Question

y'''+y''=8x^2 metodo anulador

Answer 1

Answer on Question #59789 - Math - Differential Equations

Question

Use the annihilator method to solve

$$y ^ {\prime \prime \prime} + y ^ {\prime \prime} = 8 x ^ {2}$$

Solution

Equation $y^{\prime \prime \prime} + y^{\prime \prime} = 8x^{2}$ is equivalent to $(D^{3} + D^{2})y = 8x^{2}$.

The homogeneous equation is

$$(D ^ {3} + D ^ {2}) y = 0,$$ $$D ^ {2} (D + 1) y = 0.$$

Its solution is $y_{h} = c_{1}e^{-x} + c_{2} + c_{3}x$.

$D^3$ annihilates $x^{2}$, $D^{3}$ annihilates $8x^{2}$:

$$D ^ {3} (x ^ {2}) = 0, D ^ {3} (8 x ^ {2}) = 0.$$

Applying $D^3$ to both sides of

$$(D ^ {3} + D ^ {2}) y = 8 x ^ {2}$$

gives us

$$D ^ {3} (D ^ {3} + D ^ {2}) y = D ^ {3} 8 x ^ {2} = 0,$$ $$D ^ {5} (D + 1) y = 0.$$

The general solution to equation (1) is

$$y = y _ {h} + y _ {p} = c _ {1} e ^ {- x} + c _ {2} + c _ {3} x + c _ {4} x ^ {2} + c _ {5} x ^ {3} + c _ {6} x ^ {4},$$

where $y_{h} = c_{1}e^{-x} + c_{2} + c_{3}x$, $y_{p} = c_{4}x^{2} + c_{5}x^{3} + c_{6}x^{4}$.

Putting

$$y _ {p} = c _ {4} x ^ {2} + c _ {5} x ^ {3} + c _ {6} x ^ {4},$$ $$y _ {p} ^ {\prime} = 2 c _ {4} x + 3 c _ {5} x ^ {2} + 4 c _ {6} x ^ {3},$$ $$y _ {p} ^ {\prime \prime} = 2 c _ {4} + 6 c _ {5} x + 12 c _ {6} x ^ {2},$$ $$y _ {p} ^ {\prime \prime \prime} = 6 c _ {5} + 24 c _ {6} x$$

into the original differential equation $y^{\prime \prime \prime} + y^{\prime \prime} = 8x^{2}$ gives us

$$y _ {p} ^ {\prime \prime \prime} + y _ {p} ^ {\prime \prime} = (6 c _ {5} + 24 c _ {6} x) + (2 c _ {4} + 6 c _ {5} x + 12 c _ {6} x ^ {2}) = (6 c _ {5} + 2 c _ {4}) + (24 c _ {6} + 6 c _ {5}) x + 12 c _ {6} x ^ {2} = 8 x ^ {2},$$

hence

$$12 c _ {6} = 8,$$ $$24 c _ {6} + 6 c _ {5} = 0,$$ $$6 c _ {5} + 2 c _ {4} = 0.$$

Next,

$$c _ {6} = \frac {8}{12} = \frac {2}{3},$$ $$c _ {5} = - \frac {24 c _ {6}}{6} = -4 c _ {6} = -4 \cdot \frac {2}{3} = - \frac {8}{3},$$ $$c _ {4} = \frac {-6 c _ {5}}{2} = -3 c _ {5} = -3 \cdot \left(- \frac {8}{3}\right) = 8.$$

Thus, $y_{p} = c_{4}x^{2} + c_{5}x^{3} + c_{6}x^{4} = 8x^{2} - \frac{8}{3} x^{3} + \frac{2}{3} x^{4}$ and

$$y = y _ {h} + y _ {p} = c _ {1} e ^ {- x} + c _ {2} + c _ {3} x + 8 x ^ {2} - \frac {8}{3} x ^ {3} + \frac {2}{3} x ^ {4}, c _ {1}, c _ {2}, c _ {3} \in \mathbb {R}.$$

Answer: $y = c_{1}e^{-x} + c_{2} + c_{3}x + 8x^{2} - \frac{8}{3} x^{3} + \frac{2}{3} x^{4}.$

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La respuesta al Problema #59789 – Matemática – Ecuaciones diferenciales

Problema

Resuelva

$$y''' + y'' = 8x^2$$

metodo anulador.

Solución

Una ecuación diferencial como $y''' + y'' = 8x^2$ se puede escribir en la forma $(D^3 + D^2)y = 8x^2$.

La ecuación homogénea es

$$(D^3 + D^2)y = 0,$$ $$D^2(D + 1)y = 0.$$

La solución de la ecuación es

$$y_h = c_1 e^{-x} + c_2 + c_3 x.$$

$D^3$ anula a $x^2$, $D^3$ anula a $8x^2$:

$$D^3(x^2) = 0, \quad D^3(8x^2) = 0.$$

Aplicamos operador $D^3$ a ambos lados de

$$(D^3 + D^2)y = 8x^2$$

tenemos

$$D^3(D^3 + D^2)y = D^3 8x^2 = 0,$$ $$D^5(D + 1)y = 0.$$

La solución general de la ecuación (1) es

$$y = y_h + y_p = c_1 e^{-x} + c_2 + c_3 x + c_4 x^2 + c_5 x^3 + c_6 x^4,$$

donde $y_h = c_1 e^{-x} + c_2 + c_3 x$, $y_p = c_4 x^2 + c_5 x^3 + c_6 x^4$.

Sustituimos

$$y_p = c_4 x^2 + c_5 x^3 + c_6 x^4,$$ $$y_p' = 2c_4 x + 3c_5 x^2 + 4c_6 x^3,$$ $$y_p'' = 2c_4 + 6c_5 x + 12c_6 x^2,$$ $$y_p''' = 6c_5 + 24c_6 x$$

en la ecuación $y''' + y'' = 8x^2$ y simplificamos:

$$y_p''' + y_p'' = (6c_5 + 24c_6 x) + (2c_4 + 6c_5 x + 12c_6 x^2) = (6c_5 + 2c_4) + (24c_6 + 6c_5) x + 12c_6 x^2 = 8x^2,$$

Igualamos los coeficientes y obtenemos las ecuaciones

$$12c_6 = 8,$$ $$24c_6 + 6c_5 = 0,$$ $$6c_5 + 2c_4 = 0,$$

cuyas soluciones son

$$\begin{array}{l} c_6 = \frac{8}{12} = \frac{2}{3}, \\ c_5 = -\frac{24c_6}{6} = -4c_6 = -4 \cdot \frac{2}{3} = -\frac{8}{3}, \\ c_4 = \frac{-6c_5}{2} = -3c_5 = -3 \cdot \left(-\frac{8}{3}\right) = 8. \end{array}$$

Por lo tanto, $y_p = c_4x^2 + c_5x^3 + c_6x^4 = 8x^2 - \frac{8}{3}x^3 + \frac{2}{3}x^4$

y la solución general de $y''' + y'' = 8x^2$ es

$$y = y_h + y_p = c_1e^{-x} + c_2 + c_3x + 8x^2 - \frac{8}{3}x^3 + \frac{2}{3}x^4, \quad c_1, c_2, c_3 \in \mathbb{R}.$$

La respuesta al problema: $y = c_1e^{-x} + c_2 + c_3x + 8x^2 - \frac{8}{3}x^3 + \frac{2}{3}x^4.$