Question #155051

y(x2+1)=cx


Expert's answer

y=cxx2+1y=cx2+c2cx2(x2+1)2y=c(1x2)(x2+1)2yy=c(1x2)(x2+1)2×x2+1cxy=y(1x2)x2+1y=\frac{cx}{x^2+1}\\ y'=\frac{cx^2+c-2cx^2}{(x^2+1)^2}\\ y'=\frac{c(1-x^2)}{(x^2+1)^2}\\ \frac{y'}{y}=\frac{c(1-x^2)}{(x^2+1)^2}\times \frac{x^2+1}{cx}\\ y'=\frac{y(1-x^2)}{x^2+1}


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