Partial Differential Equations Questions and Answers for Freshers focuses on “Solution of PDE by Variable Separation Method”.
1. Solve (frac{∂u}{∂x}=6 frac{∂u}{∂t}+u) using the method of separation of variables if u(x,0) = 10 e-x.
a) 10 e-x e-t/3
b) 10 ex e-t/3
c) 10 ex/3 e-t
d) 10 e-x/3 e-t
View Answer
Answer: a
Explanation: u(x,t) = X(x) T(t)
Substituting in the given equation, X’T = 6 T’X + XT
(frac{X’-X}{6X}=frac{T’}{T}=k)
(frac{X’}{X} = 1+6k ) which implies X = ce(1+6k)x
(frac{T’}{T} = k ) which implies T = c’ ekt
Therefore, u(x,t) = cc’ e(1+6k)x ekt
Now, u(x,0) = 10 e-x = cc’e(1+6k)x
Therefore, cc’ = 10 and k = -1⁄3
Therefore, u(x,t) = 10 e-x e-t/3.
2. Find the solution of (frac{∂u}{∂x}=36 frac{∂u}{∂t}+10u ) if ( frac{∂u}{∂x} (t=0)=3e^{-2x} ) using the method of separation of variables.
a) (frac{-3}{2} e^{-2x} e^{-t/3})
b) (3e^x e^{-t/3} )
c) (frac{3}{2} e^{2x} e^{-t/3} )
d) (3e^{-x} e^{-t/3} )
View Answer
Answer: a
Explanation: (u(x,t) = X(x) T(t) )
Substituting in the given equation, (X’T = 36T’X + 10XT )
(frac{X’}{X} = k ) which implies (X = c e^{kx} )
(frac{T’}{T} = frac{(k-10)}{36} ) which implies (T = c’e^{frac{k-10}{36} t} )
(frac{∂u}{∂x} (t=0)=3e^{-2x}= cc’ke^{kx} )
Therefore k = -2 and cc’ = (frac{-3}{2} )
Hence, (u(x,t) = frac{-3}{2} e^{-2x} e^{-t/3}. )
3. Solve the partial differential equation (x^3 frac{∂u}{∂x} +y^2 frac{∂u}{∂y} = 0 ) using method of separation of variables if (u(0,y) = 10 , e^{frac{5}{y}}.)
a) (10e^{frac{5}{2x^2}} e^{frac{5}{y}} )
b) (10e^{frac{-5}{2y^2}} e^{frac{5}{x}} )
c) (10e^{frac{-5}{2y^2}} e^{frac{-5}{x}} )
d) (10e^{frac{-5}{2x^2}} e^{frac{5}{y}} )
View Answer
Answer: d
Explanation: (u(x,t) = X(x) T(t) )
(x^3 X’Y+y^2 Y’X=0)
(frac{X’}{ X} = frac{k}{x^3} ) which implies (X = ce^{frac{k}{2x^2}})
(frac{Y’}{Y} = frac{-k}{y^2} ) which implies (Y = c’ e^{frac{k}{y}})
(u(x,t) = cc’e^{frac{k}{2x^2}} e^{frac{k}{y}} )
(u(0,y) = 10e^{frac{5}{y}}= cc’e^{frac{k}{y}} )
Therefore k = 5 and cc’ = 10
Hence, (u(x,t) = 10e^{frac{-5}{2x^2}} e^{frac{5}{y}}. )
4. Solve the differential equation (5 frac{∂u}{∂x}+3 frac{∂u}{∂y}=2u ) using the method of separation of variables if (u(0,y) = 9e^{-5y}.)
