1. **Problem 1: Solve the differential operator equation** $$(D^2 - DD' - D)z = \sin(3x + 4y)$$ Here, $D$ and $D'$ are differential operators with respect to $x$ and $y$ respectively. The problem is to find a function $z(x,y)$ satisfying this equation. 2. **Understanding the operators:** - $D = \frac{\partial}{\partial x}$ - $D' = \frac{\partial}{\partial y}$ The operator equation expands to: $$D^2 z - D D' z - D z = \sin(3x + 4y)$$ which means: $$\frac{\partial^2 z}{\partial x^2} - \frac{\partial}{\partial x} \left( \frac{\partial z}{\partial y} \right) - \frac{\partial z}{\partial x} = \sin(3x + 4y)$$ 3. **Rewrite the equation:** $$\frac{\partial^2 z}{\partial x^2} - \frac{\partial^2 z}{\partial x \partial y} - \frac{\partial z}{\partial x} = \sin(3x + 4y)$$ 4. **Try a particular solution of the form:** $$z_p = A \sin(3x + 4y) + B \cos(3x + 4y)$$ 5. **Calculate derivatives:** - $$\frac{\partial z_p}{\partial x} = 3A \cos(3x + 4y) - 3B \sin(3x + 4y)$$ - $$\frac{\partial^2 z_p}{\partial x^2} = -9A \sin(3x + 4y) - 9B \cos(3x + 4y)$$ - $$\frac{\partial^2 z_p}{\partial x \partial y} = \frac{\partial}{\partial y} \left( 3A \cos(3x + 4y) - 3B \sin(3x + 4y) \right) = 3A (-4) \sin(3x + 4y) - 3B (4) \cos(3x + 4y) = -12A \sin(3x + 4y) - 12B \cos(3x + 4y)$$ 6. **Substitute into the equation:** $$(-9A \sin + -9B \cos) - (-12A \sin - 12B \cos) - (3A \cos - 3B \sin) = \sin(3x + 4y)$$ Simplify: $$(-9A + 12A) \sin + (-9B + 12B) \cos - 3A \cos + 3B \sin = \sin$$ $$ (3A + 3B) \sin + (3B - 3A) \cos = \sin$$ 7. **Equate coefficients:** - For $\sin(3x + 4y)$: $$3A + 3B = 1$$ - For $\cos(3x + 4y)$: $$3B - 3A = 0$$ 8. **Solve the system:** From second equation: $$3B = 3A \Rightarrow B = A$$ Substitute into first: $$3A + 3A = 1 \Rightarrow 6A = 1 \Rightarrow A = \frac{1}{6}$$ Then $$B = \frac{1}{6}$$ 9. **Particular solution:** $$z_p = \frac{1}{6} \sin(3x + 4y) + \frac{1}{6} \cos(3x + 4y)$$ --- 10. **Problem 2: Find the distance of point $(2,-4,5)$ from the plane $$4x + 5y + 6z - 11 = 0$$ measured parallel to the line $$\frac{x}{2} = \frac{y}{1} = \frac{z}{-2}$$** 11. **Direction vector of the line:** $$\vec{d} = (2,1,-2)$$ 12. **Normal vector of the plane:** $$\vec{n} = (4,5,6)$$ 13. **Distance measured parallel to the line is the scalar projection of the vector from point to plane along the direction vector.** 14. **Find a point on the plane:** Set $x=0, y=0$: $$4(0) + 5(0) + 6z - 11 = 0 \Rightarrow 6z = 11 \Rightarrow z = \frac{11}{6}$$ Point on plane: $$P_0 = (0,0,\frac{11}{6})$$ 15. **Vector from point to plane point:** $$\vec{v} = P_0 - P = (0-2, 0+4, \frac{11}{6} - 5) = (-2,4, \frac{11}{6} - \frac{30}{6}) = (-2,4, -\frac{19}{6})$$ 16. **Distance along direction vector:** $$d = \frac{|\vec{v} \cdot \vec{n}|}{|\vec{n} \cdot \vec{d}|}$$ Calculate dot products: $$\vec{v} \cdot \vec{n} = (-2)(4) + 4(5) + \left(-\frac{19}{6}\right)(6) = -8 + 20 - 19 = -7$$ $$\vec{n} \cdot \vec{d} = 4(2) + 5(1) + 6(-2) = 8 + 5 - 12 = 1$$ 17. **Distance:** $$d = \frac{|-7|}{|1|} = 7$$ **Final answers:** - Solution to differential equation: $$z = \frac{1}{6} \sin(3x + 4y) + \frac{1}{6} \cos(3x + 4y) + z_h$$ where $z_h$ is the homogeneous solution. - Distance from point to plane measured parallel to the line is $$7$$ units.