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Question Number 120050 by bramlexs22 last updated on 29/Oct/20

(i) y^{″} - 4 y^{'} + 5 y = 4 \sin^{2} 4 x

(i i) \frac{x}{2} + 1 = \sqrt{∣ 1 - x^{2} ∣}

Answered by bemath last updated on 29/Oct/20

(∙) \frac{x + 2}{2} = \sqrt{∣ 1 - x^{2} ∣} d e f i n e d o n x ⩾ - 2

\Leftrightarrow s q u a r i n g b o t h s i d e s

\frac{{(x + 2)}^{2}}{4} = ∣ 1 - x^{2} ∣

c a s e (1) f o r - 1 ⩽ x ⩽ 1 \Rightarrow \frac{x^{2} + 4 x + 4}{4} = 1 - x^{2}

\Rightarrow x^{2} + 4 x + 4 = 4 - 4 x^{2}

\Rightarrow 5 x^{2} + 4 x = 0; x (5 x + 4) = 0 {\begin{cases} x_{1} = 0 \\ x_{2} = - \frac{4}{5} \end{cases}

c a s e (2) f o r x ⩽ - 1 \cup x ⩾ 1 \Rightarrow \frac{x^{2} + 4 x + 4}{4} = x^{2} - 1

\Rightarrow x^{2} + 4 x + 4 = 4 x^{2} - 4; 3 x^{2} - 4 x - 8 = 0

x = \frac{4 \pm \sqrt{16 - 4.3 . (- 8)}}{6} = \frac{4 \pm \sqrt{112}}{6}

x_{3.4} = \frac{4 \pm 4 \sqrt{7}}{6} = \frac{2}{3} \pm \frac{2 \sqrt{7}}{3}

T h e s o l u t i o n s e t i s {- \frac{4}{5}, \frac{2 - 2 \sqrt{7}}{3}, 0, \frac{2 + 2 \sqrt{7}}{3}}

Answered by Lordose last updated on 29/Oct/20

i

\frac{d^{2} (y)}{dx} - 4 \frac{dy}{dx} + 5 y = {4 \sin}^{2} 4 x

y (x) = y_{c} (x) + y_{p} (x)

y_{c} (x) = \frac{d^{2} (y)}{dx} - 4 \frac{dy}{dx} + 5 y = 0

set y = e^{kx}

k^{2} e^{kx} - {4 ke}^{kx} + {5 e}^{kx} = 0

e^{kx} (k^{2} - 4 k + 5) = 0

e^{kx} \neq 0 \forall k \in R

k^{2} - 4 k + 5 = 0

k = 2 \pm i where i = \sqrt{(- 1)}

y_{1} (x) = c_{1} e^{(2 + i) x} and y_{2} (x) = c_{2} e^{(2 + i) x}

y_{c} (x) = y_{1} (x) + y_{2} (x)

Recall : e^{x + iy} = e^{x} cosy + {ie}^{x} siny

y_{c} (x) = c_{1} (e^{2 x} \cos (x) + {ie}^{2 x} \sin (x)) + c_{2} (e^{2 x} \cos (x) - {ie}^{2 x} \sin (x))

y_{c} (x) = (c_{1} + c_{2}) e^{2 x} \cos (x) + i (c_{1} - c_{2}) e^{2 x} \sin (x)

let c_{1} + c_{2} = k_{1} and i (c_{1} - c_{2}) = k_{2}

y_{c} (x) = k_{1} e^{2 x} \cos (x) + k_{2} e^{2 x} \sin (x)

Next, We find y_{p} (x) by method of undetermined coefficient

y_{p} (x) = \frac{d^{2} y_{p} (x)}{{dx}^{2}} - 4 \frac{{dy}_{p} (x)}{dx} + {5 y}_{p} (x) = 2 - 2 \cos (8 x) {{4 \sin}^{2} (4 x) = 2 - \cos (8 x)}

The particular solution of

\frac{d^{2} y}{{dx}^{2}} - 4 \frac{dy}{dx} + 5 y = 2 is in the form y_{p 1} (x) = a_{1}

And the particular solution of \frac{d^{2} y}{dx} - 4 \frac{dy}{dx} + 5 y = - 2 \cos (8 x) is in the form

y_{p 2} (x) = a_{2} \cos (8 x) + a_{3} \sin (8 x)

y_{p} (x) = y_{p 1} (x) + y_{p 2} (x)

y_{p} (x) = a_{1} + a_{2} \cos (8 x) + a_{3} \sin (8 x)

\frac{{dy}_{p} (x)}{dx} = - {8 a}_{2} \sin (8 x) + {8 a}_{3} \cos (8 x)

\frac{d^{2} y_{p} (x)}{dx} = - {64 a}_{2} \cos (8 x) - {64 a}_{3} \sin (8 x)

Sub for y_{p} (x)

- {64 a}_{2} \cos (8 x) - {64 a}_{3} \sin (8 x) - 4 ({8 a}_{3} \cos (8 x) - {8 a}_{2} \sin (8 x)) + 5 (a_{1} + a_{2} \cos (8 x) + a_{3} \sin (8 x))

factorising, we have :

