solve-y-y-sin-x-

Question Number 100344 by bemath last updated on 26/Jun/20

solve y^{″} + y = \sin x

Answered by bobhans last updated on 26/Jun/20

homogenous solution

η^{2} + 1 = 0 \Rightarrow η = \pm i

y_{h} = Asin x + Bcos x

particular solution

y_{p} = x (A \sin x + Bcos x)

y_{p}^{'} = Asin x + Bcos x + x (A \cos x - Bsin x)

y_{p}^{″} = 2 Acos x - 2 Bsin x + x (- A \sin x - Bcos x)

A = 0 \land B = \frac{1}{2}

y_{p} = \frac{1}{2} x \cos x

G enerall solution

y_{g} = A \sin x + Bcos x + \frac{1}{2} x \cos x

Answered by mathmax by abdo last updated on 26/Jun/20

y^{^{″}} + y = sinx (he) \to r^{2} + 1 = 0 \Rightarrow r = \bar{+} i \Rightarrow y_{h} = {ae}^{ix} + {be}^{- ix} = α cosx + β sinx

= α u_{1} + β u_{2}

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

W_{1} = | \begin{matrix} 0 sinx \\ sinx cosx \end{matrix} | = - \sin^{2} x

W_{2} = | \begin{matrix} cosx 0 \\ - sinx sinx \end{matrix} | = cosx sinx

v_{1} = \int \frac{w_{1}}{w} dx = - \int \frac{\sin^{2} x}{1} dx = - \int \frac{1 - \cos (2 x)}{2} = - \frac{x}{2} + \frac{1}{4} \sin (2 x)

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

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

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

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

= - \frac{x}{2} cosx + \frac{1}{4} sinx the general solution is

y = y_{h} + y_{p} = α cosx + β sinx - \frac{x}{2} cosx + \frac{1}{4} sinx