## Differential Equations and Linear Algebra (4th Edition)

$y=\frac{1}{4}x^4+2x^2-x+1$
Integrate to turn $y'''$ into $y''$. $$y'''=6x$$ $$y''=3x^2+C_1$$ Keep integrating until the left side of the equation becomes $y$. $$y'=x^3+C_1x+C_2$$ $$y=\frac{1}{4}x^4+\frac{1}{2}C_1x^2+C_2x+C_3$$ Find the values at $x=0$. $$y(0)=\frac{1}{4}(0)+\frac{1}{2}C_1(0)+C_2(0)+C_3=C_3$$ $$y'(0)=0^3+C_1(0)+C_2=C_2$$ $$y''(0)=3(0)^2+C_1=C_1$$ Since $y(0)=1$ and $y(0)=C_3$, $C_3=1$. Since $y'(0)=-1$ and $y'(0)=C_2$, $C_2=-1$. Since $y''(0)=C_1$ and $y''(0)=4$, $C_1=4$. Substituting these values yields $$y=\frac{1}{4}x^4+\frac{1}{2}(4)x^2-x+1$$ $$y=\frac{1}{4}x^4+2x^2-x+1$$