1. Problem i) Solve $$\frac{\partial^2 z}{\partial x^2} = xy$$ by direct integration. Step 1: Integrate once with respect to $x$: $$\frac{\partial z}{\partial x} = \int xy \, dx = y \int x \, dx = y \frac{x^2}{2} + f(y)$$ where $f(y)$ is an arbitrary function of $y$. Step 2: Integrate again with respect to $x$: $$z = \int \left(y \frac{x^2}{2} + f(y)\right) dx = y \frac{x^3}{6} + f(y) x + g(y)$$ where $g(y)$ is another arbitrary function of $y$. --- 2. Problem ii) Solve $$\frac{\partial^2 u}{\partial x \partial y} = \cos y \cos x$$. Step 1: Integrate with respect to $x$: $$\frac{\partial u}{\partial y} = \int \cos y \cos x \, dx = \cos y \sin x + h(y)$$ where $h(y)$ is a function of $y$. Step 2: Integrate with respect to $y$: $$u = \int (\cos y \sin x + h(y)) \, dy = \sin x \sin y + H(y) + k(x)$$ where $H(y) = \int h(y) dy$ and $k(x)$ is a function of $x$. --- 3. Problem v) Solve $$\frac{\partial^2 u}{\partial x \partial y} = 6x + 12 y^2$$ with conditions $$u(1,y) = y^2 - 2y$$ and $$u(x,2) = 5x - 5$$. Step 1: Integrate with respect to $x$: $$\frac{\partial u}{\partial y} = \int (6x + 12 y^2) dx = 3x^2 + 12 y^2 x + m(y)$$ where $m(y)$ is a function of $y$. Step 2: Integrate with respect to $y$: $$u = \int (3x^2 + 12 y^2 x + m(y)) dy = 3x^2 y + 4 x y^3 + M(y) + n(x)$$ where $M(y) = \int m(y) dy$ and $n(x)$ is a function of $x$. Step 3: Use boundary conditions: - At $x=1$, $$u(1,y) = 3(1)^2 y + 4(1) y^3 + M(y) + n(1) = y^2 - 2y$$ which simplifies to $$3 y + 4 y^3 + M(y) + n(1) = y^2 - 2 y$$ - At $y=2$, $$u(x,2) = 3 x^2 (2) + 4 x (2)^3 + M(2) + n(x) = 5 x - 5$$ which simplifies to $$6 x^2 + 32 x + M(2) + n(x) = 5 x - 5$$ Step 4: From the first condition, $$M(y) = y^2 - 2 y - 3 y - 4 y^3 - n(1) = y^2 - 5 y - 4 y^3 - n(1)$$ Step 5: From the second condition, $$n(x) = 5 x - 5 - 6 x^2 - 32 x - M(2) = -6 x^2 - 27 x - 5 - M(2)$$ Step 6: Evaluate $M(2)$ using expression for $M(y)$: $$M(2) = 2^2 - 5(2) - 4(2)^3 - n(1) = 4 - 10 - 32 - n(1) = -38 - n(1)$$ Step 7: Substitute back into $n(x)$: $$n(x) = -6 x^2 - 27 x - 5 + 38 + n(1) = -6 x^2 - 27 x + 33 + n(1)$$ Step 8: The solution is $$u = 3 x^2 y + 4 x y^3 + y^2 - 5 y - 4 y^3 - n(1) + n(x)$$ Substitute $n(x)$: $$u = 3 x^2 y + 4 x y^3 + y^2 - 5 y - 4 y^3 - n(1) - 6 x^2 - 27 x + 33 + n(1) = 3 x^2 y + 4 x y^3 + y^2 - 5 y - 4 y^3 - 6 x^2 - 27 x + 33$$ --- 4. Problem vi) Solve $$\frac{\partial^2 u}{\partial x \partial y} = \sin x \sin y$$ with $$\frac{\partial u}{\partial y} = -2 \sin y$$ and $$z=0$$ at $x=0$. Step 1: Integrate with respect to $x$: $$\frac{\partial u}{\partial y} = \int \sin x \sin y \, dx = -\cos x \sin y + p(y)$$ where $p(y)$ is a function of $y$. Step 2: Use given condition $$\frac{\partial u}{\partial y} = -2 \sin y$$: $$-\cos x \sin y + p(y) = -2 \sin y$$ At $x=0$, $$-1 \cdot \sin y + p(y) = -2 \sin y \Rightarrow p(y) = - \sin y$$ Step 3: So, $$\frac{\partial u}{\partial y} = -\cos x \sin y - \sin y = -\sin y (\cos x + 1)$$ Step 4: Integrate with respect to $y$: $$u = \int -\sin y (\cos x + 1) dy = (\cos x + 1) \cos y + q(x)$$ Step 5: Use condition $z=0$ at $x=0$: $$u(0,y) = (\cos 0 + 1) \cos y + q(0) = 2 \cos y + q(0) = 0$$ So, $$q(0) = -2 \cos y$$ Since $q(0)$ is a constant in $y$, this implies $q(0)$ must be zero and the constant term is zero. Hence, $$u = (\cos x + 1) \cos y + C$$ where $C$ is a constant. --- 5. Problem iii) Solve $$\frac{\partial^2 u}{\partial x \partial y} = e^{-y} \cos x$$. Step 1: Integrate with respect to $x$: $$\frac{\partial u}{\partial y} = \int e^{-y} \cos x \, dx = e^{-y} \sin x + r(y)$$ Step 2: Integrate with respect to $y$: $$u = \int (e^{-y} \sin x + r(y)) dy = -e^{-y} \sin x + R(y) + s(x)$$ where $R(y) = \int r(y) dy$ and $s(x)$ is a function of $x$. --- 6. Problem iv) Solve $$\frac{\partial^3 z}{\partial x^2 \partial y} = \cos(2x + 3y)$$. Step 1: Integrate with respect to $y$: $$\frac{\partial^2 z}{\partial x^2} = \int \cos(2x + 3y) dy = \frac{1}{3} \sin(2x + 3y) + t(x)$$ Step 2: Integrate twice with respect to $x$: First integration: $$\frac{\partial z}{\partial x} = \int \left( \frac{1}{3} \sin(2x + 3y) + t(x) \right) dx = -\frac{1}{6} \cos(2x + 3y) + T(x) + u(y)$$ where $T(x) = \int t(x) dx$ and $u(y)$ is a function of $y$. Second integration: $$z = \int \left(-\frac{1}{6} \cos(2x + 3y) + T(x) + u(y) \right) dx = -\frac{1}{12} \sin(2x + 3y) + \int T(x) dx + x u(y) + v(y)$$ where $v(y)$ is another function of $y$. --- Final answers summarized: 1) $$z = \frac{y x^3}{6} + f(y) x + g(y)$$ 2) $$u = \sin x \sin y + H(y) + k(x)$$ 3) $$u = 3 x^2 y + 4 x y^3 + y^2 - 5 y - 4 y^3 - 6 x^2 - 27 x + 33$$ 4) $$u = (\cos x + 1) \cos y + C$$ 5) $$u = - e^{-y} \sin x + R(y) + s(x)$$ 6) $$z = -\frac{1}{12} \sin(2x + 3y) + \int T(x) dx + x u(y) + v(y)$$