Mathematics - Differential Equations
Exam Duration: 45 Mins Total Questions : 30
The degree and the order of the differential equation \(\frac { { d }^{ 2 }y }{ x{ d }^{ 2 } } +\frac { 2 }{ \left( \frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } \right) } \)+ 4y = 0 are
- (a)
2, 2
- (b)
4, 2
- (c)
2, 4
- (d)
2, 1
A spherical rain drop evaporates at the rate proportional to its surface area. The differential equation involving the rate of change of its radius r, is
- (a)
\(\frac { dr }{ dt } =k\)
- (b)
\(\frac { dr }{ dt } =-k\)
- (c)
\(\left( \frac { dr }{ dt } \right) =-kt\)
- (d)
\(\frac { dr }{ dt } =kt\)
The solution of the differential equation 3ex tan y dx + (1 - ex) sec2 y dy, is
- (a)
tan y = c\({ \left( 1-{ e }^{ x } \right) }^{ 3 }\)
- (b)
tan y = c\({ \left( 1+{ e }^{ x } \right) }^{ 3 }\)
- (c)
cot y = c\({ \left( 1-{ e }^{ x } \right) }^{ 3 }\)
- (d)
NONE OF THESE
The order and degree of the differential equation \({ x }^{ 2 }=\frac { \left[ 1+{ \left( \frac { dy }{ dx } \right) }^{ 2 } \right] ^{ 3/2 } }{ { \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } } } \) are respectively
- (a)
2, 3
- (b)
2, 2
- (c)
3, 4
- (d)
2, 4
The solution of the differential equation \(X\cos { X } \left( \frac { dY }{ dX } \right) +Y\left( X\sin { X+\cos { X } } \right) =1\) is
- (a)
xy = sin x + C cos x
- (b)
xy = cos x + C sin x = 0
- (c)
xy = sec x + C sin x = 0
- (d)
None of the above
Let the population of rabbits surviving at a time t be governed by the differential equation \(\frac { dp(t) }{ dt } =\frac { 1 }{ 2 } p(t)-200\) . If p(0) = 100, then p(t) equals
- (a)
\(600-500{ e }^{ t/2 }\)
- (b)
\(400 - 300 { e }^{ -t/2 }\)
- (c)
\(400 - 300 { e }^{ t/2 }\)
- (d)
\(300 - 200 { e }^{ -t/2 }\)
If \(\frac { dY }{ dX } =Y+3>0\) and Y(0) = 2, then Y(log 2) is equal to
- (a)
5
- (b)
13
- (c)
- 2
- (d)
7
Newton's law of cooling states that the rate at which a substance cools in moving air is proportional to the difference between the temperatures of the substance and that of the air. If the temperature of the air is 290K.
We can write it as dT = - k(T - 290), k > 0 constants. where T is temperature of substance.
The value of k must be
- (a)
In 2
- (b)
\(\frac { In2 }{ 40 } \)
- (c)
\(\frac { In2 }{ 20 } \)
- (d)
\(\frac { In2 }{ 10 } \)
Newton's law of cooling states that the rate at which a substance cools in moving air is proportional to the difference between the temperatures of the substance and that of the air. If the temperature of the air is 290K.
We can write it as dT = - k(T - 290), k > 0 constants. where T is temperature of substance.
The time taken when temperature be 295 K is
- (a)
10 min
- (b)
20 min
- (c)
40 min
- (d)
80 min
If y = f(x) passing through (1, 2) satisfies the differential equation y (1 + xy) dx - x dy = 0, then
- (a)
\(f(x)=\frac { 2x }{ 2-x^{ 2 } } \)
- (b)
\(f(x)=\frac { x+1 }{ x^{ 2 }+1 } \)
- (c)
\(f(x)=\frac { x-1 }{ 4-x^{ 2 } } \)
- (d)
\(f(x)=\frac { 4x }{ 1-2x^{ 2 } } \)
Solution of 2y sinx\(\frac{dy}{dx}\)=2sinx cosx - y2 cosx, x = \(\frac{\pi}{2}\), y=1 is given by
- (a)
y2=sinx
- (b)
y=sin2x
- (c)
y2=1+cosx
- (d)
none of these
If \(\phi\) (x) = \(\int\){ \(\phi\)(x)}-2dx and \(\phi\)(1) = 0, then \(\phi\) (x) is equal to
- (a)
{2(x-1)}1/4
- (b)
{5(x-2)}1/5
- (c)
{3 (x - 1)}1/3
- (d)
none of these
The order of differential equation whose solution is y = acosx + bsinx + ce-x is
- (a)
