STUPIDSID
1 (a)
Choose the correct answer:
(i) The general solution of the equation x2p2+3xyp+2y2=0 is _______ \[ (a) \ (y^2 x-c) (xy-c)=0 \\ (b) \ (x-y-c)(x^2 + y^2 - c)=0 \\ (c) \ (xy - c) (x^2 y-c)=0 \\ (d) \ (y-x-c) (x^2 + y^2 + c) = 0 \]
ii) The given differential equation is solvable for y, if it is possible to express y in terms of _______
(a) y and p
(b) x and p
(c) x and y
(d) y and x
iii) The singular solution of Clairaut's equation is ________
(a) y=xg(x)+f[g(x)]
(b) y=cx+f(c)
(c) cy + f(c)
(d) y g2 (x) + f [g(x)]
iv) The singular solution of the equation y=px-log p is ________ \[ (a) \ y^2 = 4ax \\ (b) \ x=1-\log x \\ (c) \ y=1 - \log \left ( \dfrac {1}{x} \right ) \\ (d) \ x^2 = y \log x \]
(i) The general solution of the equation x2p2+3xyp+2y2=0 is _______ \[ (a) \ (y^2 x-c) (xy-c)=0 \\ (b) \ (x-y-c)(x^2 + y^2 - c)=0 \\ (c) \ (xy - c) (x^2 y-c)=0 \\ (d) \ (y-x-c) (x^2 + y^2 + c) = 0 \]
ii) The given differential equation is solvable for y, if it is possible to express y in terms of _______
(a) y and p
(b) x and p
(c) x and y
(d) y and x
iii) The singular solution of Clairaut's equation is ________
(a) y=xg(x)+f[g(x)]
(b) y=cx+f(c)
(c) cy + f(c)
(d) y g2 (x) + f [g(x)]
iv) The singular solution of the equation y=px-log p is ________ \[ (a) \ y^2 = 4ax \\ (b) \ x=1-\log x \\ (c) \ y=1 - \log \left ( \dfrac {1}{x} \right ) \\ (d) \ x^2 = y \log x \]
4 M
1 (b)
Solve p2-2p sin h x-1=0
4 M
1 (c)
Solve y=2px+ tan-1 (xp2)
6 M
1 (d)
Obtain the general solution and singular solution of Clairaut's equation is (y-px)(p-1)=p
6 M
2 (a)
Choose the correct answer:
(i)The complementary function of [D4+4]x=0 is ________ \[ (a)\ x= e^{-4}[c_1 \cos t + c_2 \sin t] + e^1 [c_3 \cot t+ c_4 \sin t] \\ (b) \ x= [c_1 \cos t + c_2 \sin t] + [c_3 \cos t + c_4 \sin t] \\(c) \ x= [ c_1 + c_2 t]e^{-t} \\(d) \ x= [c_1 + c_2 t]e^{t} \]
Find the particular integral of (D3-3D2+4)y=e2x is _______ \[ (a)\ \dfrac {x^2 e^{2x}}{6} \\ (b) \ \dfrac {x^2 e^{3x}}{6} \\ (c) \ \dfrac {x^2e^x}{6} \\ (d) \ \dfrac {x^2 e^{4x}}{6} \]
\[ iii) \ Roots \ of \ \dfrac {d^2 y}{dx^2} + 4 \dfrac{dy}{dx} + 5y =0 \ are \ \_\_\_\_\_\_\_\_ \\ (a) \ 2\pm i \\ (b) \ 3\pm i \\ (c) \ 2 \pm 2i \\ (d) \ -2 \pm i \] iv) Find the particular integral of (D3 + 4D)y=sin 2x is ________ \[ (a) \ \dfrac {x \sin x}{8} \\ (b) \ \dfrac {-x \sin x}{8} \\ (c) \ \dfrac {-x \sin 2x}{8} \\ (d) \ \dfrac {x \sin 2x}{8} \]
