y(x2+1)=cx
y=cxx2+1y′=cx2+c−2cx2(x2+1)2y′=c(1−x2)(x2+1)2y′y=c(1−x2)(x2+1)2×x2+1cxy′=y(1−x2)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}y=x2+1cxy′=(x2+1)2cx2+c−2cx2y′=(x2+1)2c(1−x2)yy′=(x2+1)2c(1−x2)×cxx2+1y′=x2+1y(1−x2)
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
Comments