STUPIDSID
Choose the correct answers for the following:
1 (a) (i)
A differential equation of the first order but of higher degree, solvable for y, has the solution is:
(a) f(x,y,c)=0
(b) f(x,c1,c2)=0
(c) f(x,p,c)=0
(d) f1(x,y,c)·f2(x,y,c)=0
(a) f(x,y,c)=0
(b) f(x,c1,c2)=0
(c) f(x,p,c)=0
(d) f1(x,y,c)·f2(x,y,c)=0
1 M
1 (a) (ii)
If c2x2+1=2cy is the general solution of a differential equation then its singular solution is:
(a) y = x
(b) y = -x
(c) both (a) and (b)
(d) None of these
(a) y = x
(b) y = -x
(c) both (a) and (b)
(d) None of these
1 M
1 (a) (iii)
The general solution of the differential equation p=log(px-y) is:
(a) y = px+ep
(b) y = px-ep
(c) y = px-ec
(d) y = cx-ec
(a) y = px+ep
(b) y = px-ep
(c) y = px-ec
(d) y = cx-ec
1 M
1 (a) (iv)
The differential equation xp2 + x = 2yp can be solvable for:
(a) p
(b) y
(c) x
(d) All of these
(a) p
(b) y
(c) x
(d) All of these
1 M
1 (b)
Solve xyp2 + p(3x2-2y2) - 6xy = 0
5 M
1 (c)
Solve y = psinp + cosp
5 M
1 (d)
Solve y2logy = xyp + p2
6 M
Choose the correct answers for the following:
2 (a) (i)
\(\dfrac {1}{f(D)} (e^{3x} x^2) = \ \cdots \ \cdots\)
| \(\text{(a) } e^{3x} \dfrac {1}{f(D-3)}x^2\) | \(\text{(b) } e^{3x} \dfrac {1}{f(D+3)}x^2\) | |
| \(\text{(c) } x^2 \dfrac {1}{f(D-3)} e^{3x}\) | \(\text{(d) } x^2 \dfrac {1}{f(D+3)} e^{3x}\) |
1 M
2 (a) (ii)
The roots of auxiliary equation of (D4 + 2D3 - 5D2 - 6D)y = 0 are:
(a) -1, -1, 2, -3
(b) 0, -1, 2, -3
(c) 0, 1, -2, 3
(d) 0, -1, 2, 3
(a) -1, -1, 2, -3
(b) 0, -1, 2, -3
(c) 0, 1, -2, 3
(d) 0, -1, 2, 3
1 M
2 (a) (iii)
The particular integral of (-D+2)3y = 3e2x is:
(a) (x3e2x)/3
(b) (x3e2x)/2
(c) -(x3e2x)/2
(d) -(x3e2x)/6
(a) (x3e2x)/3
(b) (x3e2x)/2
(c) -(x3e2x)/2
(d) -(x3e2x)/6
1 M
2 (a) (iv)
If dx/dt - 2y = 0, dy/dt - 2x = 0 then y is a function of:
(a) e2t and e-2t
(b) e2it and e-2it
(c) et and e-2t
(d) none of these
(a) e2t and e-2t
(b) e2it and e-2it
(c) et and e-2t
(d) none of these
1 M
2 (b)
Solve (D3 - 6D2 + 11D - 6)y = 2x + cos2x
5 M
2 (c)
Solve: (D2 - 4D + 4)y = 8x2e2xsin(2x)
5 M
2 (d)
Solve dx/dt + dy/dt + 2x + y = 0; dy/dt + 5x + 3y = 0.
6 M
Choose the correct answers for the following:
3 (a) (i)
The complementary function of x2y" + 4xy' + 2y = ex is:
(a) c1e-x + c2e-2x
(b) c1(-x) + c2(-2x)
(c) c1e-2 + c2e2z
(d) c1/x + c2/x2
(a) c1e-x + c2e-2x
(b) c1(-x) + c2(-2x)
(c) c1e-2 + c2e2z
(d) c1/x + c2/x2
1 M
3 (a) (ii)
If y = u(x)·1 + v(x)·e2x is a particular integral of y" + y = cosecx in the method of variation of parameters then v(x) = ?
(a) e-x
(b) e-2x
(c) e2x
(d) -e-x
(a) e-x
(b) e-2x
(c) e2x
(d) -e-x
1 M
3 (a) (iii)
The roots of the auxiliary equation of the transformed equation of: (2x+1)2y" - 2(2x+1)y' - 12y = 6x+5 are:
(a) 3, -1
(b) -3, 1
(c) 12, -4
(d) None of these
(a) 3, -1
(b) -3, 1
(c) 12, -4
(d) None of these
1 M
3 (a) (iv)
Indicial equation is related to:
(a) Singular point
(b) Regular singular point
(c) Ordinary point
(d) None of these
(a) Singular point
(b) Regular singular point
(c) Ordinary point
(d) None of these
1 M
3 (b)
Solve (D2+1)y = tanx by method of variation of parameters.
