画像 (1-x^2)y'' 2xy'=0 y1=1 175216-(1-x^2)y''-2xy'+12y=0
Simplify x=y1 x=y1 Add y to both sides of the equation y^ {2}\left (2x\right)yx^ {2}1=0 All equations of the form ax^ {2}bxc=0 can be solved using the quadratic formula \frac {b±\sqrt {b^ {2}4ac}} {2a} The quadratic formula gives two solutions,Solution for ( 1 x2 ) y'' 2xy' 2y = 0 ;Find the specific solution for the following differential equations a) (dy/dx)^2 x^4 = 0 given the initial value y(1) = 0 b) dy/dx = ysinx given the initial value y(pi) = 1 calculus Sorry to post this again, but I am still unable to understand it and need help Please help1) Using 3(x3)(x^26x23)^2 as the answer to differentiating f(x The Solution Of Differential Equation X 2 Y 2 Dy Xy Dx Is Y Y X If Y 1 1 And Y X0 E Then X0 Is (1-x^2)y''-2xy'+12y=0