a) (9e^{frac{17}{5} x} e^{-5y} )
b) (9e^{frac{13}{5} x} e^{-5y} )
c) (9e^{frac{-17}{5} x} e^{-5y} )
d) (9e^{frac{-13}{5} x} e^{-5y} )
View Answer
Answer: a
Explanation: ( u(x,t) = X(x) T(t) )
(5X’Y + 3Y’X = 2XY )
(frac{X’}{X} = frac{k}{5} ) which implies ( X= c e^{frac{k}{5} x} )
(frac{Y’}{Y} = 2-k / 3) which implies ( Y = c’ , e^{frac{2-k}{3} y} )
Therefore (u(x,t) = cc’ e^{frac{k}{5} x} e^{frac{2-k}{3} y} )
(u(0,y) = cc’ = e^{frac{2-k}{3} y} , 9 , e^{-5y} )
Hence cc’ = 9 and k = 17
Therefore, (u(x,t) = 9 , e^{frac{17}{5} x} e^{-5y}. )
5. Solve the differential equation (x^2 frac{∂u}{∂x}+y^2 frac{∂u}{∂y}=u ) using the method of separation of variables if (u(0,y) = e^{frac{2}{y}} ).
a) (e^{frac{-3}{y}} e^{frac{2}{x}} )
b) (e^{frac{3}{y}} e^{frac{2}{x}} )
c) (e^{frac{-3}{x}} e^{frac{2}{y}} )
d) (e^{frac{3}{x}} e^{frac{2}{y}} )
View Answer
Answer: c
Explanation: (u(x,t) = X(x) T(t) )
(x^2 X’Y+y^2 Y’X=XY)
(X = ce^{frac{-k}{x}} )
(Y = c’e^{frac{k-1}{y}} )
(u(x,t) = cc’ e^{frac{-k}{x}} e^{frac{k-1}{y}} )
(u(0,y) = e^{frac{2}{y}}= cc’ e^{frac{k-1}{y}} )
k = 3 and cc’ = 1
Therefore (u(x,t) = e^{frac{-3}{x}} e^{frac{2}{y}}. )
6. While solving a partial differential equation using a variable separable method, we assume that the function can be written as the product of two functions which depend on one variable only.
a) True
b) False
View Answer
Answer: a
Explanation: If we have a function u(x,t), then the function u depends on both x and t. For using the variable separable method we assume that it can be written as u(x,t) = X(x).T(t) where X depends only on x and T depends only on t.
7. While solving a partial differential equation using a variable separable method, we equate the ratio to a constant which?
a) can be positive or negative integer or zero
b) can be positive or negative rational number or zero
c) must be a positive integer
d) must be a negative integer
View Answer
Answer: b
Explanation: The constant can be any rational number. For example, we use a positive rational number to solve a 1-Dimensional wave equation, we use a negative rational number to solve 1-Dimensional heat equation, 0 when we have steady state. The choice of constant depends on the nature of the given problem.
8. When solving a 1-Dimensional wave equation using variable separable method, we get the solution if _____________
a) k is positive
b) k is negative
c) k is 0
d) k can be anything
View Answer
Answer: b
Explanation: Since the given problem is 1-Dimensional wave equation, the solution should be periodic in nature. If k is a positive number, then the solution comes out to be (c7 epx⁄c+e-px⁄cc8)(c7 ept+e-ptc8) and if k is positive the solution comes out to be (ccos(px/c) + c’sin(px/c))(c’’cospt + c’’’sinpt). Now, since it should be periodic, the solution is (ccos(px/c) + c’sin(px/c))(c’’cospt + c’’’sinpt).
9. When solving a 1-Dimensional heat equation using a variable separable method, we get the solution if ______________
a) k is positive
b) k is negative
c) k is 0
d) k can be anything
View Answer
Answer: b
Explanation: Since this is a heat equation, the solution must be a transient solution, that is it should decay as time increases. This happens only when k is negative and the solution comes out to be (c’’cospx + c’’’sinpx) (c e-c2 p2 t).
10. While solving any partial differentiation equation using a variable separable method which is of order 1 or 2, we use the formula of fourier series to find the coefficients at last.
a) True
b) False
View Answer
Answer: a
Explanation: After using the boundary conditions, when we are left with only one constant and one boundary condition, then we use Fourier series coefficient formula to find the constant.
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