Solution of dx dy = − y x at x = 1 and y = √3 is
A. x - y2 = -2
B. x + y2 = 4
C. x2 - y2 = -2
D. x2 + y2 = 4
Select an option to see the answer and solution.
The solution to the ordinary differential equation d x 2 d 2 y + dx dy − 6 y = 0 is
A. y = c1 e3x + c2 e-2x
B. y = c1 e3x + c2 e2x
C. y = c1 e-3x + c2 e2x
D. y = c1 e-3x + c2 e-2x
Select an option to see the answer and solution.
Consider the differential equation dx dy = 1 + y 2 .
Which one of the following can be a particular solution of this differential equation?
A. y = tan(x + 3)
B. y = tan x + 3
C. x = tan(y + 3)
D. x = tan y + 3
Select an option to see the answer and solution.
Which ONE of the following is a linear non-homogeneous differential equation, where x and y are the independent and dependent variables respectively?
A. dx dy + xy = e − x
B. dx dy + xy = 0
C. dx dy + xy = e − y
D. dx dy + e − y = 0
Select an option to see the answer and solution.
A body originally at 60°C cools down to 40°C in 15 minutes when kept in air at a temperature of 25°C. What will be the temperature of the body at the end of 30 minutes?
A. 35.2°C
B. 31.5°C
C. 28.7°C
D. 15°C
Select an option to see the answer and solution.
The figure shows the plot of y as a function of x
The function shown is the solution of the differential equation (assuming all initial conditions to be zero) is
A. d x 2 d 2 y = 1
B. dx dy = x
C. dx dy = − x
D. dx dy = ∣ x ∣
Select an option to see the answer and solution.
The following differential equation has 3 ( d t 2 d 2 y ) + 4 ( dt dy ) 3 + y 2 + 2 = x
A. degree = 2, order = 1
B. degree = 1, order = 2
C. degree = 4, order = 3
D. degree = 2, order = 3
Select an option to see the answer and solution.
The boundary-value problem y'' + λy = 0, y(0) = y(π) = 0 will have non-zero solutions if and only if the values of λ are
A. 0, ±1, ±2, .....
B. 1, 2, 3, .....
C. 1, 4, 9, .....
D. 1, 9, 25, .....
Select an option to see the answer and solution.
The general solution of the differential equation dx dy = 1 − cos 2 x 1 + cos 2 y is
A. tan y - cot x = c (c is a constant)
B. tan x - cot y = c (c is a constant)
C. tan y + cot x = c (c is a constant)
D. tan x + cot y = c (c is a constant)
Select an option to see the answer and solution.
The solution of the differential equation dx dy + 2 xy = e − x 2 with y(0) = 1 is
A. (1 + x)e+x2
B. (1 + x)e-x2
C. (1 - x)e+x2
D. (1 - x)e-x2
Select an option to see the answer and solution.
The solution of d x 2 d 2 y + 2 dx dy + 17 y = 0 ; y(0) = 1, dx dy ( 4 π ) = 0 in the range 0 < x < 4 π is given by
A. e − x ( cos 4 x + 4 1 sin 4 x )
B. e x ( cos 4 x − 4 1 sin 4 x )
C. e − 4 x ( cos x − 4 1 sin x )
D. e − 4 x ( cos 4 x − 4 1 sin 4 x )
Select an option to see the answer and solution.
The differential equation
d x 2 d 2 y + 16 y = 0 for y(x) with the two boundary conditions
dx dy x = 0 = 1 and
dx dy x = 2 π = − 1 has
A. no solution
B. exactly two solutions
C. exactly one solution
D. infinitely many solutions
Select an option to see the answer and solution.
The solution of the differential equation x 2 d x 2 d 2 y − x dx dy + y = log x is
A. y = (c1 + c2 x) log x + 2 log x + 3
B. y = (c1 + c2 x2 ) log x + log x + 2
C. y = (c1 + c2 x) log x + log x + 2
D. y = (c1 + c2 log x) x + log x + 2
Select an option to see the answer and solution.
Solution of the differential equation 3 y dx dy + 2 x = 0 represents a family of
A. ellipses
B. circles
C. parabolas
D. hyperbolas
Select an option to see the answer and solution.
The solution of differential equation dx dy − y 2 = 1 satisfying condition y = 0 is
A. y = e x 2
B. C. y = cot ( x + 4 π )
D. y = tan ( x + 4 π )
Select an option to see the answer and solution.
The solution of the initial value problem dx dy = − 2 xy ; y(0) = 2 is
A. 1 + e-x2
B. 2e-x2
C. 1 + ex2
D. 2ex2
Select an option to see the answer and solution.
An ordinary differential equation is given below.
( dx dy ) ( x ln x ) = y
The solution for the above equation is
(Note: K denotes a constant in the options)
A. y = Kxex
B. y = Kxe-x
C. y = Klnx
D. y = Kxlnx
Select an option to see the answer and solution.
The solution to the differential equation d x 2 d 2 ⋅ u − k dx du = 0 is where k is constant, subjected to the boundary conditions u(0) = 0 and u(L) = U, is
A. u = U L x
B. u = U ( 1 − e kL 1 − e kx )
C. u = U ( 1 − e − kL 1 − e − kx )
D. u = U ( 1 + e kL 1 + e kx )
Select an option to see the answer and solution.
The matrix form of the linear system dt dx = 3 x − 5 y and dt dy = 4 x + 8 y is
A. \frac{{\text{d}}}{{{\text{dt}}}}\left\{ {\begin{array}{*{20}{c}}
{\text{x}} \\
{\text{y}}
\end{array}} \right\} = \left[ {\begin{array}{*{20}{c}}
3&{ - 5} \\
4&8
\end{array}} \right]\left\{ {\begin{array}{*{20}{c}}
{\text{x}} \\
{\text{y}}
\end{array}} \right\}
B. \frac{{\text{d}}}{{{\text{dt}}}}\left\{ {\begin{array}{*{20}{c}}
{\text{x}} \\
{\text{y}}
\end{array}} \right\} = \left[ {\begin{array}{*{20}{c}}
3&8 \\
4&{ - 5}
\end{array}} \right]\left\{ {\begin{array}{*{20}{c}}
{\text{x}} \\
{\text{y}}
\end{array}} \right\}
C. \frac{{\text{d}}}{{{\text{dt}}}}\left\{ {\begin{array}{*{20}{c}}
{\text{x}} \\
{\text{y}}
\end{array}} \right\} = \left[ {\begin{array}{*{20}{c}}
4&{ - 5} \\
3&8
\end{array}} \right]\left\{ {\begin{array}{*{20}{c}}
{\text{x}} \\
{\text{y}}
\end{array}} \right\}
D. \frac{{\text{d}}}{{{\text{dt}}}}\left\{ {\begin{array}{*{20}{c}}
{\text{x}} \\
{\text{y}}
\end{array}} \right\} = \left[ {\begin{array}{*{20}{c}}
4&8 \\
3&{ - 5}
\end{array}} \right]\left\{ {\begin{array}{*{20}{c}}
{\text{x}} \\
{\text{y}}
\end{array}} \right\}
Select an option to see the answer and solution.
The solution of the equation dt dQ + Q = 1 with Q = 0 at t = 0 is
A. Q(t) = e-t - 1
B. Q(t) = 1 + e-t
C. Q(t) = 1 - et
D. Q(t) = 1 - e-t
Select an option to see the answer and solution.