Solve-y-2y-2y-secax-

Question Number 104034 by mohammad17 last updated on 19/Jul/20

S o l v e : y^{^{″}} + 2 y^{^{'}} + 2 y = s e c a x

Answered by bramlex last updated on 19/Jul/20

H E : ℓ^{2} + 2 ℓ + 2 = 0

{(ℓ + 1)}^{2} + 1 = 0 \to ℓ = - 1 \pm i

y_{h} = e^{- x} (C_{1} \cos x + C_{2} \sin x)

Commented by mohammad17 last updated on 19/Jul/20

s i r i w a n t y p ?

Commented by bobhans last updated on 19/Jul/20

y_{1} = e^{- x} \cos x \to y_{1}^{'} = - e^{- x} \cos x - e^{- x} \sin x

\to y_{1}^{'} = - e^{- x} (\cos x + \sin x)

y_{2} = e^{- x} \sin x \to y_{2}^{'} = - e^{- x} \sin x + e^{- x} \cos x

\to y_{2}^{'} = - e^{- x} (\sin x - \cos x)

W (y_{1}, y_{2}) = | \begin{matrix} e^{- x} \cos x e^{- x} \sin x \\ - e^{- x} (\cos x + \sin x) - e^{- x} (\sin x - \cos x \end{matrix} |

= - e^{- 2 x} (\cos x \sin x - \cos^{2} x) + e^{- 2 x} (\sin x

\cos x + \sin^{2} x) = e^{- 2 x}

I_{1} = e^{2 x} \int e^{- x} \cos x . \sec x d x = - e^{x}

I_{2} = e^{2 x} \int e^{- x} \sin x . \sec x d x

I_{2} = \frac{1}{2} e^{2 x} \int e^{- x} \sin 2 x d x

y_{p} = - y_{1} I_{2} + y_{2} I_{1}

Answered by mathmax by abdo last updated on 19/Jul/20

y^{^{″}} + {2 y}^{^{'}} + 2 y = \frac{1}{\cos (x)}

h \to r^{2} + 2 r + 2 = 0 \to Δ^{^{'}} = - 1 \Rightarrow r_{1} = - 1 + i and r_{2} = - 1 - i \Rightarrow

y_{h} = {ae}^{(- 1 + i) x} + b e^{(- 1 - i) x} = e^{- x} {α cosx + β sinx} = α u_{1} + β u_{2}

W (u_{1}, u_{2}) = | \begin{matrix} e^{- x} cosx e^{- x} sinx \\ - (cosx + sinx) e^{- x} (cosx - sinx) e^{- x} \end{matrix} |

= e^{- 2 x} {\cos^{2} x - cosx sinx} + e^{- 2 x} {\sin^{2} x + sinx cosx} = e^{- 2 x} \neq 0

W_{1} = | \begin{matrix} o e^{- x} sinx \\ \frac{1}{\cos (ax)} (cosx - sinx) e^{- x} \end{matrix} | = - e^{- x} \frac{sinx}{\cos (x)}

W_{2} = | \begin{matrix} e^{- x} cosx 0 \\ - (cosx + sinx) e^{- x} \frac{1}{cosx} \end{matrix} | = e^{- x}

v_{1} = \int \frac{w_{1}}{w} dx = - \int \frac{e^{- x} sinx}{e^{- 2 x} cosx} dx = - \int e^{x} \frac{sinx}{cosx} dx

v_{2} = \int \frac{w_{2}}{w} dx = \int \frac{e^{- x}}{e^{- 2 x}} dx = \int e^{x} dx = e^{x} \Rightarrow

y_{p} = u_{1} v_{1} + u_{2} v_{2} = - e^{- x} cosx \int \frac{e^{x} sinx}{cosx} dx + sinx

the general solution is y = y_{h} + y_{p}