[10000印刷√] (1-x^2)y''-2xy' 2y=0 199585-(1-x^2)y''-2xy'+2y=0

 y = 3/2e^(x^2 1) 1/2 We have dy/dx 2xy = x Which we can write as dy/dx = 2xy x dy/dx = (2y 1)x 1/(2y 1) dy/dx = x Which is a first order separable differential equation, so we can "separate the variables" to get int \\ 1/(2y 1) \\ dy = int \\ x \\ dx Integrating we get, the General Solution 1/2ln2y1 = 1/2x^2 C Applying the initial condition y(1)=1 we find 1/2ln3 =

(1-x^2)y''-2xy'+2y=0-The equation is exact and is the total differential dF (x,y) =0 with F_x = P = 1 y^2/x^2 F_y = Q = 2y/x Integrating this equation gives F (x,y) = y^2/x G (x) The function G (x) is obtained from the first equation which gives G' (x) = 1 ,then G (x) = x C and F (x,y) = y^2/x x = C , or y^22 Montrer que lescourbes intégrales onttoutesune asymptote quandxtendvers∞ 3

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2xy9x^2 (2yx^21) (dy)/ (dx)=0, y (0)=3 \square! Songhttps//youtube/kkhWDeevA(1x^2)y" 2xy' 2y =0 Power Series Solution of Differential Equation,Power series solution,(1x^2)y" 2xy' 2y =0 power ser

Incoming Term: (1-x^2)y''-2xy'+2y=0, (1-x^2)y''-2xy'+2y=0 power series solution, (1-x^2)y''-2xy'+2y=0 y1=x, (1-x 2)y''-2xy'+2y=0, (1-x^2)y''-xy'+a^2y=0, x^2-2y^2-xy+2x-y+1=0, (1+x^2+y^2+x^2y^2)^1/2+xy dy/dx=0,

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