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Question Number 101419 by bemath last updated on 02/Jul/20

y^{″} - {4 y}^{'} + 3 y = \frac{e^{x}}{1 + e^{x}}

Answered by john santu last updated on 02/Jul/20

AE : λ^{2} - 4 λ + 3 = 0

(λ - 1) (λ - 3) = 0

y_{h} = C_{1} e^{x} + C_{2} e^{3 x}

\to y_{1} = e^{x} \land y_{2} = e^{3 x}

W = | \begin{matrix} y_{1} y_{2} \\ y_{1}^{^{'}} y_{2}^{'} \end{matrix} | = | \begin{matrix} e^{x} e^{3 x} \\ e^{x} 3 e^{3 x} \end{matrix} |

W = {3 e}^{4 x} - e^{4 x} = {2 e}^{3 x}

particular integral

= - \frac{e^{x}}{2} \int \frac{e^{x}}{e^{x} (1 + e^{x})} dx + \frac{e^{3 x}}{2} \int \frac{e^{x}}{e^{4 x}} . \frac{e^{x}}{(1 + e^{x})} dx

= - \frac{e^{x}}{2} \int \frac{dx}{1 + e^{x}} + \frac{e^{3 x}}{2} \int \frac{dx}{e^{2 x} (1 + e^{x})}

I_{1} = \int \frac{d x}{1 + e^{x}} [s e t e^{x} = z, d x = \frac{d z}{e^{x}}]

I_{1} = \int \frac{d z}{z (1 + z)} = \int \frac{d z}{z} - \int \frac{d z}{1 + z}

I_{1} = \ln (\frac{z}{1 + z}) = \ln (\frac{e^{x}}{1 + e^{x}})

I_{2} = \int \frac{d x}{e^{2 x} (1 + e^{x})} [s e t e^{x} = p]

I_{2} = \int \frac{d p}{p^{3} (1 + p)} = \int \frac{d p}{p} - \int \frac{d p}{p^{2}} + \int \frac{d p}{p^{3}} - \int \frac{d p}{(1 + p)}

I_{2} = \ln (p) + \frac{1}{p} - \frac{1}{2 p^{2}} - \ln (1 + p)

I_{2} = \ln (\frac{e^{x}}{1 + e^{x}}) + \frac{1}{e^{x}} - \frac{1}{2 e^{2 x}}

particular solution

y_{p} = - \frac{e^{x}}{2} \ln (\frac{e^{x}}{1 + e^{x}}) + \frac{e^{3 x}}{2} (\frac{1}{e^{x}} - \frac{1}{2 e^{2 x}} + \ln (\frac{e^{x}}{1 + e^{x}}))

G enerall solution

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

Answered by mathmax by abdo last updated on 02/Jul/20

(he) \to y^{^{″}} - {4 y}^{^{'}} + 3 y = 0 \to r^{2} - 4 r + 3 = 0

Δ^{^{'}} = 4 - 3 = 1 \Rightarrow r_{1} = 2 + 1 = 3 and r_{2} = 2 - 1 = 1

\Rightarrow y_{h} = {ae}^{3 x} + {be}^{x} = {au}_{1} + {bu}_{2}

W (u_{1}, u_{2}) = | \begin{matrix} e^{3 x} e^{x} \\ {3 e}^{3 x} e^{x} \end{matrix} | = e^{4 x} - {3 e}^{4 x} = - {2 e}^{4 x}

W_{1} = | \begin{matrix} o e^{x} \\ \frac{e^{x}}{1 + e^{x}} e^{x} \end{matrix} | = - \frac{e^{2 x}}{1 + e^{x}}

W_{2} = | \begin{matrix} e^{3 x} 0 \\ {3 e}^{3 x} \frac{e^{x}}{1 + e^{x}} \end{matrix} | = \frac{e^{4 x}}{1 + e^{x}}

v_{1} = \int \frac{w_{1}}{w} dx = \int - \frac{e^{2 x}}{1 + e^{x}} \times \frac{1}{- {2 e}^{4 x}} = \int \frac{e^{- 2 x}}{2 (1 + e^{x})} =_{e^{x} = t} \frac{1}{2} \int \frac{1}{t^{2} (1 + t)} \times \frac{dt}{t}

= \frac{1}{2} \int \frac{dt}{t^{3} (1 + t)} let decompose F (t) = \frac{1}{(t + 1) t^{3}}

F (t) = \frac{a}{t + 1} + \frac{b}{t} + \frac{c}{t^{2}} + \frac{d}{t^{3}}

a = - 1, d = 1 \Rightarrow F (t) = - \frac{1}{t + 1} + \frac{b}{t} + \frac{c}{t^{2}} + \frac{1}{t^{3}}

\lim_{t \to + \infty} tF (t) = 0 = - 1 + b \Rightarrow b = 1 \Rightarrow F (t) = - \frac{1}{t + 1} + \frac{1}{t} + \frac{c}{t^{2}} + \frac{1}{t^{3}}

F (1) = \frac{1}{2} = - \frac{1}{2} + 1 + c + 1 = \frac{3}{2} + c \Rightarrow c = \frac{1}{2} - \frac{3}{2} = - 1 \Rightarrow

F (t) = - \frac{1}{t + 1} + \frac{1}{t} - \frac{1}{t^{2}} + \frac{1}{t^{3}} \Rightarrow v_{1} = \frac{1}{2} (- \ln ∣ t + 1 ∣ + \ln ∣ t ∣ + \frac{1}{t} - \frac{1}{{2 t}^{2}})

= \frac{1}{2} (\ln (\frac{e^{x}}{1 + e^{x}}) + e^{- x} - \frac{1}{2} e^{- 2 x})

v_{2} = \int \frac{w_{2}}{w} dx = \int \frac{e^{4 x}}{(1 + e^{x}) (- {2 e}^{4 x})} dx = - \frac{1}{2} \int \frac{dx}{1 + e^{x}} =_{e^{x} = t} - \frac{1}{2} \int \frac{dt}{t (1 + t)}

= - \frac{1}{2} \int (\frac{1}{t} - \frac{1}{t + 1}) dt = - \frac{1}{2} \ln ∣ \frac{t}{t + 1} ∣ = - \frac{1}{2} \ln (\frac{e^{x}}{1 + e^{x}})

\Rightarrow y_{p} = u_{1} v_{1} + u_{2} v_{2} = e^{3 x} \times \frac{1}{2} {\ln (\frac{e^{x}}{1 + e^{x}}) + e^{- x} - \frac{1}{2} e^{- 2 x}}

+ e^{x} \times (- \frac{1}{2} \ln (\frac{e^{x}}{1 + e^{x}})) = \frac{1}{2} e^{3 x} \ln (\frac{e^{x}}{1 + e^{x}}) + \frac{1}{2} e^{2 x} - \frac{1}{4} e^{x}

- \frac{1}{2} e^{x} \ln (\frac{e^{x}}{1 + e^{x}}) = \frac{1}{2} (e^{3 x} - e^{x}) \ln (\frac{e^{x}}{1 + e^{x}}) + \frac{e^{2 x}}{2} - \frac{e^{x}}{4}

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