STUPIDSID
1 (a)
Choose the correct answer for the following:
( i) Suppose the equation to be solved is of the form, y=f(x, φ) then differentiating x we get equation of the form,
\[ (a) \ \phi \left (x,p, \dfrac{dp}{dy} \right )= 0 \\(b) \ \phi \left ( y, p , \dfrac {dp}{dx} \right )= 0 \\(c) \ \phi (x,yp)=0 \\(d)\ \phi (x,y,0)= 0 \]
(ii) The general solution of the equation p2-3p+2=0 is,
(a) (y+x-c)y+2x-c)
(b) (y-x-c)(y-2x-c)=0
(c) (-y-x-c)(y-2x-c)=0
(y-x-c)(y+x-c)=0
(iii) Clairaut's equation is of the form,
(a) x=py+f(p)
(b) y=p2+f(p)
(c) y=px+f(p)
(d) None of these
(iv) Singular solution of y=px+2p2 is,
(a) y2+8y=0
(b) x2-8y=0
(c) x2+8y-c=0
(d) x2+8y=0
( i) Suppose the equation to be solved is of the form, y=f(x, φ) then differentiating x we get equation of the form,
\[ (a) \ \phi \left (x,p, \dfrac{dp}{dy} \right )= 0 \\(b) \ \phi \left ( y, p , \dfrac {dp}{dx} \right )= 0 \\(c) \ \phi (x,yp)=0 \\(d)\ \phi (x,y,0)= 0 \]
(ii) The general solution of the equation p2-3p+2=0 is,
(a) (y+x-c)y+2x-c)
(b) (y-x-c)(y-2x-c)=0
(c) (-y-x-c)(y-2x-c)=0
(y-x-c)(y+x-c)=0
(iii) Clairaut's equation is of the form,
(a) x=py+f(p)
(b) y=p2+f(p)
(c) y=px+f(p)
(d) None of these
(iv) Singular solution of y=px+2p2 is,
(a) y2+8y=0
(b) x2-8y=0
(c) x2+8y-c=0
(d) x2+8y=0
4 M
1 (b)
Solve p2+2p cosh x+1=0.
4 M
1 (c)
Find singular solution of p=sin(y-xp).
6 M
1 (d)
Solve the equation y2(y-xp)=x4p2 using substitution \[ X=\dfrac {1}{x} and Y=\dfrac {1}{y} \]
6 M
2 (a)
Choose the correct answer for the following:
(i) A second order linear differential equation has,
(a) two arbitary solution
(b) One arbitary solution
(c) no arbitary solution
(d) None of these
(ii) If 2, 4i and -4i are the roots of A.E of a homogeneous linear differential equation then its solution is,
\[ (a) \ e^x+ e^x (\cos 4x+\sin 4x) \\ (b) \ C_1 e^{2x}+ C_2 \cos 4x + C_3 \sin 4x \\(c) \ C_1 e^{2x} + C_2 e^x \cos 4x+C_3 e^x \sin 4x \\(d) \ C_1 e^{2x}\cos 4x+ C_2e^{2x}\sin 4x \]
(iii) P.I. of (D+1)2 y=e-x+3 \[ (a)\ \dfrac {x^2}{2} \\ (b) \ x^3 e^x \\ (c) \ \dfrac {x^3}{3} e^{-x=3} \\ (d)\ \dfrac {x^2}{2}e^{-x+3}\]
(iv) Particular integral of f(D)y=eax V(x) is, \[ (a)\ \dfrac {e^{ax}V(x)}{f(D)} \\ (b) \ e^{ax}= \dfrac {1}{f(D)}[V(x)] \\ (c) \ e^{ax} \dfrac {1}{f(D+a)}[V(x)] \\ (d) \ \dfrac {1}{f(D+a)} [e^{ax}V(x)] \]
(i) A second order linear differential equation has,
(a) two arbitary solution
(b) One arbitary solution
(c) no arbitary solution