(i)The complementary function of [D4+4]x=0 is ________ \[ (a)\ x= e^{-4}[c_1 \cos t + c_2 \sin t] + e^1 [c_3 \cot t+ c_4 \sin t] \\ (b) \ x= [c_1 \cos t + c_2 \sin t] + [c_3 \cos t + c_4 \sin t] \\(c) \ x= [ c_1 + c_2 t]e^{-t} \\(d) \ x= [c_1 + c_2 t]e^{t} \]
Find the particular integral of (D3-3D2+4)y=e2x is _______ \[ (a)\ \dfrac {x^2 e^{2x}}{6} \\ (b) \ \dfrac {x^2 e^{3x}}{6} \\ (c) \ \dfrac {x^2e^x}{6} \\ (d) \ \dfrac {x^2 e^{4x}}{6} \]
\[ iii) \ Roots \ of \ \dfrac {d^2 y}{dx^2} + 4 \dfrac{dy}{dx} + 5y =0 \ are \ \_\_\_\_\_\_\_\_ \\ (a) \ 2\pm i \\ (b) \ 3\pm i \\ (c) \ 2 \pm 2i \\ (d) \ -2 \pm i \] iv) Find the particular integral of (D3 + 4D)y=sin 2x is ________ \[ (a) \ \dfrac {x \sin x}{8} \\ (b) \ \dfrac {-x \sin x}{8} \\ (c) \ \dfrac {-x \sin 2x}{8} \\ (d) \ \dfrac {x \sin 2x}{8} \]
4 M
2 (b)
\[ Solve \ \dfrac {d^2 y}{dx^3} + 6 \dfrac {d^2 y}{dx^2}+ 11 \dfrac {dy}{dx} + 6y = e^x +1 \]
4 M
2 (c)
\[ Solve \ \dfrac {d^2 y}{dx^2} - 4y = \cos h (2x-1)+3^x \]
6 M
2 (d)
\[ Solve \ \dfrac {dy}{dx} + y = z e^{x}, \ \dfrac {dz}{dx}+ z = y+e^x \]
6 M
3 (a)
Choose the correct answer:
(i) The Wronskian x and x ex is ________
(a) ex
(b) e2x
(c) e-2x
(d) e-x
(ii) The complementary function of x2y"-xy'-3y=x2 log x is ________
(a) c1 cos (log x) + c2 sin (log x)
(b) c1 x-1+c2x
(c) c1x+c2x3
(d) c1 cosx +c2 sin x
iii) To transform (1+x)2y"+ (1+x)y'+y=2 sin log(1+x) into a linear differential equation with constant coefficient ________
(a) (1+x)=et
(b) (1+x)=e-t
(c) (1+x)2=et
(d) (1-x)2=et
iv) The equation a0(ax+b)2y"+a1(ax+b)y'+a2y=ϕ(x) is _________
(a) Simulataneous equation
(b) Cauchy's linear equation
(c) Legendre linear equation
(d) Euler's equation
(i) The Wronskian x and x ex is ________
(a) ex
(b) e2x
(c) e-2x
(d) e-x
(ii) The complementary function of x2y"-xy'-3y=x2 log x is ________
(a) c1 cos (log x) + c2 sin (log x)
(b) c1 x-1+c2x
(c) c1x+c2x3
(d) c1 cosx +c2 sin x
iii) To transform (1+x)2y"+ (1+x)y'+y=2 sin log(1+x) into a linear differential equation with constant coefficient ________
(a) (1+x)=et
(b) (1+x)=e-t
(c) (1+x)2=et
(d) (1-x)2=et
iv) The equation a0(ax+b)2y"+a1(ax+b)y'+a2y=ϕ(x) is _________
(a) Simulataneous equation
(b) Cauchy's linear equation
(c) Legendre linear equation
(d) Euler's equation
4 M
3 (b)
Using the variation of parameters method to solve the equation y"+2y'+y=e-x log x.
