Solve-differential-equation-the-method-of-variation-of-parameters-d-2-y-dx-2-4y-4-cosec-2x-

Question Number 112740 by Mr.D.N. last updated on 09/Sep/20

Solve differential equation the method of

variation of parameters :

\frac{d^{2} y}{{dx}^{2}} + 4 y = 4 cosec 2 x

Answered by mathmax by abdo last updated on 09/Sep/20

y^{^{″}} + 4 y = \frac{4}{\sin (2 x)}

h \to r^{2} + 4 = 0 \Rightarrow r = \bar{+} 2 i \Rightarrow y_{h} = {ae}^{2 ix} + {be}^{- 2 ix} = α \cos (2 x) + β \sin (2 x)

= α u_{1} + β u_{2}

W (u_{1}, u_{2}) = | \begin{matrix} \cos (2 x) \sin (2 x) \\ - 2 \sin 2 x 2 \cos (2 x) \end{matrix} | = 2 (\cos^{2} 2 x + \sin^{2} (2 x)) = 2 \neq 0

W_{1} = | \begin{matrix} 0 \sin (2 x) \\ \frac{4}{\sin (2 x)} 2 \cos (2 x) \end{matrix} | = - 4

W_{2} = | \begin{matrix} \cos (2 x) 0 \\ - 2 \sin (2 x) \frac{4}{\sin (2 x)} \end{matrix} | = 4 cotan (2 x)

V_{1} = \int \frac{W_{1}}{W} dx = \int \frac{- 4}{2} dx = - 2 x

V_{2} = \int \frac{W_{2}}{W} dx = \int \frac{4 cotan (2 x)}{2} dx = 2 \int \frac{\cos (2 x)}{\sin (2 x)} dx = \ln (∣ \sin (2 x) ∣)

\Rightarrow y_{p} = u_{1} v_{1} + u_{2} v_{2} = \cos (2 x) (- 2 x) + \sin (2 x) \ln ∣ \sin (2 x) ∣

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

y = α \cos (2 x) + β \sin (2 x) - 2 x \cos (2 x) + \sin (2 x) \ln ∣ \sin (2 x) ∣