Задача №1852

\[ \begin{aligned} & y'=y^3-x;\;y'(0)=1^3-0=1.\\ & y''=3y^2y'-1;\;y''(0)=3\cdot{1^2}\cdot{1}-1=2.\\ & y'''=6y(y')^2+3y^2y'';\;y'''(0)=6\cdot{1}\cdot{1^2}+3\cdot{1^2}\cdot{2}=12.\\ & y^{(4)}=6(y')^3+6y\cdot{2y'}\cdot{y''}+6y\cdot{y'}\cdot{y''}+3y^2\cdot{y'''}=6(y')^3+18yy'y''+3y^2y''';\;y^{(4)}(0)=6+36+36=78. \end{aligned} \]

\[ y(x)=1+\frac{1\cdot{x^1}}{1!}+\frac{2\cdot{x^2}}{2!}+\frac{12\cdot{x^3}}{3!}+\frac{78\cdot{x^4}}{4!} =1+x+x^2+2x^3+\frac{13x^4}{4}+... \]