Engineering Mathematics - Differential Equations
Exam Duration: 45 Mins Total Questions : 30
The solution of the first order differential equation \(x^{ \prime }(t)=-3x(t),x(0)={ x }_{ 0 }\) is
- (a)
\(x(t)={ x }_{ 0 }e^{ -3t }\)
- (b)
\(x(t)={ x }_{ 0 }e^{ -3 }\)
- (c)
\({ x }_{ 0 }e^{ -1/3 }\)
- (d)
\({ x }_{ 0 }e^{ -t }\)
For the equation x"(t)+3x'+2x(t) = 5, the solution x(t) approaches which of the following values as \(t\rightarrow \infty \)?
- (a)
0
- (b)
5/2
- (c)
5
- (d)
10
The order and degree of the differential equation \(\frac { { d }^{ 3 }y }{ dx^{ 2 } } +4\sqrt { \left( \frac { dy }{ dx } \right) ^{ 3 }+{ y }^{ 2 }=0 } \) are respectively
- (a)
3 and 2
- (b)
2 and 3
- (c)
3 and 3
- (d)
3 and 1
The degree of the differential equation \(\frac { { d }^{ 2 }x }{ d{ t }^{ 2 } } +2{ x }^{ 3 }=0\) is
- (a)
0
- (b)
1
- (c)
2
- (d)
3
The solution for the differential equation \(\frac { dy }{ dx } ={ x }^{ 2 }y\) with the condition that y = 1 at x = 0 is
- (a)
\(y={ e }^{ 1/2x }\)
- (b)
\(logy=\frac { { x }^{ 3 } }{ 3 } +4\)
- (c)
\(logy=\frac { { x }^{ 2 } }{ 2 } \)
- (d)
\(y=e^{ x^{ 3 }/3 }\)
The solution of the differential equation \(\frac { dy }{ dx } +\frac { y }{ x } =x\) with the condition that y = 1 at x = 1 is
- (a)
\(y=\frac { 2 }{ { 3x }^{ 2 } } +\frac { x }{ 3 } \)
- (b)
\(y=\frac { x }{ 2 } +\frac { 1 }{ 2x } \)
- (c)
\(y=\frac { 2 }{ 3 } +\frac { x }{ 3 } \)
- (d)
\(y=\frac { 2 }{ 3x } +\frac { { x }^{ 2 } }{ 3 } \)
The solution of the ordinary differential equation \(\frac { d^{ 2 }y }{ dx^{ 2 } } +\frac { dy }{ dx } -6y=0\) is
- (a)
\(y=c_{ 1 }e^{ 3x }+c_{ 2 }e^{ -2x }\)
- (b)
\(y=c_{ 1 }e^{ 3x }+c_{ 2 }e^{ 2x }\)
- (c)
\(c_{ 1 }e^{ -3x }+c_{ 2 }e^{ 2x }\)
- (d)
\(c_{ 1 }e^{ -3x }+c_{ 2 }e^{ -2x }\)
Th solution of differential equation \(\frac { dy }{ dx } =-\frac { x }{ y } \) at x=1 and \(y=\sqrt { 3 } \) is
- (a)
\(x-{ y }^{ 2 }=-2\)
- (b)
\(x+{ y }^{ 2 }=-4\)
- (c)
\({ x }^{ 2 }-{ y }^{ 2 }=2\)
- (d)
\({ x }^{ 2 }+{ y }^{ 2 }=4\)
The solution of the differential equation \({ x }^{ 2 }\frac { dy }{ dx } +2xy-x+1=0\) given that at x=1, y=0 is
- (a)
\(\frac { 1 }{ 2 } -\frac { 1 }{ x } +\frac { 1 }{ { 2x }^{ 2 } } \)
- (b)
\(\frac { 1 }{ 2 } -\frac { 1 }{ x } -\frac { 1 }{ { 2x }^{ 2 } } \)
- (c)
\(\frac { 1 }{ 2 } +\frac { 1 }{ x } +\frac { 1 }{ { 2x }^{ 2 } } \)