(d) None of these
(ii) If 2, 4i and -4i are the roots of A.E of a homogeneous linear differential equation then its solution is,
\[ (a) \ e^x+ e^x (\cos 4x+\sin 4x) \\ (b) \ C_1 e^{2x}+ C_2 \cos 4x + C_3 \sin 4x \\(c) \ C_1 e^{2x} + C_2 e^x \cos 4x+C_3 e^x \sin 4x \\(d) \ C_1 e^{2x}\cos 4x+ C_2e^{2x}\sin 4x \]
(iii) P.I. of (D+1)2 y=e-x+3 \[ (a)\ \dfrac {x^2}{2} \\ (b) \ x^3 e^x \\ (c) \ \dfrac {x^3}{3} e^{-x=3} \\ (d)\ \dfrac {x^2}{2}e^{-x+3}\]
(iv) Particular integral of f(D)y=eax V(x) is, \[ (a)\ \dfrac {e^{ax}V(x)}{f(D)} \\ (b) \ e^{ax}= \dfrac {1}{f(D)}[V(x)] \\ (c) \ e^{ax} \dfrac {1}{f(D+a)}[V(x)] \\ (d) \ \dfrac {1}{f(D+a)} [e^{ax}V(x)] \]
4 M
2 (b)
\[ Solve \ \dfrac {d^3y}{dx^3}- 3 \dfrac {d^2y}{dx^2}+ 3 \dfrac {dy}{dx}- y = 0 \]
4 M
2 (c)
Solve y"-3y'+2y=2 sin x cos x
6 M
2 (d)
Solve the system of equation, \[ \dfrac {dx}{dt}- 2y = \cos 2t, \ \dfrac{dy}{dt} + 2x =\sin 2t \]
6 M
3 (a)
Choose the correct answer for the following:
(i) In x2y"+ xy'-y=0 if et=x then we get x2y" as,
(a) (D-1)y
(b) (D+1)y
(c) D(D+1)y
(d) None of these
(ii) In second order homogeneous differential equation P0(x)y"+P1(x)y'+P2(x)y=0 x=a is a singular point if,
(a) P0(a)>0
(b) P0(a)?0
(c) P0(a)=0
(d) P0(a)<0
(iii) The general solution of \[ x^2 \dfrac {d^2 y}{dx^2}+ x\dfrac{dy}{dx}-y = 0 \ is, \\ (a) \ y=C_1x-C_2 \dfrac {1}{x} \\ (b) \ C_1x + C_2 \dfrac {1}{x} \\ (c) \ C_1x+C_2 x \\ (d) \ C_1 x- C_2 x \]
(iv) Frobenius series solution of second order linear differential equation is of the form,
\[ (a) \ x^{m} \sum^{\infty}_{r=0}a_rx^r \\ (b) \ \sum^{\infty}_{r=0}a_rx^r \\ (c) \ \sum^{\infty}_{r=a}a_rx^{m-r} \\ None \ of \ these \]
(i) In x2y"+ xy'-y=0 if et=x then we get x2y" as,
(a) (D-1)y
(b) (D+1)y
(c) D(D+1)y
(d) None of these
(ii) In second order homogeneous differential equation P0(x)y"+P1(x)y'+P2(x)y=0 x=a is a singular point if,
(a) P0(a)>0
(b) P0(a)?0
(c) P0(a)=0
(d) P0(a)<0
(iii) The general solution of \[ x^2 \dfrac {d^2 y}{dx^2}+ x\dfrac{dy}{dx}-y = 0 \ is, \\ (a) \ y=C_1x-C_2 \dfrac {1}{x} \\ (b) \ C_1x + C_2 \dfrac {1}{x} \\ (c) \ C_1x+C_2 x \\ (d) \ C_1 x- C_2 x \]
(iv) Frobenius series solution of second order linear differential equation is of the form,
\[ (a) \ x^{m} \sum^{\infty}_{r=0}a_rx^r \\ (b) \ \sum^{\infty}_{r=0}a_rx^r \\ (c) \ \sum^{\infty}_{r=a}a_rx^{m-r} \\ None \ of \ these \]
4 M
3 (b)
Solve y"+a2y=sec ax by the method of variation of parameters.
