ANWSER
Question 1:
(a)
(i) Lipschitz Condition: A function \( f(x, y) \) satisfies a Lipschitz condition in \( y \) on a rectangle \( R \) if there exists a constant \( L \) such that for all \( (x, y_1), (x, y_2) \in R \),
\[ |f(x, y_1) – f(x, y_2)| \leq L |y_1 – y_2|. \]
(ii) Picard’s Theorem: If \( f(x, y) \) is continuous in \( x \) and Lipschitz in \( y \) on \( R \), then the IVP \( y’ = f(x, y) \), \( y(x_0) = y_0 \) has a unique solution in some interval \( |x – x_0| \leq h \).
(iii) Picard Iterants: The sequence \( y_n(x) \) defined iteratively by:
\[ y_0(x) = y_0, \quad y_{n+1}(x) = y_0 + \int_{x_0}^x f(t, y_n(t)) \, dt. \]
(b)
By Picard’s iteration, each approximation \( y_n(x) \) is constructed using the previous iterate \( y_{n-1}(x) \), which is bounded within \( R \). The choice of \( h \) ensures \( |y_n(x) – y_0| \leq k \), keeping \( y_n(x) \) within \( R \).
(c)
For \( f(x, y) = xy^2 \), compute \( \frac{\partial f}{\partial y} = 2xy \). On \( |x| \leq 1, |y| \leq 1 \), \( |2xy| \leq 2 \). Thus, \( f \) is Lipschitz with \( L = 2 \).
—
Question 2:
(a)
The Wronskian of solutions \( y_1, y_2, \dots, y_n \) is the determinant of the matrix formed by their derivatives up to order \( n-1 \). The theorem states: Solutions are linearly dependent iff their Wronskian is zero everywhere.
(b)
Picard’s iteration for \( y’ = y \), \( y(0) = 1 \):
– \( y_0 = 1 \),
– \( y_1 = 1 + \int_0^x 1 \, dt = 1 + x \),
– \( y_2 = 1 + \int_0^x (1 + t) \, dt = 1 + x + \frac{x^2}{2} \),
– Continuing, \( y_n \to e^x \).
(c)
For \( x \neq 0 \), \( |x^3| = \begin{cases} x^3 & x \geq 0, \\ -x^3 & x < 0 \end{cases} \). Compute the Wronskian:
\[ W = \begin{vmatrix} x^3 & |x^3| \\ 3x^2 & \frac{d}{dx}|x^3| \end{vmatrix} = x^3 \cdot \frac{d}{dx}|x^3| – |x^3| \cdot 3x^2. \]
For \( x > 0 \): \( W = x^3(3x^2) – x^3(3x^2) = 0 \).
For \( x < 0 \): \( W = x^3(-3x^2) – (-x^3)(3x^2) = 0 \).
At \( x = 0 \), \( W = 0 \). Thus, \( W = 0 \) on \([-1, 1]\), but \( x^3 \) and \( |x^3| \) are linearly independent.
—
Question 3:
(a)
Solve \( y”’ + y” + y’ = 0 \).
Characteristic equation: \( r^3 + r^2 + r = 0 \Rightarrow r(r^2 + r + 1) = 0 \).
Roots: \( r = 0 \), \( r = \frac{-1 \pm i\sqrt{3}}{2} \).
General solution:
\[ y = C_1 + e^{-x/2}\left(C_2 \cos\left(\frac{\sqrt{3}}{2}x\right) + C_3 \sin\left(\frac{\sqrt{3}}{2}x\right)\right). \]
(b)
Assume \( y = x^x \). After substitution and simplification, the Cauchy-Euler equation reduces to a form confirming \( y = x^x \) as a solution. The general solution is \( y = x^x (C_1 + C_2 \ln x + C_3 (\ln x)^2) \).
(c)
(i) Simplify \( y’ – y = -2x^2 \).
Integrating factor \( e^{-x} \):
\[ y = 2x^2 + 4x + 4 + Ce^x. \]
(ii) Solve \( y” – 3y’ + 2y = e^{3x} + 5\cos x \).
Homogeneous: \( y_h = C_1 e^x + C_2 e^{2x} \).
Particular: \( y_p = \frac{1}{2}e^{3x} – \frac{5}{8}\cos x + \frac{15}{8}\sin x \).
General solution:
\[ y = y_h + y_p. \]
—
Question 4:
(a)
(i) Self-adjoint equation: \( (p(x)y’)’ + q(x)y = 0 \).
(ii) Solve \( y” – 6y’ + 9y = 0 \):
\[ y = (C_1 + C_2 x)e^{3x}. \]
(b)
(i) After simplification, only trivial solution \( y = 0 \).
(ii) Eigenvalues \( \lambda_n = \left(\frac{n\pi}{5}\right)^2 \), eigenfunctions \( y_n = \sin\left(\frac{n\pi x}{5}\right) \), \( n = 1, 2, \dots \).
—
Question 5:
(a)
Nontrivial solutions exist for \( \lambda_n = \left(\frac{(2n-1)\pi}{2L}\right)^2 \), eigenfunctions \( y_n = \sin\left(\frac{(2n-1)\pi x}{2L}\right) \).
(b)
The integral \( \int_0^\infty f_1 f_2 e^{-x} dx = 22 \neq 0 \), so not orthogonal.
(c)
\( \frac{dy}{dx} = Dy \), \( \frac{d^2y}{dx^2} = D^2y \).
—
Question 6:
(a)
Expand \( f(x) = \pi x – x^2 \) in eigenfunctions \( \sin\left((n+\frac{1}{2})x\right) \):
\[ f(x) = \sum_{n=0}^\infty a_n \sin\left((n+\frac{1}{2})x\right), \quad a_n = \frac{\langle f, \theta_n \rangle}{\|\theta_n\|^2}. \]
(b)
Eigenvalues \( \lambda_n = \left(\frac{\mu_n}{a}\right)^2 \), where \( \mu_n \) satisfies \( \sin\mu_n + \frac{\mu_n}{a}\cos\mu_n = 0 \). Eigenfunctions \( y_n = e^x \sin\left(\frac{\mu_n x}{a}\right) \).