6 M
3 (c)
\[ Solve \ x^2 \dfrac {d^2 y}{dx^2}- (2m-1) x \dfrac {dy}{dx} + (m^2 + n^2) y = n^2 x^m \log x \]
6 M
3 (d)
Obtain the Frobenius method solve the equation \[ x\dfrac {d^2y}{dx^2} + \dfrac {dy}{dx}- y =0 \]
6 M
4 (a)
i) Partial differential equation by eliminating a and b from the relation Z=(x-a)2+(y-b)2 is ________
(a) p2q2=4z
(b) pq=4z
(c) r=4z
(d) t=4
ii) The Lagrange's linear partial differential equation Pp+Qq=R the subsidiary equation is _______ \[ (a) \ \dfrac {dx}{R} = \dfrac {dy}{P} = \dfrac {dz}{Q} \\ (b) \ \dfrac {dx}{P} = \dfrac {dy}{Q} = \dfrac {dz}{R} \\ (c) \ \dfrac {dx}{Q} = \dfrac {dy}{R} = \dfrac {dz}{P} \\ (d) \ \dfrac {dx}{P} + \dfrac {dy}{Q} + \dfrac {dz}{R} \] iii) By the method mof separation of variable we seek a solution in the form is _______
(a) x=x+y
(b) z= x2+y2
(c) x=z+y
(d) x=x(x)y(y)
iv) The solution of \[ \dfrac{\partial^2 z}{\partial x^2}= \sin (xy) \ is \\(a) \ z= -x^2 \sin (xy) + y\ f(x)+ \phi (x) \\ (b) \ \dfrac {-\sin (xy)}{y}+ x \ f(y)+ \phi (y) \\(c) \ z= \dfrac {-\sin xy}{x^2} + y f(x) + \phi (x) \\(d) \ None \ of \ these \]
(a) p2q2=4z
(b) pq=4z
(c) r=4z
(d) t=4
ii) The Lagrange's linear partial differential equation Pp+Qq=R the subsidiary equation is _______ \[ (a) \ \dfrac {dx}{R} = \dfrac {dy}{P} = \dfrac {dz}{Q} \\ (b) \ \dfrac {dx}{P} = \dfrac {dy}{Q} = \dfrac {dz}{R} \\ (c) \ \dfrac {dx}{Q} = \dfrac {dy}{R} = \dfrac {dz}{P} \\ (d) \ \dfrac {dx}{P} + \dfrac {dy}{Q} + \dfrac {dz}{R} \] iii) By the method mof separation of variable we seek a solution in the form is _______
(a) x=x+y
(b) z= x2+y2
(c) x=z+y
(d) x=x(x)y(y)
iv) The solution of \[ \dfrac{\partial^2 z}{\partial x^2}= \sin (xy) \ is \\(a) \ z= -x^2 \sin (xy) + y\ f(x)+ \phi (x) \\ (b) \ \dfrac {-\sin (xy)}{y}+ x \ f(y)+ \phi (y) \\(c) \ z= \dfrac {-\sin xy}{x^2} + y f(x) + \phi (x) \\(d) \ None \ of \ these \]
4 M
4 (b)
From the partial differential equation of all sphere of radius 3 units having their centre in the xy-plane.
4 M
4 (c)
\[ Solve \ x(y^2 + z) p-y (x^2 +z) = z (x^2 - y^2) \]
6 M
4 (d)
Use the method of seperation of variable to solve \[ y^3 \dfrac {\partial z}{\partial x} + x^2 \dfrac {\partial z}{\partial y}=0 \]
6 M
5 (a)
Choose the correct answer:
(i) The value of \[ \int^1_0 \int^{x^2}{0} e^{y/x}dydx \ is \ \_\_\_\_\_\_ \] (a) 0
(b) 1
(c) 3
(d) 1/2
ii) The value of ? (1/2) is _______ \[ (a) \ 2\sqrt{\pi} \\ (b) \ \pi /2 \\ (c) \ \sqrt{\pi} \\ (d) \ \sqrt{2x} \]
(iii) The integral \[ \int^a_0 \int_y^a\dfrac {x}{x^2 + y^2}dxdy \] after changing the order of integration is _______ \[ (a) \ \int^a_0 \int^x_0 dydx \\ (b) \ \int^a_0 \int^x_0 \dfrac{x}{x^2 + y^2}dxdy \\(c) \ \int^x_0 \int^x_0 \dfrac{x}{x^2 + y^2}dxdy \\ (d) \ \int^x_0 \int^a_0 \dfrac {x}{x^2 + y^2}dxdy \]
iv) The value of β (3, 1/2) is ________ \[ (a) \ \dfrac {15}{16} \\(b) \ \dfrac {16}{15} \\ (c) \ \dfrac {16}{5}\\ (d) \ \dfrac {16}{3} \]
(i) The value of \[ \int^1_0 \int^{x^2}{0} e^{y/x}dydx \ is \ \_\_\_\_\_\_ \] (a) 0
(b) 1
(c) 3
(d) 1/2
ii) The value of ? (1/2) is _______ \[ (a) \ 2\sqrt{\pi} \\ (b) \ \pi /2 \\ (c) \ \sqrt{\pi} \\ (d) \ \sqrt{2x} \]
(iii) The integral \[ \int^a_0 \int_y^a\dfrac {x}{x^2 + y^2}dxdy \] after changing the order of integration is _______ \[ (a) \ \int^a_0 \int^x_0 dydx \\ (b) \ \int^a_0 \int^x_0 \dfrac{x}{x^2 + y^2}dxdy \\(c) \ \int^x_0 \int^x_0 \dfrac{x}{x^2 + y^2}dxdy \\ (d) \ \int^x_0 \int^a_0 \dfrac {x}{x^2 + y^2}dxdy \]
iv) The value of β (3, 1/2) is ________ \[ (a) \ \dfrac {15}{16} \\(b) \ \dfrac {16}{15} \\ (c) \ \dfrac {16}{5}\\ (d) \ \dfrac {16}{3} \]
4 M
5 (b)
Change the order of integration in \[ \int^{4a}_0 \int^{\sqrt{az}}_{\frac {x^2}{4x}}dydx \] and hence evaluate the same.
