This set of Engineering Mathematics Multiple Choice Questions & Answers (MCQs) 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
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
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
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
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
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
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
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
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
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
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.
Sanfoundry Global Education & Learning Series – Partial Differential Equations.
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