1. **Classify the PDEs and find their types** - For equation $U_{xx} - U_{yy} = 0$ - This fits the form $aU_{xx} + bU_{xy} + cU_{yy} = 0$ with $a=1, b=0, c=-1$ - The discriminant is $\Delta = b^2 - 4ac = 0^2 - 4(1)(-1) = 4 > 0$ - Since $\Delta > 0$, this PDE is **hyperbolic**. - For equation $U_{xx} + U_{yy} = 0$ - Here $a=1, b=0, c=1$ - The discriminant is $\Delta = 0^2 - 4(1)(1) = -4 < 0$ - Since $\Delta < 0$, this PDE is **elliptic**. - For equation $U_{xxx} + U_{xy} + U_{yy} = 0$, this involves third order derivatives and mixed derivatives unlike standard linear PDE classification, so it does not fit directly into ellipse, hyperbola, or parabola classification. 2. **Classify and sketches** - Hyperbolic PDE ($U_{xx} - U_{yy} = 0$) resembles wave equation: solution characteristics form intersecting straight lines (saddle shape). - Elliptic PDE ($U_{xx} + U_{yy}=0$) resembles Laplace equation: solutions are smooth, potential-like functions. 3. **Find the general solution of $xC_x + y C_y = z$** - Given PDE: $x \frac{\partial C}{\partial x} + y \frac{\partial C}{\partial y} = z$ - Let $C=C(x,y)$, treat as first order linear PDE. - The associated characteristic equations are: $$ \frac{dx}{x} = \frac{dy}{y} = \frac{dC}{z} $$ - From $\frac{dx}{x} = \frac{dy}{y}$, integrating gives $\frac{y}{x} = k =$ constant - Thus solution depends on $w=\frac{y}{x}$, and $C$ satisfies ODE along characteristic curves. - The general solution form is: $$ C = xF\left( \frac{y}{x} \right) + G(z) $$ where $F$ and $G$ are arbitrary functions determined by boundary conditions. 4. **Solve dy/dx = -sin(cosy)** - Given ODE: $\frac{dy}{dx} = -\sin(\cos y)$ - Separate variables: $$ \int dy = \int - \sin(\cos y) dx $$ - This ODE is non-linear and not easily separable; implicit solution or numerical methods may be needed. 5. **Solve $x^2 y_{11} - 3 c y_1 = \frac{1}{2}$** - Rewriting: $$ x^2 \frac{\partial^2 y}{\partial x^2} - 3 c \frac{\partial y}{\partial x} = \frac{1}{2} $$ - Treat as ODE in $x$ with parameter $c$: Let $y' = y_1$, then $$ x^2 y'' -3 c y' = \frac{1}{2} $$ - This is a non-homogeneous Cauchy-Euler equation. - Solve homogeneous equation: $$ x^2 y'' - 3 c y' = 0 $$ Assuming solution $y = x^m$, then $$ m(m-1) x^m - 3 c m x^m = 0 \implies m(m-1) - 3 c m = 0 $$ - Simplify: $$ m^2 - m - 3 c m = 0 \implies m^2 - (1 + 3 c) m = 0 $$ - Roots: $$ m = 0, \quad m = 1 + 3 c $$ - Homogeneous solution: $$ y_h = C_1 + C_2 x^{1 + 3 c} $$ - For particular solution, guess $y_p = A \ln x$ or constant since RHS is constant. Try $y_p = K$, then $$ y_p'' = 0, y_p' = 0 \implies LHS = 0 \neq \frac{1}{2} $$ Try $y_p = A x^r$, where $r$ to be determined. Try $y_p = A$, constant; no. Try variation of parameters or method of undetermined coefficients for Cauchy-Euler: Try $y_p = B x^{m}$ with $m=1$: Substitute: $$ y_p = B x, y_p' = B, y_p'' = 0 $$ Plug in: $$ x^2 \cdot 0 - 3 c B = \frac{1}{2} \implies -3 c B = \frac{1}{2} \implies B = -\frac{1}{6 c} $$ Thus, $$ y_p = - \frac{1}{6 c} x $$ - General solution: $$ y = C_1 + C_2 x^{1 + 3 c} - \frac{1}{6 c} x $$ 6. **Given boundary conditions**: - $u(C_x, 0) = 2 C_x$ - $u(C_0, y) = y^2$ - These can be used to determine arbitrary constants given the solution form of $u(x,y)$ if provided. --- **Summary:** - $U_{xx} - U_{yy} = 0$ is hyperbolic. - $U_{xx} + U_{yy} = 0$ is elliptic. - General solution of $x C_x + y C_y = z$ involves characteristics with $y/x$ constant. - ODE $dy/dx = -\sin(\cos y)$ is nonlinear; explicit solution difficult. - PDE $x^2 y_{11} -3 c y_1 = 1/2$ has solution $y = C_1 + C_2 x^{1 + 3 c} - \frac{1}{6 c} x$.