4 M
5 (c)
\[ Evaluate \ \int^c_{-c} \int^{b}_{-b} \int^{a}_{-a}(x^2 + y^2 + z^2) dx \ dy \ dz \]
6 M
5 (d)
\[ Prove \ that \ \int^1_0 \dfrac {x^2}{\sqrt{1-x^4}}dx^{-x} \int^1_0 \dfrac {1}{\sqrt{1+x}}dx = \dfrac {\pi}{4 \sqrt{2}} \]
6 M
6 (a)
Let S be the closed boundary surface of a region of volume V then for a vector field of difined in V and S ∫s f.nds is ________
(a) ∫v curl of dv
(b) ∫v f dv
(c) ∫v grad dv
(d) None of these
\[ If \ \int_c f.dr \ where \ f=3xy\widehat{i}-y^2 \widehat{j}\] and C is the part of the parabola y=2x2 from the region (0, 0) to the point (1, 2) is _______
(a) 7/6
(b) -7/6
(c) 3x+3y
(d) -35
iii) In the Green's theorem in the plane \[ \oint_c Mdx+Ndy = \_\_ \_\_ \_\_ \_\_ \] \[ (a) \ \iint_R \left [ \dfrac {\partial M}{\partial y} + \dfrac {\partial N}{\partial x} \right ]dxdy \\ (b) \ \iint_R \left [ \dfrac {\partial M}{\partial y} - \dfrac {\partial N} { \partial x}\right ]dxdy \\ (c) \ \iint_R \left [ \dfrac {\partial N} { \partial x} - \dfrac {\partial M}{\partial y} \right ]dxdy \\ (d) \iint_R \left [ \dfrac {\partial N}{\partial x} + \dfrac {\partial M}{\partial y} \right ]dxdy \]
iv) A necessary and sufficient condition that the line integral \[ \int_c \overrightarrow{F}.d\overrightarrow{r} \] for any closed curve C is _______ \[ (a) \ div \overrightarrow{F}=0 \\(b) \ div \overrightarrow{F} e 0 \\(c) \ curl \overrightarrow{F} =0 \\(d) \ grad \overrightarrow{F}=0 \]
(a) ∫v curl of dv
(b) ∫v f dv
(c) ∫v grad dv
(d) None of these
\[ If \ \int_c f.dr \ where \ f=3xy\widehat{i}-y^2 \widehat{j}\] and C is the part of the parabola y=2x2 from the region (0, 0) to the point (1, 2) is _______
(a) 7/6
(b) -7/6
(c) 3x+3y
(d) -35
iii) In the Green's theorem in the plane \[ \oint_c Mdx+Ndy = \_\_ \_\_ \_\_ \_\_ \] \[ (a) \ \iint_R \left [ \dfrac {\partial M}{\partial y} + \dfrac {\partial N}{\partial x} \right ]dxdy \\ (b) \ \iint_R \left [ \dfrac {\partial M}{\partial y} - \dfrac {\partial N} { \partial x}\right ]dxdy \\ (c) \ \iint_R \left [ \dfrac {\partial N} { \partial x} - \dfrac {\partial M}{\partial y} \right ]dxdy \\ (d) \iint_R \left [ \dfrac {\partial N}{\partial x} + \dfrac {\partial M}{\partial y} \right ]dxdy \]
iv) A necessary and sufficient condition that the line integral \[ \int_c \overrightarrow{F}.d\overrightarrow{r} \] for any closed curve C is _______ \[ (a) \ div \overrightarrow{F}=0 \\(b) \ div \overrightarrow{F} e 0 \\(c) \ curl \overrightarrow{F} =0 \\(d) \ grad \overrightarrow{F}=0 \]
4 M
6 (b)
Using the divergence theorem, evaluate\[ \int_c f.nds \ where \ f=4xz\widehat{i}- y^2 \widehat{j}+ yz\widehat{k} \] and S is the surface of the cube bounded by x=0, x=1, y=0, y=1, z=0, z=1.
