Question

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

Answer 1

Answer on Question #65075 – Math – Differential Equations

Solve the following differential equations

Question

i)

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

Solution

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

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

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

Then

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

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

$(D^4 - D'^4 - 2D^2 D'^2)$ 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

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

Answer: $z = \sum Ae^{hx + ky}$ ; $h^4 - k^4 - 2h^2k^2 = 0$ .

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:

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

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

Then partial integral:

$$\begin{array}{l} p = \frac {1}{D ^ {2} - 2 D D ^ {\prime} + D ^ {\prime 2}} (1 2 x y) = \frac {1}{(D - D ^ {\prime}) ^ {2}} (1 2 x y) = \frac {1}{D ^ {2} \left(1 - \frac {D ^ {\prime}}{D}\right) ^ {2}} (1 2 x y) = \\ = \frac {1 2}{D ^ {2}} \left(1 - \frac {D ^ {\prime}}{D}\right) ^ {- 2} (x y) = \frac {1 2}{D ^ {2}} \left(1 + \frac {2 D ^ {\prime}}{D} + \frac {3 D ^ {\prime}}{D ^ {2}} + \dots\right) (x y) = \frac {1 2}{D ^ {2}} \left(x y + \frac {2}{D} (x) + 0 \dots\right) = \\ = \frac {1 2}{D ^ {2}} (x y) + \frac {2 4}{D ^ {3}} (x) = 1 2 y \left(\frac {x ^ {3}}{6}\right) + 2 4 \left(\frac {x ^ {4}}{2 4}\right) = x ^ {4} + 2 x ^ {3} y; \\ \end{array}$$ $$z = u + p = \varphi_ {1} (y + x) + x \varphi_ {2} (y + x) + x ^ {4} + 2 x ^ {3} y,$$

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

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