Question

solve: (D2+2DD'+D'2)z=12xy .

Answer 1

Answer on Question #40479 – Math – Differential Calculus | Equations

solve: $(D2 + 2DD' + D'2)z = 12xy$.

**Soluiton:**

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

The auxiliary equation of the given equation is $k^2 - 2k + 1 = 0$,

whence $k = 1,1$.

Then the complementary function of the given equation is $u = \varphi_1(y + 1,x) + x\varphi_2(y + 1,x) = \varphi_1(y + x) + x\varphi_2(y + x)$.

Then partial integral $p = \frac{1}{D^2 - 2DD' + D'^2} (12xy) = \frac{1}{(D - D')^2} (12xy) = \frac{1}{D^2\left(1 - \frac{D'}{D}\right)^2} (12xy) =$

$$= \frac{12}{D^2} \left(1 - \frac{D'}{D}\right)^{-2} (xy) = \frac{12}{D^2} \left(1 + \frac{2D'}{D} + \frac{3D'}{D^2} + \cdots\right) (xy) = \frac{12}{D^2} \left(xy + \frac{2}{D}(x) + 0\ldots\right) =$$ $$= \frac{12}{D^2} (xy) + \frac{24}{D^3} (x) = 12y \left(\frac{x^3}{6}\right) + 24 \left(\frac{x^4}{24}\right) = x^4 + 2x^3 y.$$

Then the general solution of the given equation is $z = u + p$, such that

$z = \varphi_1(y + x) + x\varphi_2(y + x) + x^4 + 2x^3 y$, where $\varphi_1, \varphi_2$ are arbitrary functions.

**Answer:** $z = \varphi_1(y + x) + x\varphi_2(y + x) + x^4 + 2x^3 y$.