の X(0)=0 X6) = 0 2. ニ
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Chapter2: Second-order Linear Odes
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Why is lambda= (nip/2 )^2and not (npi)^2? Can you please explain it to me?
![### Separation of Variables in Partial Differential Equations
Given the partial differential equation (PDE):
\[ a^2 U_{xx} = U_{tt} \]
Assume the solution \( U(x, t) \) can be written as a product of two functions, one depending on \( x \) and the other on \( t \):
\[ U(x, t) = X(x)T(t) \]
Substituting \( U(x, t) = X(x)T(t) \) into the PDE, we obtain:
\[ a^2 X'' T = X T'' \]
Rearranging the terms gives:
\[ \frac{X''}{X} = \frac{T''}{a^2 T} \]
Since the left-hand side is a function of \( x \) alone and the right-hand side is a function of \( t \) alone, both sides must be equal to a constant, say \( -\lambda \). Thus we set:
\[ \frac{X''}{X} = -\lambda \quad \text{and} \quad \frac{T''}{a^2 T} = -\lambda \]
This leads to the two ordinary differential equations (ODEs):
\[ X'' + \lambda X = 0 \]
\[ T'' + a^2 \lambda T = 0 \]
#### Boundary Conditions
Given boundary conditions:
\[ U(0, t) = U(2, t) = 0 \]
This implies:
\[ X(0) = 0 \]
\[ X(2) = 0 \]
For certain values of \( \lambda \), solutions to the first ODE are:
\[ \lambda = \lambda_n = \left(\frac{n \pi}{2}\right)^2, \quad n \in \mathbb{N} \]
The corresponding solutions are:
\[ X_n = C_n \sin \left( \frac{n \pi}{2} x \right) \]
#### Solutions for \( T(t) \)
The second ODE becomes:
\[ T_n'' + a^2 \left(\frac{n \pi}{2}\right)^2 T_n = 0 \]
The solutions to this ODE are:
\[ T_n(t) = d_n \cos \left( \frac{n \pi a}{2} t \right) + e_n \sin \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fca708747-373f-4dbe-b127-10492ca0e68f%2F41783ba4-9c5f-444c-a486-bbbb82a1c8e8%2Fhg6fume_processed.png&w=3840&q=75)
Transcribed Image Text:### Separation of Variables in Partial Differential Equations
Given the partial differential equation (PDE):
\[ a^2 U_{xx} = U_{tt} \]
Assume the solution \( U(x, t) \) can be written as a product of two functions, one depending on \( x \) and the other on \( t \):
\[ U(x, t) = X(x)T(t) \]
Substituting \( U(x, t) = X(x)T(t) \) into the PDE, we obtain:
\[ a^2 X'' T = X T'' \]
Rearranging the terms gives:
\[ \frac{X''}{X} = \frac{T''}{a^2 T} \]
Since the left-hand side is a function of \( x \) alone and the right-hand side is a function of \( t \) alone, both sides must be equal to a constant, say \( -\lambda \). Thus we set:
\[ \frac{X''}{X} = -\lambda \quad \text{and} \quad \frac{T''}{a^2 T} = -\lambda \]
This leads to the two ordinary differential equations (ODEs):
\[ X'' + \lambda X = 0 \]
\[ T'' + a^2 \lambda T = 0 \]
#### Boundary Conditions
Given boundary conditions:
\[ U(0, t) = U(2, t) = 0 \]
This implies:
\[ X(0) = 0 \]
\[ X(2) = 0 \]
For certain values of \( \lambda \), solutions to the first ODE are:
\[ \lambda = \lambda_n = \left(\frac{n \pi}{2}\right)^2, \quad n \in \mathbb{N} \]
The corresponding solutions are:
\[ X_n = C_n \sin \left( \frac{n \pi}{2} x \right) \]
#### Solutions for \( T(t) \)
The second ODE becomes:
\[ T_n'' + a^2 \left(\frac{n \pi}{2}\right)^2 T_n = 0 \]
The solutions to this ODE are:
\[ T_n(t) = d_n \cos \left( \frac{n \pi a}{2} t \right) + e_n \sin \
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