Question

(D²+3DD'+2D'²)z=12xy

Answer 1

ANSWER on Question #77405 – Math – Differential Equations

QUESTION

Solve the partial differential equation

$$(D^2 + 3DD' + 2D'^2)z = 12xy$$

SOLUTION

Let us use some known facts.

Let the given differential equation be

$$F(D, D') = f(x, y).$$

Factorize $F(D, D')$ into linear factors. Then use the following results:

**Rule I.** Corresponding to each non-repeated factor $(bD - aD' - c)$, the part of C.F. is taken as

$$e^{\left(\frac{cx}{b}\right)} \varphi(by + ax), \quad \text{if } b \neq 0$$

We now have three particular cases of Rule I:

**Rule IA.** Take $c = 0$ in Rule I. Hence corresponding to each linear factor $(bD - aD')$, the part of C.F. is

$$\varphi(by + ax), \quad \text{if } b \neq 0.$$

**Rule IB.** Take $a = 0$ in Rule I. Hence corresponding to each linear factor $(bD - c)$, the part of C.F. is

$$e^{\left(\frac{cx}{b}\right)} \varphi(by), \quad \text{if } b \neq 0.$$

**Rule IC.** Take $a = c = 0$ and $b = 1$ in Rule I. Hence corresponding to each linear factor $(1 \cdot D)$, the part of C.F. is

$$\varphi(y).$$

In our case,

$$(D^{2} + 3DD' + 2D'^{2})z = 12xy \rightarrow z(x,y) = C.F. + P.I.$$

0 STEP: We factor the expression

$$\begin{aligned} D^{2} + 3DD' + 2D'^{2} &= (D^{2} + DD') + (2DD' + 2D'^{2}) = \\ &= D(D + D') + 2D'(D + D') = (D + D')(D + 2D') \end{aligned}$$

Conclusion,

$$\boxed{(D^{2} + 3DD' + 2D'^{2})z = (D + D')(D + 2D')z}$$

1 STEP: Let find C.F.

$$\begin{aligned} \left\{(D + D')z \atop (bD - aD' - c)z \right. &\rightarrow \left\{ \begin{array}{l} b = 1 \\ a = -1 \\ c = 0 \end{array} \right. \rightarrow (C.F.)_1 = e^{\left(\frac{0 \cdot x}{1}\right)} \cdot \varphi_1(1 \cdot y + (-1) \cdot x) \rightarrow \\ &\quad \boxed{(C.F.)_1 = \varphi_1(y - x), \text{ where } \varphi_1 \text{ is arbitrary function}} \end{aligned}$$ $$\begin{aligned} \left\{(D + 2D')z \atop (bD - aD' - c)z \right. &\rightarrow \left\{ \begin{array}{l} b = 1 \\ a = -2 \\ c = 0 \end{array} \right. \rightarrow (C.F.)_2 = e^{\left(\frac{0 \cdot x}{1}\right)} \cdot \varphi_2(1 \cdot y + (-2) \cdot x) \rightarrow \\ &\quad \boxed{(C.F.)_2 = \varphi_2(y - 2x), \text{ where } \varphi_2 \text{ is arbitrary function}} \end{aligned}$$

Then,

$$C.F. = (C.F.)_1 + (C.F.)_2 \rightarrow \boxed{C.F. = \varphi_1(y - x) + \varphi_2(y - 2x)}$$

2 STEP: Let find P.I.

$$P.I. = \frac{1}{(D + D')(D + 2D')} (12xy) = \frac{1}{D\left(1 + \frac{D'}{D}\right)D\left(1 + \frac{2D'}{D}\right)} (12xy) =$$ $$\begin{aligned} & = \frac{1}{D^{2}} \cdot \left[ 1 + \frac{D'}{D} \right]^{-1} \left[ 1 + \frac{2D'}{D} \right]^{-1} (12xy) = \left[ \frac{1}{1 + x} = 1 - x + x^{2} - x^{3} + x^{4} + \cdots, |x| < 1 \right] = \\ & = \frac{1}{D^{2}} \cdot \left[ 1 - \frac{D'}{D} + \cdots \right] \left[ 1 - \frac{2D'}{D} + \cdots \right] (12xy) = \\ & = \frac{1}{D^{2}} \cdot \left[ 1 - \frac{D'}{D} - \frac{2D'}{D} + \frac{2D'^{2}}{D^{2}} \cdots \right] (12xy) = \frac{1}{D^{2}} \cdot \left[ 1 - \frac{3D'}{D} + \frac{2D'^{2}}{D^{2}} \cdots \right] (12xy) = \\ & = \frac{1}{D^{2}} \cdot \left[ 12xy - \frac{3}{D} \cdot \left(\frac{\partial}{\partial y} (12xy)\right) + \frac{2}{D^{2}} \cdot \left(\frac{\partial^{2}}{\partial y^{2}} (12xy)\right) + \cdots \right] = \\ & = \left[ \frac{1}{D} \equiv \int dx \right] = \frac{1}{D^{2}} \cdot \left[ 12xy - \frac{3}{D} \cdot 12x + \frac{2}{D^{2}} \cdot 0 + \cdots \right] = \frac{1}{D^{2}} \cdot \left[ 12xy - (3 \cdot 12) \cdot \int x \, dx \right] = \\ & = \frac{1}{D^{2}} \cdot \left[ 12xy - 36 \cdot \frac{x^{2}}{2} \right] = \frac{1}{D} \cdot \left( \int (12xy - 18x^{2}) \, dx \right) = \frac{1}{D} \cdot \left( \frac{12yx^{2}}{2} - \frac{18x^{3}}{3} \right) = \\ & = \frac{1}{D} \cdot (6yx^{2} - 6x^{3}) = \int (6yx^{2} - 6x^{3}) \, dx = \frac{6yx^{3}}{3} - \frac{6x^{4}}{4} = 2yx^{3} - \frac{3x^{4}}{2} \end{aligned}$$

Then,

$$\boxed{P.I. = 2yx^{3} - \frac{3x^{4}}{2}}$$

Conclusion,

$$z(x, y) = C.F. + P.I. = \varphi_{1}(y - x) + \varphi_{2}(y - 2x) + 2yx^{3} - \frac{3x^{4}}{2}$$ $$\boxed{\begin{cases} z(x, y) = \varphi_{1}(y - x) + \varphi_{2}(y - 2x) + 2yx^{3} - \frac{3x^{4}}{2} \\ \text{where } \varphi_{1} \text{ and } \varphi_{2} \text{ are arbitrary functions} \end{cases}}$$

ANSWER: $\begin{cases} z(x, y) = \varphi_{1}(y - x) + \varphi_{2}(y - 2x) + 2yx^{3} - \frac{3x^{4}}{2} \\ \text{where } \varphi_{1} \text{ and } \varphi_{2} \text{ are arbitrary functions} \end{cases}$

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