# Chapter 10 - Differential Equations - 10.1 Solutions of Elementary and Separable Differential Equations - 10.1 Exercises - Page 535: 24

$${y^2} - y = \frac{{{x^3}}}{3} + 5x + 110$$

#### Work Step by Step

\eqalign{ & \frac{{dy}}{{dx}} = \frac{{{x^2} + 5}}{{2y - 1}};\,\,\,\,\,\,\,\,\,\,\,y\left( 0 \right) = 11 \cr & {\text{Separating variables leads to}} \cr & \left( {2y - 1} \right)dy = \left( {{x^2} + 5} \right)dx \cr & {\text{To solve this equation}}{\text{, determine the antiderivative of each side}} \cr & \int {\left( {2y - 1} \right)dy} = \int {\left( {{x^2} + 5} \right)} dx \cr & {\text{integrate by using the power rule}} \cr & {y^2} - y = \frac{{{x^3}}}{3} + 5x + C\,\,\,\,\,\,\,\,\,\,\,\,\left( {\bf{1}} \right) \cr & \cr & {\text{we can find the constant }}C{\text{ using the initial value problem}}\,\,\,y\left( 0 \right) = 1 \cr & \,y\left( 0 \right) = 11{\text{ implies that }}y = 11{\text{ when }}x = 0 \cr & {\text{substituting these values into }}\left( {\bf{1}} \right) \cr & {\left( {11} \right)^2} - 11 = \frac{{{{\left( 0 \right)}^3}}}{3} + 5\left( 0 \right) + C \cr & C = 110 \cr & {\text{substitute }}C = 110{\text{ into }}\left( {\bf{1}} \right){\text{ to find the particular solution}} \cr & {y^2} - y = \frac{{{x^3}}}{3} + 5x + 110 \cr}

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.