y-y-1-cosx-

Question Number 107108 by Ar Brandon last updated on 08/Aug/20

y^{″} + y = \frac{1}{cosx}

Commented by mohammad17 last updated on 08/Aug/20

(m^{2} - i^{2}) = 0 \Rightarrow m = \mp i

Y c = c_{1} c o s x + c_{2} s i n x

l e t : Y p = u_{1} v_{1} + u_{2} v_{2}

u_{1} = c o s x, u_{2} = s i n x \Rightarrow u_{1}^{'} = - s i n x, u_{2}^{'} = c o s x

D = u_{1} u_{2}^{'} - u_{2} u_{1}^{'} = {c o s}^{2} x + {s i n}^{2} x = 1

v_{1} = \int \frac{- s i n x}{c o s x} d x = l n ∣ c o s x ∣

v_{2} = \int \frac{c o s x}{c o s x} d x = x

∴ Y p = c o s x l n ∣ c o s x ∣ + x s i n x

Y = Y c + Y p = c_{1} c o s x + c_{2} s i n x + c o s x l n ∣ c o s x ∣ + x s i n x

M . S . M o h a m m a d T a h a

Commented by Ar Brandon last updated on 08/Aug/20

Thanks��

Commented by mohammad17 last updated on 08/Aug/20

w e l c o m e

Answered by mathmax by abdo last updated on 08/Aug/20

y^{^{″}} + y = \frac{1}{cosx}

h \to y^{^{″}} + y = 0 \Rightarrow r^{2} + 1 = 0 \Rightarrow r = \bar{+} 1 \Rightarrow y_{h} = acosx + bsinx = {au}_{1} + {bu}_{2}

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

W_{1} = | \begin{matrix} 0 sinx \\ \frac{1}{cosx} cosx \end{matrix} | = - \frac{sinx}{cosx}

W_{2} = | \begin{matrix} cosx 0 \\ - sinx \frac{1}{cosx} \end{matrix} | = 1

v_{1} = \int \frac{w_{1}}{w} dx = - \int \frac{sinx}{cosx} dx = \ln ∣ cosx ∣

v_{2} = \int \frac{w_{2}}{w} dx = \int dx = x \Rightarrow y_{p} = u_{1} v_{1} + u_{2} v_{2} = cosxln ∣ cosx ∣ + xsinx

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

\Rightarrow y = cosxln ∣ cosx ∣ + xsinx + acosx + bsinx