- (d)
\(-\frac { 1 }{ 2 } +\frac { 1 }{ x } +\frac { 1 }{ { 2x }^{ 2 } } \)
Biotransformation of an organic compound having concentration x can be modeled using an ordinary differential equation \(\frac { dx }{ dt } +k{ x }^{ 2 }=0\), where k is the reaction rate constant. If x = a at t = 0, the solution of the equation is
- (a)
\(x=a{ e }^{ -kt }\)
- (b)
\(\frac { 1 }{ x } =\frac { 1 }{ a } +kt\)
- (c)
\(x=a(1-e^{ -kt })\)
- (d)
\(x=a+kt\)
The Blasius equation \(\frac { d^{ 3 }f }{ d\eta ^{ 3 } } +\frac { f }{ 2 } \frac { d^{ 2 }f }{ d\eta ^{ 2 } } =0\) is a
- (a)
second order non-linear differential equation
- (b)
third order non-linear ordinary differential equation
- (c)
third order linear ordinary differential equation
- (d)
mixed order non-linear ordinary differential equation
The solution of the differential equation \(\frac { dy }{ dx } +2xy={ e }^{ -x^{ 2 } }\) with y(0) = 1 is
- (a)
\((1+x)e^{ x^{ 2 } }\)
- (b)
\((1+x)e^{ -x^{ 2 } }\)
- (c)
\((1-x)e^{ x^{ 2 } }\)
- (d)
\((1-x)e^{ -x^{ 2 } }\)
The solution of the differential equation \(\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } +\frac { dy }{ dx } +y=0\) is
- (a)
\(A{ e }^{ x }+Be^{ -x }\)
- (b)
\(e^{ x }(Ax+B)\)
- (c)
\(e^{ x }\left\{ Acos\left( \frac { \sqrt { 3 } }{ 2 } \right) x+Bcos\left( \frac { \sqrt { 3 } }{ 2 } \right) x \right\} \)
- (d)
\(e^{ -x/2 }\left\{ Acos\left( \frac { \sqrt { 3 } }{ 2 } \right) x+Bsin\left( \frac { \sqrt { 3 } }{ 2 } \right) x \right\} \)
For \(\frac { { d }^{ 2 }y }{ dx^{ 2 } } +4\frac { dy }{ dx } +3y=3e^{ 2x }\) , the particular integral is
- (a)
\(\frac { 1 }{ 15 } e^{ 2x }\)
- (b)
\(\frac { 1 }{ 5 } e^{ 2x }\)
- (c)
\(3e^{ 2x }\)
- (d)
\(c_{ 1 }e^{ -x }+c_{ 1 }e^{ -3x }\)
The complete solution for the ordinary differential equation
\(\frac { d^{ 2 }y }{ dx^{ 2 } } +p\frac { dy }{ dx } +qy=0\quad is\quad y={ c }_{ 1 }e^{ -x }+{ c }_{ 2 }e^{ -3x }\)
Then, p and q are
- (a)
p = 3 and q = 3
- (b)
p = 3 and q = 4
- (c)
p = 4 and q = 3
- (d)
p = 4 and q = 4
The complete solution for the ordinary differential equation
\(\frac { d^{ 2 }y }{ dx^{ 2 } } +p\frac { dy }{ dx } +qy=0\quad is\quad y={ c }_{ 1 }e^{ -x }+{ c }_{ 2 }e^{ -3x }\)
Which of the following is a solution of the differential equation \(\frac { d^{ 2 }y }{ dx^{ 2 } } +p\frac { dy }{ dx } +(q+1)y=0\) ?
