## Calculus with Applications (10th Edition)

$$y = {x^2} - {x^3} + 5$$
\eqalign{ & \frac{{dy}}{{dx}} + 3{x^2} = 2x;\,\,\,\,\,\,\,y\left( 0 \right) = 5 \cr & {\text{subtracting }}3{x^2}{\text{to both sides of the equation}}{\text{. That is}}{\text{,}} \cr & \frac{{dy}}{{dx}} = 2x - 3{x^2} \cr & {\text{Separating variables leads to}} \cr & dy = \left( {2x - 3{x^2}} \right)dx \cr & {\text{To solve this equation}}{\text{, determine the antiderivative of each side}} \cr & \int {dy} = \int {\left( {2x - 3{x^2}} \right)dx} \cr & {\text{Integrating by using the power rule }} \cr & y = 2\left( {\frac{{{x^2}}}{2}} \right) - 3\left( {\frac{{{x^3}}}{3}} \right) + C \cr & y = {x^2} - {x^3} + C\,\,\,\left( {\bf{1}} \right) \cr & \cr & {\text{We can find the constant }}C{\text{ using the initial value problem}}\,\,y\left( 0 \right) = 5 \cr & y\left( 0 \right) = 5{\text{ implies that }}y = 5{\text{ when }}x = 0 \cr & {\text{Substituting these values into }}\left( {\bf{1}} \right) \cr & 5 = {\left( 0 \right)^2} - {\left( 0 \right)^3} + C\, \cr & C = 5 \cr & {\text{Substitute }}C = 5{\text{ into }}\left( {\bf{1}} \right){\text{ to find the particular solution}} \cr & y = {x^2} - {x^3} + 5 \cr}