4 M
3 (c)
\[ Solve \ x^2 \dfrac {d^2 y}{dx^2}+ 4x \dfrac {dy}{dx}+ 2 y = e^x \]
6 M
3 (d)
Obtain the series solution of \[ \dfrac {dy}{dx}- 2xy=0 \]
6 M
4 (a)
Choose the correct answer for the following:
(i) PDE of az+b=a2x+y is, \[ (a) \ \dfrac {\partial z}{\partial x} \cdot \dfrac{\partial z}{\partial y}= 1 \\ (b) \dfrac {\partial z} {\partial x} \cdot \dfrac {\partial z}{\partial y} = 0 \\ (c) \ \dfrac {\partial z}{\partial x} + \dfrac {\partial z}{\partial y} = 0 \\ (d)\ \dfrac{\partial z}{\partial x}+ \dfrac {\partial z}{\partial y}=1 \]
(ii) The solution of PDE Zxx=2 y2 is,
(a) z=x2+xf(y)+ g(y)
(b) z=x2y2+xf(y)+g(y)
(c) z=x2y2+f(x)+g(x)
(d) z=y2+xf(y)+g(y)
iii) The subsidiary equations of (y2+z2)p+x(yq-z)=0 are, \[ (a)\ \dfrac {dx}{p}= \dfrac {dy}{q} = \dfrac {dz}{R} \\ (b) \ \dfrac {dx}{y^2 + z^2} = \dfrac {dy}{x} = \dfrac {dz}{xz} \\ (c) \ \dfrac {dx}{y^2 + z^2} = \dfrac {dy}{xy} = \dfrac {dz}{xz} \\ (d) \ None \ of \ these \] (iv) In the method of seperation of variable to solve xzn+zt=0 the assumed solution is of the form,
(a) X(x)Y(x)
(b) X(y)Y(y)
(c) X(t)Y(t)
(d) X(x)T(t)
(i) PDE of az+b=a2x+y is, \[ (a) \ \dfrac {\partial z}{\partial x} \cdot \dfrac{\partial z}{\partial y}= 1 \\ (b) \dfrac {\partial z} {\partial x} \cdot \dfrac {\partial z}{\partial y} = 0 \\ (c) \ \dfrac {\partial z}{\partial x} + \dfrac {\partial z}{\partial y} = 0 \\ (d)\ \dfrac{\partial z}{\partial x}+ \dfrac {\partial z}{\partial y}=1 \]
(ii) The solution of PDE Zxx=2 y2 is,
(a) z=x2+xf(y)+ g(y)
(b) z=x2y2+xf(y)+g(y)
(c) z=x2y2+f(x)+g(x)
(d) z=y2+xf(y)+g(y)
iii) The subsidiary equations of (y2+z2)p+x(yq-z)=0 are, \[ (a)\ \dfrac {dx}{p}= \dfrac {dy}{q} = \dfrac {dz}{R} \\ (b) \ \dfrac {dx}{y^2 + z^2} = \dfrac {dy}{x} = \dfrac {dz}{xz} \\ (c) \ \dfrac {dx}{y^2 + z^2} = \dfrac {dy}{xy} = \dfrac {dz}{xz} \\ (d) \ None \ of \ these \] (iv) In the method of seperation of variable to solve xzn+zt=0 the assumed solution is of the form,
(a) X(x)Y(x)
(b) X(y)Y(y)
(c) X(t)Y(t)
(d) X(x)T(t)
4 M
4 (b)
\[ Solve \ \dfrac {\partial ^3 z}{\partial x^2 \partial y}= cos (2x+3y)\]
4 M
4 (c)
Solve xp-yq=y2-x2
6 M
4 (d)
Solve 3ux+2uy=0 by the seperation of variable method given that u=4e-x when y=0
6 M
5 (a)
Choose the correct answer for the following:
\[ \int^{1}_0 \int^{x^2}_0 e^{y/x}dy dx = \_\_\_\_\_\_\_ \\ (a) \ 1 \ \ (b) \ -1/2 \ \ (c) \ 1/2 \ \ (d) \ None\ of \ these \] (ii) The integral \[ \iint_R f(x,y) dxdy \] by changing to polar form becomes, \[ (a) \ \iint_R \phi (r, \theta) drd\theta \\ (b) \ \iint_R f(r, \theta)drd\theta \\ (c) \ \iint_R f(r,\theta)rdrd\theta \\ (d)\ \iint_R \phi (r, \theta)rdrd \theta \] (iii) For a real positive number n, the Gamma function ?(n)= _________ \[ (a) \ \int^{\infty}_0 x^{n-1}e^{-x}dx \\ (b) \ \int^1_0 x^{n-1}e^{-x}dx \\ (c) \ \int^{x}_0 x^ne^{-x}dx \\ (d) \ \int^1 _0 x^n e^{-x}dx \]
(iv) The Beta and Gamma functions relation for B(,n)= _______ \[ (a) \ \dfrac {\Gamma (m )\Gamma (n)} {\Gamma (m-n)} \\ (b) \ \dfrac {\Gamma (m)\Gamma (n)}{\Gamma (m+n)} \\ (c) \ \Gamma (m)\Gamma(n) \\ (d) \ \Gamma(mn) \]
\[ \int^{1}_0 \int^{x^2}_0 e^{y/x}dy dx = \_\_\_\_\_\_\_ \\ (a) \ 1 \ \ (b) \ -1/2 \ \ (c) \ 1/2 \ \ (d) \ None\ of \ these \] (ii) The integral \[ \iint_R f(x,y) dxdy \] by changing to polar form becomes, \[ (a) \ \iint_R \phi (r, \theta) drd\theta \\ (b) \ \iint_R f(r, \theta)drd\theta \\ (c) \ \iint_R f(r,\theta)rdrd\theta \\ (d)\ \iint_R \phi (r, \theta)rdrd \theta \] (iii) For a real positive number n, the Gamma function ?(n)= _________ \[ (a) \ \int^{\infty}_0 x^{n-1}e^{-x}dx \\ (b) \ \int^1_0 x^{n-1}e^{-x}dx \\ (c) \ \int^{x}_0 x^ne^{-x}dx \\ (d) \ \int^1 _0 x^n e^{-x}dx \]
(iv) The Beta and Gamma functions relation for B(,n)= _______ \[ (a) \ \dfrac {\Gamma (m )\Gamma (n)} {\Gamma (m-n)} \\ (b) \ \dfrac {\Gamma (m)\Gamma (n)}{\Gamma (m+n)} \\ (c) \ \Gamma (m)\Gamma(n) \\ (d) \ \Gamma(mn) \]
4 M
5 (b)
By changing the order of integration evaluate, \[ \int^a_0 \int^{\sqrt{x/a}}_{x/a}(x^2 + y^2)dydx, \ a>0 \]
4 M
5 (c)
\[ \displaystyle Evaluate \ \int^a_0 \int^{x}_0 \int^{x-y}_0 e^{x+y+z}dzdydx \]
6 M
5 (d)
Express the integral \[ \int^1_0 \dfrac{dx}{\sqrt{1-x^n}}\] in terms of the Gamma function, Hence evaluate \[ \int^1_0 \dfrac {dx}{\sqrt{1-x^{2/3}}} \]
6 M
6 (a)
Choose the correct answer for the following:
(i) The scalar surface integral of \[ \overrightarrow{f} \] over s, where s is a surface in a three-dimensional region R is given by, \[ \int \overrightarrow{f}.nds= \_\_\_\_\_\_\_ \] by using Gauss divergence theorem \[ (a) \ \iiint_v \nabla\cdot \overrightarrow{f}dV \\ (b)\ \iint_s \nabla\cdot \overrightarrow{t}dx dy \\ (c) \ \iiint_v \nabla \cdot \overrightarrow{F}dV \\ (d) \ None \ of \ these \] (ii) If all the surface are closed in a region containing volume V then the following theorem is applicable.