- (a)
\(e^{ -3x }\)
- (b)
\(xe^{ -x }\)
- (c)
\(xe^{ -2x }\)
- (d)
\(x^{ 2 }e^{ -2x }\)
The solution of the differential equation \(\frac { dy }{ dx } +{ y }^{ 2 }=0\) is
- (a)
\(y=\frac { 1 }{ x+c } \)
- (b)
\(y=-\frac { { x }^{ 3 } }{ 3 } +c\)
- (c)
\(ce^{ x }\)
- (d)
Unsolvable as equation is non-linear
The solution of the differential equation \(\frac { dy }{ dx } =ky,\) y(0) = c is
- (a)
\(x=ce^{ -ky }\)
- (b)
\(x=ke^{ cy }\)
- (c)
\(y=ce^{ ky }\)
- (d)
\(y=ce^{ -kx }\)
A function n(x) satisfied the differential equation \(\frac { { d }^{ 2 }n(x) }{ dx^{ 2 } } -\frac { n(x) }{ { L }^{ 2 } } =0\), where L is a constant. The boundary conditions are n(0) = k and \(n(\infty )=0\). The solution to this equation is
- (a)
\(n(x)=ke^{ x/L }\)
- (b)
\(n(x)=ke^{ -x/\sqrt { L } }\)
- (c)
\(n(x)=k^{ 2 }e^{ -x/L }\)
- (d)
\(n(x)=ke^{ -x/L }\)
The order of the differential equation \(\frac { { d }^{ 2 }y }{ dt^{ 2 } } +\left( \frac { dy }{ dt } \right) ^{ 3 }+{ y }^{ 4 }=e^{ -t }\)
- (a)
1
- (b)
2
- (c)
3
- (d)
4
Which of the following is a solution to the differential equation \(\frac { dx(t) }{ dt } +3x(t)=0?\)
- (a)
\(x(t)=3{ e }^{ -t }\)
- (b)
\(x(t)=2{ e }^{ -3t }\)
- (c)
\(x(t)=-\frac { 3 }{ 2 } { t }^{ 2 }\)
- (d)
\(x(t)=3{ t }^{ 2 }\)
The following differential equation ha \(3\left( \frac { { d }^{ 2 }y }{ d{ t }^{ 2 } } \right) +4\left( \frac { dy }{ dt } \right) ^{ 3 }+{ y }^{ 2 }+2=x\)
- (a)
degree = 2 and order = 1
- (b)
degree = 1 and order = 2
- (c)
degree = 4 and order = 3
- (d)
degree = 2 and order = 3
A solution of the following differential equation is given by \(\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } -5\frac { dy }{ dx } +6y=0\)
- (a)
\(y={ e }^{ 2x }+{ e }^{ -3x }\)
- (b)
\(y={ e }^{ 2x }+{ e }^{ 3x }\)
- (c)
\(y={ e }^{ -2x }+{ e }^{ 3x }\)
- (d)
\(y={ e }^{ -2x }+{ e }^{ -3x }\)
The solution of differential equation \(\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } +\left( 9x-\frac { 20 }{ { x }^{ 2 } } \right) y=0\) in terms of Bessel's function is
- (a)
\(y=\sqrt { x } \left[ { c }_{ 1 }/3(2{ x }^{ 3/2 })+{ c }_{ 2 }Y_{ 3 }(2x^{ 3/2 }) \right] \)
- (b)
\(y=x\left[ { c }_{ 1 }/3(2{ x }^{ 3/2 })+{ c }_{ 2 }Y_{ 3 }(2x^{ 3/2 }) \right] \)
- (c)
\(y=\sqrt { x } \left[ { c }_{ 1 }/3(2{ x }^{ 3/2 })-{ c }_{ 2 }Y_{ 3 }(2x^{ 3/2 }) \right] \)
- (d)
\(y=x^{ 2 }\left[ { c }_{ 1 }/3(2{ x }^{ 3/2 })-{ c }_{ 2 }Y_{ 3 }(2x^{ 3/2 }) \right] \)
A tightly stretched string with fixed end points x = 0 and x = pi is initially at rest in its equilibrium position. It is set viberating by giving t each of its points an initial velocity \(\left( \frac { \partial y }{ \partial t } \right) _{ t=0 }=0.03sinx-0.04sin3x\) What will be then find the displacement y(x, t) at any point of string at any time t?