{5 a}_{1} + (- {59 a}_{2} - {32 a}_{3}) \cos (8 x) + ({32 a}_{2} - {59 a}_{3}) \sin (8 x) = 2 - 2 \cos (8 x)

{5 a}_{1} = 2

{59 a}_{2} + {32 a}_{3} = 2

{32 a}_{2} - {59 a}_{3} = 0

Hence, a_{1} = \frac{2}{5}, a_{2} = \frac{118}{4505}, a_{3} = \frac{64}{4505}

y_{p} (x) = a_{1} + a_{2} \cos (8 x) + a_{3} \sin (8 x)

y_{p} (x) = \frac{2}{5} + \frac{118 \cos (8 x)}{4505} + \frac{64 \sin (8 x)}{4505}

y (x) = y_{c} (x) + y_{p} (x)

y (x) = k_{1} e^{2 x} cosx + k_{2} e^{2 x} sinx + \frac{118 \cos (8 x)}{4505} + \frac{64 \sin (8 x)}{4505} + \frac{2}{5}

Answered by Bird last updated on 29/Oct/20

y^{^{″}} - 4 y^{^{'}} + 5 y = 4 \times \frac{1 - c o s (8 x)}{2} \Rightarrow

y^{^{″}} - 4 y^{^{'}} + 5 y = 2 - 2 c o s (8 x)

h \to r^{2} - 4 r + 5 = 0 \to Δ^{^{'}} = 4 - 5 = - 1

\Rightarrow r_{1} = 2 + i a n d r_{2} = 2 - i

y_{h} = {a e}^{(2 + i) x} + {b e}^{(2 - i) x}

= e^{2 x} {α c o s x + β s i n x}

= α e^{2 x} c o s x + β e^{2 x} s i n x = α u_{1} + β u_{2}

W (u_{1}, u_{2}) = | \begin{matrix} e^{2 x} c o s x e^{2 x} s i n x \\ e^{2 x} (2 c o s x - s i n x) e^{2 x} (2 s i n x + c o s x) \end{matrix} |

= e^{4 x} (2 s i n x + c o s x) c o s x - e^{4 x} (2 c o s x - s i n x) s i n x

= e^{4 x} {2 s i n x c o s x + {c o s}^{2} x - 2 c o s x s i n x + {s i n}^{2} x}

= e^{4 x} \neq 0

W_{1} = | \begin{matrix} o e^{2 x} s i n x \\ 2 - 2 c o s (8 x) e^{2 x} (2 s i n x + c o s x) \end{matrix} |

= - 2 e^{2 x} (1 - c o s (8 x) s i n x

W_{2} = | \begin{matrix} e^{2 x} c o s x 0 \\ e^{2 x} (2 c o s x - s i n x) 2 - 2 c o s (8 x) \end{matrix} |

= 2 e^{2 x} (1 - c o s (8 x)) c o s x

V_{1} = \int \frac{W_{1}}{W} d x = - \int \frac{2 e^{2 x} (1 - c o s) 8 x) s i n x}{e^{4 x}} d x

= - 2 \int e^{- 2 x} (1 - c o s (8 x)) s i n x d x

= - 2 \int e^{- 2 x} s i n x d x + 2 \int e^{- 2 x} c o s (8 x) s i n x d x

\int e^{- 2 x} s i n x d x = I m (\int e^{- 2 x + i x} d x)

\int e^{(- 2 + i) x} d x = \frac{1}{- 2 + i} e^{(- 2 + i) x}

= \frac{- 1 (2 + i) e^{- 2 x} (c o s x + i s i n x)}{5}

= - \frac{1}{5} e^{- 2 x} (2 c o s x + 2 i s i n x + i c o s x

- s i n x) \Rightarrow

\int e^{- 2 x} s i n x d x = - \frac{1}{5} e^{- 2 x} {2 s i n x + c o s x}

w e h a v e c o s (8 x) s i n x

= c o s (8 x) c o s (\frac{π}{2} - x)

= \frac{1}{2} {c o s (7 x + \frac{π}{2}) + c o s (9 x - \frac{π}{2}))

= \frac{1}{2} {- s i n (7 x) + s i n (9 x)} \Rightarrow

2 \int e^{- 2 x} c o s (8 x) s i n x d x

= \int e^{- 2 x} s i n (9 x) d x - \int e^{- 2 x} s i n (7 x) d x

= \dots . (e s z y t o f i n d)

V_{2} = \int \frac{W_{2}}{W} d x = \int \frac{2 e^{2 x} (1 - c o s (8 x) c o s x}{e^{4 x}} d x

= \int 2 e^{- 2 x} c o s x d x - 2 \int e^{2 x} c o s (8 x) c o s x d x

= \dots . (u s e s a m e w a y \dots) \Rightarrow

y_{p} = u_{1} v_{1} + u_{2} v_{2} a n d g e n e r a l s o l u t i o n

i s y = y_{p} + y_{h}

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