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(1-x^2)y''-2xy'+2y=0 y1=x

(1-x^2)y''-2xy'+2y=0 y1=x- X ^ (2) y '' 7xy ' 16y = 0, y1 = x ^ 4 1 See answer what does y1=x^4 mean I"m assuming taht y'' is 2nd derivitive of y and y' is first derivitive I need more details about this to solve it I might assume that you are trying to solve the differential equaiton but When I set , (I think this is how you solve second order with variable coefficients) I got to the point Then I let and got , separated variables and integrated to get and I am not sure if this is correct, but it seems reasonable Also, , and this didn't seem to get me anywhere but, by factoring out , I was able to make some more inferences I

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Steps Using the Quadratic Formula Steps for Completing the Square View solution steps Steps Using the Quadratic Formula (y1) { x }^ { 2 } 2xyy1=0 ( y − 1) x 2 − 2 x y y 1 = 0 All equations of the form ax^ {2}bxc=0 can be solved using the quadratic formula \frac {Answer (1 of 5) The equation \displaystyle{ (1x^2)y'' 2xy' 2y = 0 }\qquad(1) Since we have no obvious way to find any particular solution of (1) so we should try to find its general solution in the form of a power series as follows \displaystyle{ y = C_0 C_1x C_2x^2 \dots C_nx^2 find a series solution about the point x=0 of (1x^2)y"2xy'2y=0

0 y 2 Solution We look for the critical points in the interior\quad 1 \lt x \lt 1 , find a second solution By signing up, you'll getY 1(x)=x−1 Si y2(x)=u(x)y1(x) ⇒ y2 = ux−1 ⇒ y 0 2 = u x −1 − ux−2 ⇒ y 00 2

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Ejercicios EDO's de primer orden 3 1 y3 dy = dx x2 Z y−3 dy = Z x−2 dx, 1 −2 y−2 = −x−1 c 1, −1 2y2 −1 x c 1, 1 y2 2 x c, c = −2c 1 Solución implícita 1 y2 2xc x Solución explícita y = ±Sección 43 Ecuaciones lineales homogéneas con coeficientes constontes 133 16 (1 x')y" 2xy' = 0;

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