4 M
6 (c)
Use the Green's theorem, evaluate \[ \iint_c (2x^2 - y^2)dx + (x^2 + y^2)dy \] where C is the triangle formed by the lines x=0, y=0 and x+y=1.
6 M
6 (d)
Verify the Stoke's theorem for \[ f=-y^3 \widehat{i}+ x^3 \widehat{j} \] where S is the circle disc x2+y2 ≤ 1, z=0.
6 M
7 (a)
Choose the correct answer:
\[ (i)\ L\{\sinh at \}= \_\_\_\_\_\_\_ \\ (a) \ \dfrac {s}{s^2 -a^2} \\ (b) \ \dfrac {s}{s}{s^2 + a^2} \\ (c) \ \dfrac {a}{s^2 -a^2} \\ (d) \ \dfrac {a}{s^2 +a^2}\]
ii) if L{f(t)}=F(s) then L{eatf(t)} is _______ (a) F(s+a)
(b) F(s-a)
(c) F(s)
(d) None of these
\[ iii) \ L\left \{ \dfrac {e^t \sin t}{t} \right \} \\(a) \ \dfrac {\pi}{2}+ \tan^{-1}(s-1) \\(b) \ \dfrac {\pi}{2}+ \tan^{-1}s \\ (c) \ \dfrac {\pi}{2}- \cot^{-1}s \\(d) \ \cot^{-1}(s-1) \]
iv) Transform of unit step function L{u(t-a)} is, ______ \[ (a)\ \dfrac {e^{as}}{s} \\(b) \ \dfrac {e^{-s}}{s} \\(c) \ \dfrac {e^{2s}}{s} \\(d) \ \dfrac {e^{-as}}{s} \]
\[ (i)\ L\{\sinh at \}= \_\_\_\_\_\_\_ \\ (a) \ \dfrac {s}{s^2 -a^2} \\ (b) \ \dfrac {s}{s}{s^2 + a^2} \\ (c) \ \dfrac {a}{s^2 -a^2} \\ (d) \ \dfrac {a}{s^2 +a^2}\]
ii) if L{f(t)}=F(s) then L{eatf(t)} is _______ (a) F(s+a)
(b) F(s-a)
(c) F(s)
(d) None of these
\[ iii) \ L\left \{ \dfrac {e^t \sin t}{t} \right \} \\(a) \ \dfrac {\pi}{2}+ \tan^{-1}(s-1) \\(b) \ \dfrac {\pi}{2}+ \tan^{-1}s \\ (c) \ \dfrac {\pi}{2}- \cot^{-1}s \\(d) \ \cot^{-1}(s-1) \]
iv) Transform of unit step function L{u(t-a)} is, ______ \[ (a)\ \dfrac {e^{as}}{s} \\(b) \ \dfrac {e^{-s}}{s} \\(c) \ \dfrac {e^{2s}}{s} \\(d) \ \dfrac {e^{-as}}{s} \]
4 M
7 (b)
\[ Evaluate \ L\left \{ 3^t + \dfrac {\cos 2t - \cos 3t}{t}+ t \sin t \right \} \]
4 M
7 (c)
Find the Laplace transform of the triangular wave, given by\[ f(t)= \left\{\begin{matrix}t &0 & and \ f(t+2c)= f(t) \\2C-1 &C<1<2C & \ \end{matrix}\right. \]
6 M
7 (d)
\[ Express \ f(t) = \left\{\begin{matrix}\cos t &if 0<1<\pi \\\cos 2t & if \pi < t< 2\pi \\\cos 3t & if \ t> 2\pi \end{matrix}\right. \] in terms of unit step function and hence find L{f(t)}
6 M
8 (a)
Choose the correct answer:
