Solve-2x-x-2-y-2-1-x-y-2y-0-Given-y-x-2-is-a-solution-

Question Number 71570 by TawaTawa last updated on 17/Oct/19

Solve : (2 x + x^{2}) y^{″} - 2 (1 + x) y^{'} + 2 y = 0

Given y = x^{2} is a solution .

Answered by mind is power last updated on 17/Oct/19

y = {zx}^{2}

y^{'} = 2 xz + z^{'} x^{2}

y^{″} = 2 z + {4 xz}^{'} + z^{″} x^{2}

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

+ {2 x}^{2} z = 0

\Rightarrow ({2 x}^{2} + {2 x}^{2} + 4 x - 4 x - {4 x}^{2}) z + (2 x + x^{2}) x^{2} z^{″} + ({8 x}^{2} + {4 x}^{3} - {2 x}^{2} - {2 x}^{3}) z^{'} = 0

({2 x}^{3} + x^{4}) z^{″} + ({6 x}^{2} + {2 x}^{3}) z^{'} = 0

\Leftrightarrow (x^{2} + 2 x) z^{″} + 2 (3 + x) z^{'} = 0 . . \forall x \neq 0

Y = z^{'}

(x^{2} + 2 x) Y^{'} + 2 (3 + x) Y = 0

\Rightarrow \frac{Y^{'}}{Y} = \frac{6 + 2 x}{x (x + 2)} = \frac{3}{x} - \frac{1}{x + 2}

\Rightarrow \ln ∣ Y ∣= 3 \ln ∣ x ∣ - \ln ∣ x + 2 ∣

\ln ∣ Y ∣= \ln ∣ \frac{x^{3}}{x + 2} ∣ + c

\Rightarrow∣ Y ∣= k . ∣ \frac{x^{3}}{x + 2} ∣

Y = c . \frac{x^{3}}{x + 2} \dots . c = \underline{+} k

z^{'} = c \frac{x^{3}}{x + 2} = c \frac{(x + 2) (x^{2} - 2 x + 4) - 8}{x + 2}

z^{'} = c (x^{2} - 2 x + 4) - \frac{8 c}{x + 2}

\Rightarrow z = c (\frac{x^{3}}{3} - x^{2} + 4 x) - 8 cln ∣ x + 2 ∣ + k

y = {zx}^{2} = c (\frac{x^{5}}{3} - x^{4} + {4 x}^{3}) + {kx}^{2} - {8 cx}^{2} \ln ∣ x + 2 ∣

c, k constante

solution is iner space of TWo Dimensiln S = vect (\frac{x^{5}}{3} - x^{4} + {4 x}^{3} - 8 \ln ∣ x + 2 ∣, x^{2})

Commented by TawaTawa last updated on 17/Oct/19

Wow, God bless you sir

Commented by mind is power last updated on 17/Oct/19

y^{'} re welcom