3
- (b)
2
- (c)
1
- (d)
none of these
The degree of the equation satisfying the relation \(\sqrt { 1+{ x }^{ 2 } } +\sqrt { 1+{ y }^{ 2 } } =\lambda \left( x\sqrt { 1+{ y }^{ 2 } } -y\sqrt { 1+{ x }^{ 2 } } \right) \) , is
- (a)
1
- (b)
2
- (c)
3
- (d)
none of these
The differential equation of family of curves whose tangent form an angle of π/4 with the hyperbola xy = c2 is
- (a)
\(\frac { dy }{ dx } =\frac { { x }^{ 2 }+{ c }^{ 2 } }{ { x }^{ 2 }-{ c }^{ 2 } } \)
- (b)
\(\frac { dy }{ dx } =\frac { { x }^{ 2 }-{ c }^{ 2 } }{ { x }^{ 2 }+{ c }^{ 2 } } \)
- (c)
\(\frac { dy }{ dx } =-\frac { { c }^{ 2 } }{ { x }^{ 2 } } \)
- (d)
none of these
The differential equation of all non-horizontal lines in a plane is
- (a)
\(\frac { { d }^{ 2 }y }{ { d }x^{ 2 } } =0\)
- (b)
\(\frac { { d }^{ 2 }x }{ { dy }^{ 2 } } =0\)
- (c)
\(\frac { dy }{ dx } =0\)
- (d)
\(\frac { dx }{ dy } =0\)
If the solution of the differential equation \(\frac { dy }{ dx } =\frac { ax+3 }{ 2y+f } \) represents a circle, then the value of 'a' is
- (a)
2
- (b)
- 2
- (c)
3
- (d)
- 4
For the differential equation \(x\frac { dy }{ dx } +2y=xy\frac { dy }{ dx } \) ,
- (a)
order is 1 and degree is 1
- (b)
solution in ln(yx2) = C - y
- (c)
order is 1 and degree is 2
- (d)
solution is ln(xy2) = C + y
If ydx - xdy + ln x dx = 0, y(1) = -1, then
- (a)
y + 1 + ln x = 0
- (b)
y + 1 + 2 ln x =0
- (c)
2(y + 1) + ln x = 0
- (d)
y + 1 - y ln x = 0
If \(\frac { dy }{ dx } \) = y sin 2x, y(0) = 1 then solution is
- (a)
\(y={ e }^{ \sin ^{ 2 }{ x } }\)
- (b)
\(y=\sin ^{ 2 }{ x } \)
- (c)
\(y=\cos ^{ 2 }{ x } \)
- (d)
\(y={ e }^{ \cos ^{ 2 }{ x } }\)
If \(\frac { dy }{ dx } =\frac { x+y }{ x } \) , y(1) = 1, then y =
- (a)
x + ln x
- (b)
x2 + x ln x
- (c)
xex-1
- (d)
x + x ln x
If x dy = (2y + 2x4 + x2) dx, y(1) = 0, then y(e) =
- (a)
1
- (b)
e
- (c)
e2
- (d)
e4
Integrating factor of the differential equation \(\cos { x } \frac { dy }{ dx } +y\sin { x } \) = 1 is
- (a)
cos x
- (b)
tan x
- (c)
sec x
- (d)
sin x
Solution of the differential equation tan y sec2xdx + tan x sec2y dy = 0 is
- (a)
tan x + tan y = k
- (b)
tan x - tan y = k
- (c)
\(\frac { \tan { x } }{ \tan { y } } =k\)
- (d)
tanx . tany = k
Integrating factor of \(\frac { dy }{ dx } -y\) = 1,y(0) = 1 is given by
- (a)
xy = -ex
- (b)
xy = - e-x
- (c)
xy = -1
- (d)
y = 2ex - 1
tan-1x + tan-1y = c is the general solution of the differential equation
- (a)
\(\frac { dy }{ dx } =\frac { 1+{ y }^{ 2 } }{ 1+{ x }^{ 2 } } \)
- (b)
\(\frac { dy }{ dx } =\frac { 1+{ x }^{ 2 } }{ 1+{ y }^{ 2 } } \)
- (c)
(1 + x2) dy + (1 + y2) dx = 0
- (d)
(1 + x2)dx + (1 + y2)dy = 0
Integrating factor of the differential equation \(\frac { dy }{ dx } +y\tan { x } -\sec { x } =0\) is
- (a)
cos x
- (b)
sec x
- (c)
ecosx
- (d)
esecx
The differential equation of the family of curves x2 + y2 - 2ay = 0, where a is arbitrary constant, is
- (a)
\(\left( { x }^{ 2 }-{ y }^{ 2 } \right) \frac { dy }{ dx } =2xy\)
- (b)
\(2\left( { x }^{ 2 }+{ y }^{ 2 } \right) \frac { dy }{ dx } =xy\)
- (c)
\(2\left( { x }^{ 2 }-{ y }^{ 2 } \right) \frac { dy }{ dx } =xy\)
- (d)
\(\left( { x }^{ 2 }+{ y }^{ 2 } \right) \frac { dy }{ dx } =2xy\)
The general solution of the differential equation (ex + 1)ydy = (y + 1)exdx is
- (a)
(y + 1) = k(ex + 1)
- (b)
y + 1 = ex + 1 + k
- (c)
y = log {k(y + 1) (ex + 1)}
- (d)
y = \(\log { \left\{ \frac { { e }^{ x }+1 }{ y+1 } \right\} } +k\)