- (a)
\(\frac { 1 }{ c } \) [0.03 sin x sin ct - 0.0133 sin 3x sin 3ct]
- (b)
\(\frac { 1 }{ c } \)[0.03 sin 3x sin 3ct - 0.0133 sin x sin ct]
- (c)
\(\frac { 1 }{ c } \) [0.05 sin x sin ct - 0.0133 sin 3x sin 3ct]
- (d)
\(\frac { 1 }{ c } \) [0.0133 sin x sin ct - 0.03 sin 3x sin 3ct]
The ends A and B of a rod of length 20 cm are at temperature 30 degree C and 8 degree C until stady prevails. Then the temperature of the rod ends are changed to 40 degree C and 60 degree C respectively. The specific heat, density and the thermal conductivity of the material of the rod are such that the combination \(\frac { k }{ \rho \sigma } ={ c }^{ 2 }=1\) . The temperature distribution function \(\mu (x,t)\) is
- (a)
\(\mu (x,t)=40-x-\frac { 20 }{ \pi } \sum _{ n=1 }^{ \infty }{ \left\{ \frac { 2(-1)^{ n }-1 }{ n } \right\} } sin\frac { n\pi x }{ 20 } e^{ -\left( \frac { n\pi }{ 20 } \right) ^{ 2 }t }\)
- (b)
\(\mu (x,t)=40+x-\frac { 20 }{ \pi } \sum _{ n=1 }^{ \infty }{ \left\{ \frac { 2(-1)^{ n }+1 }{ n } \right\} } sin\frac { n\pi x }{ 20 } e^{ -\left( \frac { n\pi }{ 20 } \right) ^{ 2 }t }\)
- (c)
\(\mu (x,t)=40+x-\frac { 20 }{ \pi } \sum _{ n=1 }^{ \infty }{ \left\{ \frac { 2(-1)^{ n }+1 }{ n } \right\} } sin\frac { n\pi x }{ 20 } e^{ -\left( \frac { n\pi }{ 20 } \right) ^{ 2 }t }\)
- (d)
None of the above
The solution of \(x\frac { dy }{ dx } +y={ x }^{ 4 }\) with the condition \(y(1)=\frac { 6 }{ 5 } \) is
- (a)
\(y=\frac { { x }^{ 4 } }{ 5 } +\frac { 1 }{ x } \)
- (b)
\(y=\frac { { 4x }^{ 4 } }{ 5 } +\frac { 4 }{ 5x } \)
- (c)
\(y=\frac { { x }^{ 4 } }{ 5 } +1\)
- (d)
\(y=\frac { { x }^{ 5 } }{ 5 } +1\)
It is given that y'' + 2y' + y = 0, y (0) = 0, y (1) = 0. What is y (0.5)?
- (a)
0
- (b)
0.37
- (c)
0.62
- (d)
1.13
If \({ x }^{ 2 }\frac { dy }{ dx } +2xy=\frac { 2\log { x } }{ x } \)and y(1) = 0, then what
- (a)
e
- (b)
1
- (c)
\(\frac { 1 }{ e } \)
- (d)
\(\frac { 1 }{ { e }^{ 2 } } \)
The solution of the differential equation \(k^{ 2 }\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =y-y_{ 2 }\) under boundary conditions.
\((i)\quad y=y_{ 1 }\quad at\quad x=0\\ (ii)\quad y=y_{ 2 }\quad at\quad x=\infty ,\quad where\quad { k }_{ 1 }{ y }_{ 1 }\quad and\quad y_{ 2 }\quad are\quad constants\)
- (a)
\(y=({ y }_{ 1 }-{ y }_{ 2 })e^{ (-x/k^{ 2 }) }+y_{ 2 }\)
- (b)
\(y={ y }_{ 1 }-{ y }_{ 2 }\quad e^{ (-x/k^{ 2 }) }+y_{ 1 }\)
- (c)
\(y=({ y }_{ 1 }-{ y }_{ 2 })sinh\left( \frac { x }{ k } \right) +y_{ 1 }\)
- (d)
\(y=({ y }_{ 1 }-{ y }_{ 2 })e^{ (-x/k }+y_{ 2 }\)
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