This set of 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
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.
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