Y`+y^2=1 solve the diff.eq.?

May 04 [Sat], 2013, 12:46

dy/dx + y? = 1dy/dx = 1 - y?dy/dx = (1 + y)(1 - y)Separate the variables.dy / ((1 + y)(1 - y)) = dx2dy / ((1 + y)(1 - y)) = 2dxExpress the left-hand side using partial fractions.( 1 / (1 + y) + 1 / (1 - y) ) dy = 2dxIntegrate.ln(1 + y) - ln(1 - y) = 2x + constantln((1 + y)/(1 - y)) = 2x + constant(1 + y) / (1 - y) = ke^2x1 + y = (1 - y)ke^2x1 + y = ke^2x - kye^2xkye^2x + y = ke^2x - 1y(ke^2x + 1) = ke^2x - 1y = (ke^2x - 1) / (ke^2x + 1)

dy/dx =(1-y^2)dy/(y^2-1) +dx =01/2{log(y+1)/y-1}+ x =Clog{(y+1)/(y-1)} +2x =Clog{(y+1)/(y-1)} =(C-2x)(y+1)/(y-1) = e^(C-2x)Apply componando and dividendoy/2 ={e^(c-2x) +1}/{e^(c-2x)-1}Therefore y = 2*{e^(c-2x) +1}/{e^(c-2x)-1} .................Ans


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