the tangent at any point (x, y) on it as x2−2yx, then the curve also passes through the point :
- (3, 0)
- (−1, 2)
- (−√2, 1)
- (√3, 0)
Q. The equation (x−a)2+(y−b)2=k(px+qy+r)2, where (pa+qb+r)≠0 represents
- a parabola for k<1p2+q2
- an ellipse for 0<k<1p2+q2
- a hyperbola for k>1p2+q2
- a point circle for k=0
Q. A differentiable function f satisfies f(x)=x∫0(f(t)cost−cos(t−x))dt. Which of the following hold(s) good?
- f(x) has the minimum value 1−e
- f(x) has the maximum value 1−e−1
- f′′(π2)=e
- f′(0)=1
Q. Let f be a non-negative function defined on the interval [0, π2]
If x∫0(f′(t)−sin 2t)dt=0∫xf(t)tan t dt
and f(0)=1, assume y=f(x) then
If x∫0(f′(t)−sin 2t)dt=0∫xf(t)tan t dt
and f(0)=1, assume y=f(x) then
- f(0)=1 is the maximum value of f
- f(0)=1 is the minimum value of f
- π2∫0y dx=3−π2
- f(π3)=1
Q.
Which of the following equation is non-linear
- dydx=cos x
d2ydx2+y=0
dx+dy=0
xdydx+3dydx=y2
Q. Let, f(x)={[x], −2≤x≤−1|x|+1, −1<x≤2 and g(x)={[x], −π≤x≤0sinx, 0<x≤π, then the exhaustive domain of g(f(x)) is
- [–2, 0]
- [–2, 2]
- [–1, 2]
- [0, 2]
Q. The solution of (x+2y3)(dydx)=y is (where c is arbitrary constant)
- x=y3+cy
- x=y3−cy
- y=y3−cy
- y=y3+cy
Q. If y(x) is the solution of the differential equation
(x+2)dydx=x2+4x–9, x≠–2
and y(0)=0, then y(–4)is equal to :
(x+2)dydx=x2+4x–9, x≠–2
and y(0)=0, then y(–4)is equal to :
- 2
- 0
- −1
- 1
Q. Let y=y(x) be the solution curve of the differential equation, (y2−x)dydx=1, satisfying y(0)=1. This curve intersects the x-axis at a point whose abscissa is :
- 2+e
- 2
- 2−e
- −e
Q. Let dydx=yϕ′(x)−y2ϕ(x), where ϕ(x) is a specified function satisfying ϕ(1)=1, ϕ(4)=1296. If y(1)=1 then 181y(4) is equal to
Q. Given that the inverse trigonometric functions take principal values only. Then, the number of real values of x which satisfy sin−1(3x5)+sin−1(4x5)=sin−1x is equal to :
- 0
- 3
- 2
- 1
Q. If y1(x) is a solution of the differential equation dydx−f(x)y=0, then a solution of the differential equation dydx−f(x)y=r(x) is
- 1y1(x)∫r(x)y1(x)dx
- y1(x)∫r(x)y1(x)dx
- ∫r(x)y1(x)dx
- none of these
Q.
A curve passes through (2, 0) and the slope of tangent at a point P(x, y) is equal to (x+1)2+y−3/(x+1). Then equation of the curve is: [IIT 2004]
- y=x2+2x
- y=x2−2x
- y=2x2−x
- None of these
Q. Solution of the differential equation
cosxdy=y(sin x−y)dx, 0<x<π2 is
cosxdy=y(sin x−y)dx, 0<x<π2 is
- ysec x=tan x+c
- ytan x=sec x+c
- tan x=(sec x+c)y
- sec x=(tan x+c)y
Q. Let y=y(x) be the solution of the differential equation, dydx+ytanx=2x+x2tanx, x∈(−π2, π2), such that y(0)=1. Then :
- y(π4)−y(−π4)=√2
- y(π4)+y(−π4)=π22+2
- y′(π4)−y′(−π4)=π−√2
- y′(π4)+y′(−π4)=−√2
Q. A differentiable function f satisfies f(x)=x∫0(f(t)cost−cos(t−x))dt. Which of the following hold(s) good?
- f(x) has the minimum value 1−e
- f(x) has the maximum value 1−e−1
- f′′(π2)=e
- f′(0)=1
Q. If cosxdydx+ysinx=1 and y(π4)=√2, then y(−π3) is
- 1−√32
- 1+√32
- 1−√3
- 1+√3
Q. The solution of (x+2y3)(dydx)=y is (where c is arbitrary constant)
- x=y3+cy
- x=y3−cy
- y=y3−cy
- y=y3+cy