\[ L^{-1} \left \{ \cot^{-1} \left ( \dfrac {s}{a} \right ) \right \} = \_\_\_\_\_\_\_\_ \\(a) \ \dfrac {\sin t}{t} \\(b) \ \dfrac {\sin at}{t} \\(c) \ \dfrac {\sinh at}{t} \\(d) \ \dfrac {\sinh t}{t} \] \[ (ii) \ L^{-1} \left \{ \dfrac {1}{4s^2 -36} \right \}= \_\_\_\_\_\_\_\_ \\(a) \ \dfrac {\cosh 6t}{4} \\(b) \ \dfrac {\sin 3t}{12} \\(c) \ \dfrac{\sinh 3t}{12} \\(d) \ \dfrac{\cosh 3t}{6}\\ \] \[ iii) \ L^{-1}\left \{ \dfrac {1}{s(s^2 +a^2)} \right \} = \_\_\_\_\_\_\_\_ \_\ \\(a) \ \dfrac {1-\cos at}{a^2} \\(b) \ \dfrac {1+ \cos at}{a^2} \\(c) \ \dfrac {1- \sin at}{a^2} \\(d) \ \dfrac {1+ \sin 3t}{6} \] \[ (iv) \ L^{-1} \left \{ \dfrac {s^2 - 3s +4}{s^4} \right \} = \_\_\_\_\_\_\_\_ \_\ \\(a)\ 1-3t + 2t^3 \\(b) \ 1+ \dfrac {t^2}{3} \\(c) \ t+ \dfrac {3}{2}+ 1 \\(d) \ t-\dfrac {3}{2}t^2 + \dfrac {2}{3} t^3 \]
\[ L^{-1} \left \{ \cot^{-1} \left ( \dfrac {s}{a} \right ) \right \} = \_\_\_\_\_\_\_\_ \\(a) \ \dfrac {\sin t}{t} \\(b) \ \dfrac {\sin at}{t} \\(c) \ \dfrac {\sinh at}{t} \\(d) \ \dfrac {\sinh t}{t} \] \[ (ii) \ L^{-1} \left \{ \dfrac {1}{4s^2 -36} \right \}= \_\_\_\_\_\_\_\_ \\(a) \ \dfrac {\cosh 6t}{4} \\(b) \ \dfrac {\sin 3t}{12} \\(c) \ \dfrac{\sinh 3t}{12} \\(d) \ \dfrac{\cosh 3t}{6}\\ \] \[ iii) \ L^{-1}\left \{ \dfrac {1}{s(s^2 +a^2)} \right \} = \_\_\_\_\_\_\_\_ \_\ \\(a) \ \dfrac {1-\cos at}{a^2} \\(b) \ \dfrac {1+ \cos at}{a^2} \\(c) \ \dfrac {1- \sin at}{a^2} \\(d) \ \dfrac {1+ \sin 3t}{6} \] \[ (iv) \ L^{-1} \left \{ \dfrac {s^2 - 3s +4}{s^4} \right \} = \_\_\_\_\_\_\_\_ \_\ \\(a)\ 1-3t + 2t^3 \\(b) \ 1+ \dfrac {t^2}{3} \\(c) \ t+ \dfrac {3}{2}+ 1 \\(d) \ t-\dfrac {3}{2}t^2 + \dfrac {2}{3} t^3 \]
4 M
8 (b)
\[ Find \ L^{-1} \left \{ \dfrac{3s+7}{s^2 - 2s -3} \right \} \]
4 M
8 (c)
Using Convolution theorem evaluate \[ L^{-1} \left \{ \dfrac{1} {(s+1)(s^2+4)}\right \} \]
6 M
8 (d)
\[ Solve \ \dfrac {d^2y}{dt} + 5 \dfrac {dy}{dt} + 6y = 5e^{2t} \ given \ that \ y(0)= 2, \ \dfrac {dy(0)}{dt}=1 \] by using Laplace transform method.
6 M
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