Question

((D^2)-(2DD')+(D'^2))z=(e^(x+y))+(2(x^2)y) solution

Answer 1

Question #71945 - Math - Differential Equations

$$((D ^ {\wedge} 2) - (2 D D ^ {\prime}) + (D ^ {\prime \wedge} 2)) z = (e ^ {\wedge} (x + y)) + (2 (x ^ {\wedge} 2) y)$$

Solution:

$$\frac {\partial^ {2} z}{\partial x ^ {2}} - 2 \frac {\partial^ {2} z}{\partial x \partial y} + \frac {\partial^ {2} z}{\partial y ^ {2}} = e ^ {x + y} + 2 x ^ {2} y$$

$\frac{\partial^2z}{\partial x^2} - 2\frac{\partial^2z}{\partial x\partial y} + \frac{\partial^2z}{\partial y^2}$ parabolic partial differential equation $(B^{2} - AC = 0$ where $B = -1$ and $A = C = 1)$

Change $(\mathrm{x},\mathrm{y})$ to $(\alpha ,\beta)$ $\alpha = \alpha (x,y)\beta = \beta (x,y)$ $\frac{D(\alpha,\beta)}{D(x,y)} = \left| \begin{array}{cc}\frac{\partial\alpha}{\partial x} & \frac{\partial\alpha}{\partial y}\\ \frac{\partial\beta}{\partial x} & \frac{\partial\beta}{\partial y} \end{array} \right|\neq 0$

Then equation $A\frac{\partial^2z}{\partial x^2} + 2B\frac{\partial^2z}{\partial x\partial y} + C\frac{\partial^2z}{\partial y^2} = F(x,y)$ transform to $A_1\frac{\partial^2z}{\partial \alpha^2} + 2B_1\frac{\partial^2z}{\partial \alpha\partial\beta} + C_1\frac{\partial^2z}{\partial \beta^2} = F_1(\alpha,\beta)$ where

$$\left\{ \begin{array}{c} A _ {1} (\alpha , \beta) = A \frac {\partial^ {2} \alpha}{\partial x ^ {2}} + 2 B \frac {\partial^ {2} \alpha}{\partial x \partial y} + C \frac {\partial^ {2} \alpha}{\partial y ^ {2}} \\ B _ {1} (\alpha , \beta) = A \frac {\partial \alpha}{\partial x} \frac {\partial \beta}{\partial x} + B \left(\frac {\partial \alpha}{\partial x} \frac {\partial \beta}{\partial y} + \frac {\partial \alpha}{\partial y} \frac {\partial \beta}{\partial x}\right) + C \frac {\partial \alpha}{\partial y} \frac {\partial \beta}{\partial y} \\ C _ {1} (\alpha , \beta) = A \frac {\partial^ {2} \beta}{\partial x ^ {2}} + 2 B \frac {\partial^ {2} \beta}{\partial x \partial y} + C \frac {\partial^ {2} \beta}{\partial y ^ {2}} \end{array} \right.$$

If control $B^{2} - AC = (B_{1}^{2} - A_{1}C_{1})\left(\frac{\partial\alpha}{\partial x}\frac{\partial\beta}{\partial y} -\frac{\partial\alpha}{\partial y}\frac{\partial\beta}{\partial x}\right)^{2}$

$B_{1} = A_{1} = 0$ then equation

$$A \frac {\partial^ {2} \alpha}{\partial x ^ {2}} + 2 B \frac {\partial^ {2} \alpha}{\partial x \partial y} + C \frac {\partial^ {2} \alpha}{\partial y ^ {2}} = 0$$

have solution $\alpha = x + y$

Let $\beta = x^{2}$ ( $\beta$ -any function $(x,y)$ with $\frac{D(\alpha,\beta)}{D(x,y)} = \left| \begin{array}{cc} \frac{\partial\alpha}{\partial x} & \frac{\partial\alpha}{\partial y} \\ \frac{\partial\beta}{\partial x} & \frac{\partial\beta}{\partial y} \end{array} \right| \neq 0$ )

$$\frac {D (\alpha , \beta)}{D (x , y)} = \frac {\partial \alpha}{\partial x} \frac {\partial \beta}{\partial y} - \frac {\partial \alpha}{\partial y} \frac {\partial \beta}{\partial x} = 0 - 2 x$$

Equation $A_{1}\frac{\partial^{2}z}{\partial\alpha^{2}} + 2B_{1}\frac{\partial^{2}z}{\partial\alpha\partial\beta} + C_{1}\frac{\partial^{2}z}{\partial\beta^{2}} = F_{1}(\alpha ,\beta)$ transform to

$$\frac {\partial^ {2} z}{\partial \beta^ {2}} = \frac {F _ {1} (\alpha , \beta)}{C _ {1}}$$ $$F _ {1} (\alpha , \beta) = \mathrm {e} ^ {\alpha} + 2 \beta (\alpha - \sqrt {\beta})$$ $$C _ {1} = 2$$

Solve $\frac{\partial^2z}{\partial\beta^2} = \frac{\mathrm{e}^{\alpha}}{2} +\beta \big(\alpha -\sqrt{\beta}\big)$

$$z (\beta) = \operatorname {Const} _ {1} + \operatorname {Const} _ {2} \beta - \frac {4 \beta^ {\frac {7}{2}}}{3 5} + \frac {\alpha \beta^ {3}}{6} + \frac {\beta^ {2} e ^ {\alpha}}{2}$$ $$z (x, y) = \operatorname {Const} _ {1} + \operatorname {Const} _ {2} x ^ {2} - \frac {4 x ^ {7}}{3 5} + \frac {(x + y) x ^ {6}}{6} + \frac {x ^ {4} e ^ {x + y}}{2}$$

Answer:

$$z (x, y) = \operatorname {Const} _ {1} + \operatorname {Const} _ {2} x ^ {2} - \frac {4 x ^ {7}}{3 5} + \frac {(x + y) x ^ {6}}{6} + \frac {x ^ {4} e ^ {x + y}}{2}$$

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