Question

Solve the following differential equations (i) [D^3-DD'^2-D^2+DD']z (ii) [D^4-D'^4-2D^2D'^2]z =0 (iii) [D^2-2DD'+D'^2]z =12xy.

Answer 1

Answer on Question #66342 – Math – Differential Equations

Question

Solve the following differential equations

i)

$$[ D ^ {3} - D D ^ {\prime 2} - D ^ {2} + D D ^ {\prime} ] z = 0$$

Solution

$$D (D ^ {2} - D ^ {\prime 2} - D + D ^ {\prime}) z = 0$$ $$D [ (D - D ^ {\prime}) (D + D ^ {\prime}) - (D - D ^ {\prime}) ] z = 0$$ $$D (D - D ^ {\prime}) (D + D ^ {\prime} - 1) z = 0$$

Corresponding for each non-repeated factor $(bD - aD' - c)$, the part of general solution is taken as

$$e ^ {c x / b} \varphi (b y + a x) ; \ b \neq 0$$

Then

$$D ^ {\prime} \Rightarrow \varphi_ {1} (y)$$ $$(D - D ^ {\prime}) \Rightarrow \varphi_ {2} (y + x)$$ $$(D + D ^ {\prime} - 1) \Rightarrow e ^ {x} \varphi_ {3} (y - x)$$

Answer:

$$z = \varphi_ {1} (y) + \varphi_ {2} (y + x) + e ^ {x} \varphi_ {3} (y - x)$$

Reference: Ordinary And Partial Differential Equations, $18^{th}$ Edition, Dr. M. D. Raisinghania, Section 5.6

Question

ii)

$$[ D ^ {4} - D ^ {\prime 4} - 2 D ^ {2} D ^ {\prime 2} ] z = 0$$

Solution

$(D^4 - D'^4 - 2D^2D'^2)$ cannot be resolved into linear factors.

Let a trial solution be:

$$z = A e ^ {h x + k y} \quad ; \quad A, k, h - a r b i t r a r y c o n s t a n t s$$

Then:

$$D z = A h e ^ {h x + k y} \quad ; \quad D ^ {\prime} z = A k e ^ {h x + k y}$$ $$D ^ {2} z = A h ^ {2} e ^ {h x + k y} \quad ; \quad D ^ {\prime 2} z = A k ^ {2} e ^ {h x + k y}$$ $$D ^ {4} z = A h ^ {4} e ^ {h x + k y} \quad ; \quad D ^ {\prime 4} z = A k ^ {4} e ^ {h x + k y}$$ $$[ D ^ {4} - D ^ {\prime 4} - 2 D ^ {2} D ^ {\prime 2} ] z = A e ^ {h x + k y} (h ^ {4} - k ^ {4} - 2 h ^ {2} k ^ {2}) = 0$$

Answer:

$$z = \sum A e ^ {h x + k y} \quad ; \quad h ^ {4} - k ^ {4} - 2 h ^ {2} k ^ {2} = 0$$

Question

iii)

$$[ D ^ {2} - 2 D D ^ {\prime} + D ^ {\prime 2} ] z = 1 2 x y$$

Solution

The auxiliary equation:

$$k ^ {2} - 2 k + 1 = 0$$ $$k _ {1, 2} = 1$$

Then the complementary function:

$$u = \varphi_{1}(y + 1, x) + x \varphi_{2}(y + 1, x) = \varphi_{1}(y + x) + x \varphi_{2}(y + x)$$

When $f(x,y) = V$, where $V$ is function of $x$ and $y$, then

Partial integral:

$$p = \frac{1}{F(D, D')} V$$

We have: $F(D, D')$ and $V = 12xy$

Then partial integral:

$$\begin{aligned} p &= \frac{1}{D^{2} - 2 D D' + D'^{2}} (12xy) = \frac{1}{(D - D')^{2}} (12xy) = \frac{1}{D^{2} \left(1 - \frac{D'}{D}\right)^{2}} (12xy) = \\ &= \frac{12}{D^{2}} \left(1 - \frac{D'}{D}\right)^{-2} (xy) = \frac{12}{D^{2}} \left(1 + \frac{2 D'}{D} + \frac{3 D'}{D^{2}} + \cdots\right) (xy) = \frac{12}{D^{2}} \left(x y + \frac{2}{D} (x) + 0 \cdots\right) = \\ &= \frac{12}{D^{2}} (xy) + \frac{24}{D^{3}} (x) = 12y \left(\frac{x^{3}}{6}\right) + 24 \left(\frac{x^{4}}{24}\right) = x^{4} + 2x^{3}y \end{aligned}$$

Answer:

$$z = u + p = \varphi_{1}(y + x) + x \varphi_{2}(y + x) + x^{4} + 2x^{3}y$$

where $\varphi_{1}, \varphi_{2}$ are arbitrary functions

Reference: Ordinary And Partial Differential Equations, $18^{th}$ Edition, Dr. M. D. Raisinghania, Section 5.7

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