2xy-1-4x-y-2x-1-y-y-

Question Number 196846 by mokys last updated on 01/Sep/23

2 {x y}^{″} + (1 - 4 x) y^{'} + (2 x - 1) y = y

Answered by witcher3 last updated on 04/Sep/23

{2 xy}^{″} + (1 - 4 x) y^{'} + (2 x - 1) y = 0

y (x) = e^{x} . . solution

y = {ze}^{x}

\Rightarrow y^{'} = (z + z^{'}) e^{x}

y^{″} = (z^{″} + {2 z}^{'} + z) e^{x}

\Leftrightarrow 2 x (z^{″} + {2 z}^{'} + z) + (1 - 4 x) (z + z^{'}) + (2 x - 1) z = 0

{2 xz}^{″} + (4 x + 1 - 4 x) z^{'} = 0

{2 xz}^{″} + z^{'} = 0

\frac{z^{″}}{z^{'}} = - \frac{1}{2 x} \Rightarrow z^{'} = \ln (\frac{1}{\sqrt{∣ x ∣}}) + c

z^{'} = \frac{k}{\sqrt{∣ x ∣}} \Rightarrow z = 2 k \sqrt{∣ x ∣} + c =

y = {ce}^{x} + b \sqrt{∣ x ∣} e^{x}

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