5 M
3 (c)
Solve x2y" - xy' + 2y = xsin(logx)
5 M
3 (d)
Solve (1+x2)y" + xy' - y = 0 in series solution.
6 M
Choose the correct answers for the following:
4 (a) (i)
z = (x-a)2 + (y-b)2, a and b are arbitrary constants, is a solution of:
(a) z = 2p2 + 2q2
(b) 4z = p2 + q2
(c) p = 2(x-a)
(d) q = 2(y-b)
(a) z = 2p2 + 2q2
(b) 4z = p2 + q2
(c) p = 2(x-a)
(d) q = 2(y-b)
1 M
4 (a) (ii)
For z = (x+a)(x+b), z=0 is a:
(a) Singular solution
(b) General solution
(c) Particular solution
(d) Complete solution
(a) Singular solution
(b) General solution
(c) Particular solution
(d) Complete solution
1 M
4 (a) (iii)
Suitable set of multipliers to solve (y2 + z2)p + xyq = zx.
(a) 0, 1, 1
(b) x, -y, -z
(c) 1, -y/x, -z/x
(d) All of these
(a) 0, 1, 1
(b) x, -y, -z
(c) 1, -y/x, -z/x
(d) All of these
1 M
4 (a) (iv)
Taking Z=X(x)·Y(y) is a solution of a partial differential equation then this procedure is called:
(a) Separation of derivatives
(b) Lagranges method
(c) Separation of variables
(d) Partial separation of variables
(a) Separation of derivatives
(b) Lagranges method
(c) Separation of variables
(d) Partial separation of variables
1 M
4 (b)
Form a partial differential equation by eliminating arbitrary function from the relation z = f(xy/z).
5 M
4 (c)
Solve xp - yq = y2 - x2
5 M
4 (d)
Solve ∂2z/∂x2 - 2∂z/∂x + ∂z/∂y = 0 by the method of separation of variables.
6 M
Choose the correct answers for the following:
5 (a) (i)
∫01 ∫01-y (x2 - y2)dxdy = ...
(a) 0
(b) 1/12
(c) 1/6
(d) None of these
(a) 0
(b) 1/12
(c) 1/6
(d) None of these
1 M
5 (a) (ii)
∫01 ∫02 ∫0(2-x-y) dzdydx = ...
(a) 3
(b) 2
(c) 1
(d) None of these
(a) 3
(b) 2
(c) 1
(d) None of these
1 M
5 (a) (iii)
∫01[log(1/x)]1/2dx = ...
(a) Γ(1/2)
(b) Γ(3/2)
(c) Γ(5/2)
(d) None of these
(a) Γ(1/2)
(b) Γ(3/2)
(c) Γ(5/2)
(d) None of these
1 M
5 (a) (iv)
∫0π⁄2 cosmdx = ...
(a) 1/2 β((m-1)/2,1/2)
(b) β((m+1)/2,1/2)
(c) 1/2 β((m+1)/2,1/2)
(d) 2β((m+1)/2,1/2)
(a) 1/2 β((m-1)/2,1/2)
(b) β((m+1)/2,1/2)
(c) 1/2 β((m+1)/2,1/2)
(d) 2β((m+1)/2,1/2)
1 M
5 (b)
Change into polar coordinates and evaluate ∫0∞∫0∞e-(x2+y2)dydx
5 M
5 (c)
Evaluate ∫-cc ∫-bb ∫-aa (x2+y2+z2)dzdydx
5 M
5 (d)
Prove that β(m,n)=Γ(m)Γ(n)/Γ(m+n)
6 M
Choose the correct answers for the following:
6 (a) (i)
Which theorem gives a relation between surface integral and volume integral?
(a) Greens
(b) Stokes
(c) Divergence
(d) None of these
(a) Greens
(b) Stokes
(c) Divergence
(d) None of these
1 M
6 (a) (ii)
If c is x + y = 1 from (0,1) to (1,1) then∫C(y2dx + x2dy) = ?
