Answer
$$y = {x^4} - {x^3} + \frac{{{x^2}}}{2} - \frac{1}{2}$$
Work Step by Step
$$\eqalign{
& \frac{{dy}}{{dx}} = 4{x^3} - 3{x^2} + x;\,\,\,\,\,\,\,y\left( 1 \right) = 0 \cr
& {\text{Separating variables leads to}} \cr
& dy = \left( {4{x^3} - 3{x^2} + x} \right)dx \cr
& {\text{To solve this equation}}{\text{, determine the antiderivative of each side}} \cr
& \int {dy} = \int {\left( {4{x^3} - 3{x^2} + x} \right)dx} \cr
& {\text{integrating by using the power rule }} \cr
& y = 4\left( {\frac{{{x^4}}}{4}} \right) - 3\left( {\frac{{{x^3}}}{3}} \right) + \frac{{{x^2}}}{2} + C \cr
& y = {x^4} - {x^3} + \frac{{{x^2}}}{2} + C\,\,\,\,\,\left( {\bf{1}} \right) \cr
& \cr
& {\text{We can find the constant }}C{\text{ using the initial value problem}}\,\,y\left( 1 \right) = 0 \cr
& y\left( 1 \right) = 0{\text{ implies that }}y = 0{\text{ when }}x = 1 \cr
& {\text{Substituting these values into }}\left( {\bf{1}} \right) \cr
& 0 = {\left( 1 \right)^4} - {\left( 1 \right)^3} + \frac{{{{\left( 1 \right)}^2}}}{2} + C \cr
& 0 = \frac{1}{2} + C \cr
& C = - \frac{1}{2} \cr
& {\text{Substitute }}C = - \frac{1}{2}{\text{ into }}\left( {\bf{1}} \right){\text{ to find the particular solution}} \cr
& y = {x^4} - {x^3} + \frac{{{x^2}}}{2} - \frac{1}{2} \cr} $$
