Question

(1+x2)y’-xy=2x решите уравнение

Answer 1

$(1+x^2)y'-xy=2x\y=uv;y'=u'v+v'u\(1+x^2)u'v+(1+x^2)v'u-xuv=2x\(1+x^2)u'v+u((1+x^2)v'-xv)=2x\ begin{cases}(1+x^2)v'-xv=0\(1+x^2)u'v=2xend{cases} \frac{(1+x^2)dv}{dx}=xv\frac{dv}{v}=frac{xdx}{1+x^2}\intfrac{dv}{v}=frac{1}{2}intfrac{d(1+x^2)}{1+x^2}\ln|v|=frac{1}{2}ln|1+x^2|\v=sqrt{1+x^2}\frac{sqrt{(1+x^2)^3}du}{dx}=2x\du=frac{2xdx}{sqrt{(1+x^2)^3}}\int du=frac{d(1+x^2)}{sqrt{(1+x^2)^3}}\u=-frac{2}{sqrt{1+x^2}}+C\y=Csqrt{1+x^2}-2$