Question-94997

Question Number 94997 by i jagooll last updated on 22/May/20

Answered by bobhans last updated on 22/May/20

homogenous solution

υ^{2} - 2 υ + 2 = 0

υ = 1 \pm i \Rightarrow y_{h} = {Ae}^{x} \cos x + {Be}^{x} \sin x

particular integral

y_{p} = ϑ_{1} e^{x} \sin x + ϑ_{2} e^{x} \cos x

y^{'} = ϑ_{1} e^{x} \sin x + ϑ_{1} e^{x} \cos x + ϑ_{2} e^{x} \cos x - ϑ_{2} e^{x} \sin x +

(ϑ_{1}^{'} e^{x} \sin x + ϑ_{2}^{'} e^{x} \cos x)

set ϑ_{1}^{'} e^{x} \sin x + ϑ_{2}^{'} e^{x} \cos x = 0

y^{″} = {2 e}^{x} (ϑ_{1} \cos x - ϑ_{2} \sin x) + (ϑ_{1}^{'} e^{x} \sin x + ϑ_{1}^{'} e^{x} \cos x +

ϑ_{2}^{'} e^{x} \cos x - ϑ_{2}^{'} e^{x} \sin x)

set ϑ_{1}^{'} e^{x} \sin x + ϑ_{1}^{'} e^{x} \cos x + ϑ_{2}^{'} e^{x} \cos x -

ϑ_{2}^{'} e^{x} \sin x = e^{x} (1 + \sin x)

\Rightarrow ϑ_{1} = \int (\cos x + \frac{1}{2} \sin 2 x) dx = \sin x - \frac{1}{4} \cos 2 x

\Rightarrow ϑ_{2} = \int (- \sin x + \frac{1 - \cos 2 x}{2}) dx

ϑ_{2} = \cos x + \frac{1}{2} x - \frac{1}{4} \sin 2 x

complete solution y_{h} + y_{p}

Answered by mathmax by abdo last updated on 23/May/20

equation at form y^{(2)} + {py}^{^{'}} qy = f (x)

(he) \to y^{^{″}} - {2 y}^{^{'}} + 2 y = 0 \to r^{2} - 2 r + 2 = 0

Δ^{^{'}} = 1 - 2 = - 1 \Rightarrow r_{1} = 1 + i and r_{2} = 1 - i \Rightarrow

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

let u_{1} = e^{x} cosx and u_{2} = e^{x} sinx

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

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

= e^{2 x} and W_{1} = | \begin{matrix} o e^{x} sinx \\ e^{x} (1 + sinx) e^{x} sinx + e^{x} cosx \end{matrix} |

= - e^{2 x} sinx (1 + sinx)

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

v_{1}^{^{'}} = \frac{W_{1}}{W} and v_{2}^{^{'}} = \frac{W_{2}}{W} \Rightarrow v_{1}^{^{'}} = \frac{- e^{2 x} sinx (1 + sinx)}{e^{2 x}}

= - sinx (1 + sinx) \Rightarrow v_{1} = - \int sinx dx - \int \sin^{2} x dx

= cosx - \frac{1}{2} \int (1 - \cos (2 x)) dx = cosx - x + \frac{1}{4} \sin (2 x)

v_{2}^{^{'}} = \frac{e^{2 x} cosx (1 + sinx)}{e^{2 x}} = cosx + \frac{1}{2} \sin (2 x) \Rightarrow

v_{2} = \int (cosx + \frac{1}{2} \sin (2 x)) dx = sinx - \frac{1}{4} \cos (2 x) \Rightarrow

y_{p} = u_{1} v_{1} + u_{2} v_{2} = e^{x} cosx (cosx - x + \frac{1}{4} \sin (2 x))

+ e^{x} sinx (sinx - \frac{1}{4} \cos (2 x)) = e^{x} \cos^{2} x - {xe}^{x} cosx + \frac{1}{4} e^{x} cosxsin (2 x)

+ e^{x} \sin^{2} x - \frac{1}{4} e^{x} sinx \cos (2 x) = e^{x} - {xe}^{x} cosx

+ \frac{1}{4} e^{x} (cosx \sin (2 x) - sinx \cos (2 x)) \Rightarrow

y_{g} = y_{h} + y_{p}