Question

(D²+3DD'+2D') = cos (x+3y)

Answer 1

ANSWER on Question #77815 – Math – Differential Equations

QUESTION

$$(D^2 + 3DD' + 2D')z = \cos(x + 3y)$$

SOLUTION

As we know

$$F(D,D')z = f(x,y) \rightarrow z(x,y) = C.F. + P.I.$$

where

$$C.F. : F(D,D')z = 0$$

1 STEP: Let find C.F.

$$(D^2 + 3DD' + 2D')z = 0$$

Since $(D^2 + 3DD' + 2D')$ cannot be resolved into linear factors in $D$ and $D'$, hence

$$C.F. = A e^{hx + ky},$$

where $A, h, k$ are arbitrary constants.

$$D^2 (A e^{hx + ky}) = \frac{\partial^2}{\partial x^2} (A e^{hx + ky}) = h^2 \cdot A e^{hx + ky}$$ $$3DD' (A e^{hx + ky}) = 3 \cdot \frac{\partial^2}{\partial x \partial y} (A e^{hx + ky}) = 3hk \cdot A e^{hx + ky}$$ $$2D' (A e^{hx + ky}) = 2 \cdot \frac{\partial}{\partial y} (A e^{hx + ky}) = 2k \cdot A e^{hx + ky}$$

Then,

$$(D^2 + 3DD' + 2D')z = 0 \rightarrow A(h^2 + 3hk + 2k) e^{hx + ky} = 0 \rightarrow h^2 + 3hk + 2k = 0 \rightarrow$$ $$k(2 + 3h) = -h^2 \rightarrow \boxed{k = \frac{-h^2}{2 + 3h}}$$

Conclusion,

$$\left\{ \begin{array}{c} C. F. = \sum_ {h = - \infty} ^ {+ \infty} A e ^ {h x + k y} \\ k = \frac {- h ^ {2}}{2 + 3 h} \end{array} \right.$$

2 STEP: Let find P.I.

$$\begin{array}{l} P. I. = \frac {1}{D ^ {2} + 3 D D ^ {\prime} + 2 D ^ {\prime}} \cos (x + 3 y) = \frac {1}{D ^ {2} + 3 D D ^ {\prime} + 2 D ^ {\prime}} \cos (1 \cdot x + 3 \cdot y) = \\ = \frac {1}{(1 \cdot i) ^ {2} + 3 \cdot (1 \cdot i) \cdot (3 \cdot i) + 2 D ^ {\prime}} \cos (1 \cdot x + 3 \cdot y) = \frac {1}{i ^ {2} + 3 \cdot 3 i ^ {2} + 2 D ^ {\prime}} \cos (x + 3 y) = \\ = \frac {1}{- 1 + 3 \cdot (- 3) + 2 D ^ {\prime}} \cos (x + 3 y) = \frac {1}{- 1 - 9 + 2 D ^ {\prime}} \cos (x + 3 y) = \frac {1}{2 D ^ {\prime} - 1 0} \cos (x + 3 y) = \\ = \frac {1}{2 (D ^ {\prime} - 5)} \cos (x + 3 y) = \frac {(D ^ {\prime} + 5)}{2 (D ^ {\prime} - 5) (D ^ {\prime} + 5)} \cos (x + 3 y) = \frac {1}{2} \cdot \frac {D ^ {\prime} + 5}{D ^ {\prime 2} - 2 5} \cos (x + 3 y) = \\ = \frac {1}{2} \cdot \frac {D ^ {\prime} + 5}{(3 i) ^ {2} - 2 5} \cos (x + 3 y) = \frac {1}{2} \cdot \frac {D ^ {\prime} + 5}{- 9 - 2 5} \cos (x + 3 y) = \frac {1}{2} \cdot \frac {D ^ {\prime} + 5}{- 3 4} \cos (x + 3 y) = \\ = - \frac {1}{6 8} \left(\frac {\partial}{\partial y} + 5\right) \cos (x + 3 y) = - \frac {1}{6 8} \left(\frac {\partial}{\partial y} (\cos (x + 3 y)) + 5 \cdot \cos (x + 3 y)\right) = \\ = - \frac {1}{6 8} (- 3 \sin (x + 3 y) + 5 \cos (x + 3 y)) \\ \end{array}$$

Conclusion,

$$\boxed {P. I. = \frac {3}{6 8} \cdot \sin (x + 3 y) - \frac {5}{6 8} \cdot \cos (x + 3 y)}$$

Then,

$$z (x, y) = C. F. + P. I. = \sum_ {h = - \infty} ^ {+ \infty} A e ^ {h x + k y} + \frac {3}{6 8} \cdot \sin (x + 3 y) - \frac {5}{6 8} \cdot \cos (x + 3 y)$$ $$\left\{ \begin{array}{c} z (x, y) = \sum_ {h = - \infty} ^ {+ \infty} A e ^ {h x + k y} + \frac {3}{6 8} \cdot \sin (x + 3 y) - \frac {5}{6 8} \cdot \cos (x + 3 y) \\ k = \frac {- h ^ {2}}{2 + 3 h} \end{array} \right.$$

ANSWER:

$$\left\{ \begin{array}{c} z (x, y) = \sum_ {h = - \infty} ^ {+ \infty} A e ^ {h x + k y} + \frac {3}{6 8} \cdot \sin (x + 3 y) - \frac {5}{6 8} \cdot \cos (x + 3 y) \\ k = \frac {- h ^ {2}}{2 + 3 h} \end{array} \right.$$

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