(a) Stroke's theorem
(b) Green's theorem
(c) Gauss divergence theorem
(d) None of these
(iii) The value of \[ \int \left \{ (2xy-x^2)dx + (x^2 + y^2)dx \right \} \] by using Green's theorem is,
(a) Zeron (b) One (c) Two (d) Three
(iv) \[ \iint_s f.nds = \_\_\_\_\_\_\_ \] where f=xi+yj+2k and S is the surface of the sphere x2y2+z2=a2
(a) 4πa (b) 4πa2 (c) 4πa3 (d) 4π
(i) The scalar surface integral of \[ \overrightarrow{f} \] over s, where s is a surface in a three-dimensional region R is given by, \[ \int \overrightarrow{f}.nds= \_\_\_\_\_\_\_ \] by using Gauss divergence theorem \[ (a) \ \iiint_v \nabla\cdot \overrightarrow{f}dV \\ (b)\ \iint_s \nabla\cdot \overrightarrow{t}dx dy \\ (c) \ \iiint_v \nabla \cdot \overrightarrow{F}dV \\ (d) \ None \ of \ these \] (ii) If all the surface are closed in a region containing volume V then the following theorem is applicable.
(a) Stroke's theorem
(b) Green's theorem
(c) Gauss divergence theorem
(d) None of these
(iii) The value of \[ \int \left \{ (2xy-x^2)dx + (x^2 + y^2)dx \right \} \] by using Green's theorem is,
(a) Zeron (b) One (c) Two (d) Three
(iv) \[ \iint_s f.nds = \_\_\_\_\_\_\_ \] where f=xi+yj+2k and S is the surface of the sphere x2y2+z2=a2
(a) 4πa (b) 4πa2 (c) 4πa3 (d) 4π
4 M
6 (b)
Find the work done by a force f=(2y-x2)i+ 6yzj-8xz2k from the point (0, 0, 0) to the point (1, 1, 1) along the straight-line joining these points.
4 M
6 (c)
If C is a simple closed curve in the xy-plane, prove by using Green's theorem that the integral \[ \int_C \dfrac {1}{2} (xdy-ydx) \] represent the area A enclosed by . Hence evaluate \[ \dfrac {x^2}{a^2} + \dfrac {y^2}{b^2} = 1 \]
6 M
6 (d)
Verify Stoke's theorem for \[ \overrightarrow{f} = (2x-y)i - yz^2 j- y^2 zk \] for the upper half of the sphere x2+y2+z2=1
6 M
7 (a)
Choose the correct answer for the following:
(i) L[tn]= ________
\[ (a) \ \dfrac {n}{s^{n+1}} \\ (b) \ \dfrac {n}{s^{n-1}} \\ (c) \ \dfrac {n!}{s^{n-1}} \\ (d) \ \dfrac {n!}{s^{n+1}} \]
(ii) L[e-3t]= _______
\[ (a) \ \dfrac {3}{s-3} \\ (b) \ \dfrac {3}{s+3} \\ (c) \ \dfrac {1}{s+3} \\ (d) \ \dfrac {1}{s-3} \]
iii) L{f(t-a)H(t-a)} is equal to, \[ (a) \ \dfrac{3!}{(s+2)^4} \\ (b) \ \dfrac{3!}{(s-2)^4} \\ (c) \ \dfrac{3}{(s-2)^4} \\ (d) \ \dfrac{3}{(s-2)} \]
(iv) L{δ(t-1)}= _______
(a) e-s (b) e5 (c) eaS (d) e-aS
(i) L[tn]= ________
