Chapter 8 - Mathematical Modeling With Differential Equations - Chapter 8 Review Exercises - Page 594: 7

$$ - \frac{1}{{{y^4}}} + 2\ln \left| y \right| = 2\ln \left| x \right| - 1$$

$$\eqalign{ & y' = \frac{{{y^5}}}{{x\left( {1 + {y^4}} \right)}},\,\,\,\,\,\,\,\,\,{\text{initial condition }}y\left( 1 \right) = 1 \cr & \cr & {\text{write }}y'{\text{ as }}\frac{{dy}}{{dx}} \cr & \frac{{dy}}{{dx}} = \frac{{{y^5}}}{{x\left( {1 + {y^4}} \right)}} \cr & {\text{separating the variables}} \cr & \frac{{dy}}{{{y^5}}} = \frac{{dx}}{{x\left( {1 + {y^4}} \right)}} \cr & \frac{{1 + {y^4}}}{{{y^5}}}dy = \frac{1}{x}dx \cr & \cr & {\text{Integrate both sides of the equation}} \cr & \int {\frac{{1 + {y^4}}}{{{y^5}}}} dy = \int {\frac{1}{x}} dx \cr & \int {\left( {\frac{1}{{{y^5}}} + \frac{1}{y}} \right)} dy = \int {\frac{1}{x}} dx \cr & \int {\left( {{y^{ - 5}} + \frac{1}{y}} \right)} dy = \int {\frac{1}{x}} dx \cr & \frac{{{y^{ - 4}}}}{{ - 4}} + \frac{1}{y} = \int {\frac{1}{x}} dx \cr & - \frac{1}{{4{y^4}}} + \ln \left| y \right| = \ln \left| x \right| + C \cr & \cr & {\text{Use the initial condition }}y\left( 1 \right) = 1 \cr & - \frac{1}{{4{{\left( 1 \right)}^4}}} + \ln \left| 1 \right| = \ln \left| 1 \right| + C \cr & - \frac{1}{4} = C \cr & \cr & {\text{Then}}{\text{, the particular solution is}} \cr & - \frac{1}{{4{y^4}}} + \ln \left| y \right| = \ln \left| x \right| - \frac{1}{4} \cr & {\text{multiply by 4}} \cr & - \frac{1}{{{y^4}}} + 2\ln \left| y \right| = 2\ln \left| x \right| - 1 \cr} $$