(a) 0
(b) 1
(c) 2
(d) 3
(a) 0
(b) 1
(c) 2
(d) 3
1 M
6 (a) (iii)
The work done by the force F(bar) = yI+xJ+zK moves a particle from (0,0,0) to (2,1,1) along the curve x = t2, y = t, z = 0 is:
(a) 3t2
(b) 0
(c) 1
(d) None of these
(a) 3t2
(b) 0
(c) 1
(d) None of these
1 M
6 (a) (iv)
If S is any closed surface enclosing the volume V then by divergence theorem, the value of ∫SR–.dS– is:
(a) V
(b) 2V
(c) 3V
(d) None of these
(a) V
(b) 2V
(c) 3V
(d) None of these
1 M
6 (b)
Use Green's theorem to evaluate ∫C[(y - sinx)dx + cosxdy] where C is enclosed by y = 0, x = π/2, y = 2x/π
5 M
6 (c)
Use Stoke's theorem to evaluate ∫S curl F–.d(S–) where F–= yI + (x-2xz)J - xyK and S is the surface of the sphere x2y2 + z2 = a2 above the x-y plane.
5 M
6 (d)
By transforming to a triple integral, evaluate: ∫S{x3 dydz + x2y dzdx + x2z dxdy} where S is the closed surface bounded by the planes z=0, z=b and the cylinder x2 + y2 = a2
6 M
Choose the correct answers for the following:
7 (a) (i)
L(2cosh2t) = ?
(a) 4/(s2 - 4)
(b) 4s/(s2 - 4)
(c) 2s/(s2 - a2)
(d) None of these
(a) 4/(s2 - 4)
(b) 4s/(s2 - 4)
(c) 2s/(s2 - a2)
(d) None of these
1 M
7 (a) (ii)
L{sint/t} = ?
(a) cot-1s
(b) 1/(s2+1)
(c) tan-1s
(d) cot-1(s-1)
(a) cot-1s
(b) 1/(s2+1)
(c) tan-1s
(d) cot-1(s-1)
1 M
7 (a) (iii)
L(f'(t)) = ?
(a) sf(t) - f(0)
(b) sf'(s) - f(0)
(c) f(s)··f(0)
(d) None of these
(a) sf(t) - f(0)
(b) sf'(s) - f(0)
(c) f(s)··f(0)
(d) None of these
1 M
7 (a) (iv)
L(sin2t.δ(t-2)) = ?
(a) e2ssin4
(b) e-2ssin2
(c) e-4ssin2
(d) e-2ssin4
(a) e2ssin4
(b) e-2ssin2
(c) e-4ssin2
(d) e-2ssin4
1 M
7 (b)
Prove that L(tn) = n! / sn+1 if n is a positive integer.
5 M
7 (c)
Find L((e-tsint)/t) and hence find ∫0∞(e-t)sint/t dt
5 M
7 (d)
Express in terms of unit step function and hence find L{f(t)}
f(t) = t-1; 1 < t < 2
f(t) = -t-3; 2 < t < 3
0,otherwise
f(t) = t-1; 1 < t < 2
f(t) = -t-3; 2 < t < 3
0,otherwise
6 M
Choose the correct answers for the following:
8 (a) (i)
L-1(s-5/2) = ?
(a) 2t3⁄2/√π
(b) 4t3⁄2/(3√π)
(c) 8t3⁄2/(15√π)
(d) None of these
(a) 2t3⁄2/√π
(b) 4t3⁄2/(3√π)
(c) 8t3⁄2/(15√π)
(d) None of these
1 M
8 (a) (ii)
L-1(f-(s).g-(s)) = ?
(a) f(t).g(t)
(b) ∫0t f(u)g(t-u)du
(c) ∫0t f(t-u)g(u)du
(d) either (b) or (c)
(a) f(t).g(t)
(b) ∫0t f(u)g(t-u)du
(c) ∫0t f(t-u)g(u)du
(d) either (b) or (c)
1 M
8 (a) (iii)
L-1{1/(s2+5)} = ?
(a) 1/5 sin√t
(b) 1/√5 sin√5t
(c) 1/√5 sin√5t
(d) sin√5t
(a) 1/5 sin√t
(b) 1/√5 sin√5t
(c) 1/√5 sin√5t
(d) sin√5t
1 M
8 (a) (iv)
L-1(∫s∞F(s)ds) = ?
(a) t·f(t)
(b) f(t)/t
(c) f(s)/s
(d) None of these
(a) t·f(t)
(b) f(t)/t
(c) f(s)/s
(d) None of these
1 M
8 (b)
Find L-1{log (s+1)/(s-1)}
5 M
8 (c)
Find L-1[1/(4s2-9)] by using convolution theorem.
5 M
8 (d)
Solve by using Laplace transformation y''' + 2y'' - y' - 2y = 0 where y=1, dy/dt = 2 = d2y/dt2 at t=0
6 M
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