\[ (a) \ \dfrac {n}{s^{n+1}} \\ (b) \ \dfrac {n}{s^{n-1}} \\ (c) \ \dfrac {n!}{s^{n-1}} \\ (d) \ \dfrac {n!}{s^{n+1}} \]
(ii) L[e-3t]= _______
\[ (a) \ \dfrac {3}{s-3} \\ (b) \ \dfrac {3}{s+3} \\ (c) \ \dfrac {1}{s+3} \\ (d) \ \dfrac {1}{s-3} \]
iii) L{f(t-a)H(t-a)} is equal to, \[ (a) \ \dfrac{3!}{(s+2)^4} \\ (b) \ \dfrac{3!}{(s-2)^4} \\ (c) \ \dfrac{3}{(s-2)^4} \\ (d) \ \dfrac{3}{(s-2)} \]
(iv) L{δ(t-1)}= _______
(a) e-s (b) e5 (c) eaS (d) e-aS
4 M
7 (b)
Evaluate L{sin3 2t}
6 M
7 (c)
Find L{f(t)} given that \[f(t)= \begin{cases}2 &3>t>0 \\t &t>3 \end{cases}\]
6 M
7 (d)
Express \[f(t) = \begin{cases}t^2 &2>t>0 \\4t &4\ge t>2 \\8 &t>4 \end{cases}\] in terms of unit step function and hence find their Laplace transform.
4 M
8 (a)
Choose the correct answer for the following:
(i) L-1 {cos at}= _______ \[ (a)\ \dfrac {s}{s^2 + a^2} \\ (b) \ \dfrac {s}{s^2 - a^2} \\ (c) \ \dfrac {1}{s^2 + a^2} \\ (d) \ \dfrac {1}{s^2 - a^2} \] (ii) L-1 {F (s-a)}= ________
(a) etf(t)
(b) eatf(t)
(c) e-atf(t)
(d) None of these
\[ L^{-1} \left \{ \cot^{-1} \left ( \dfrac {2}{s^2} \right ) \right \} = \_\_\_\_\_\_ \\ (a) \ \dfrac {\sin t}{t} \\ (b) \ \dfrac {\sinh at}{t} \\ (c) \ \dfrac{\sin at }{t} \\ (d) \ \dfrac {\sinh t}{t} \]
(iv) For the function f(t)=1, convolution theorem condition,
(a) Not satisfied
(b) Satisfied with some condition
(c) Satisfied
(d) None of these
(i) L-1 {cos at}= _______ \[ (a)\ \dfrac {s}{s^2 + a^2} \\ (b) \ \dfrac {s}{s^2 - a^2} \\ (c) \ \dfrac {1}{s^2 + a^2} \\ (d) \ \dfrac {1}{s^2 - a^2} \] (ii) L-1 {F (s-a)}= ________
(a) etf(t)
(b) eatf(t)
(c) e-atf(t)
(d) None of these
\[ L^{-1} \left \{ \cot^{-1} \left ( \dfrac {2}{s^2} \right ) \right \} = \_\_\_\_\_\_ \\ (a) \ \dfrac {\sin t}{t} \\ (b) \ \dfrac {\sinh at}{t} \\ (c) \ \dfrac{\sin at }{t} \\ (d) \ \dfrac {\sinh t}{t} \]
(iv) For the function f(t)=1, convolution theorem condition,
(a) Not satisfied
(b) Satisfied with some condition
(c) Satisfied
(d) None of these
4 M
8 (b)
Find the inverse Laplace transform of \[ \dfrac {2s^2 - 6s + 5}{(s-1)(s-2)(s-3)} \]
4 M
8 (c)
Find \[ L^{-1} \left(\dfrac{s}{(s-1)(s^2 + 4)}\right) \] using convolution theorem
6 M
8 (d)
Solve differential equation y"(t) + y = F(t) where \[F(t)= \begin{cases} 0 & 1>t>0 \\2 &t>1 \end{cases}\] Given that y(0)=0=y'(0)
6 M
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