Chapter 6 - Differential Equations - 6.4 Exercises - Page 430: 48

$$\eqalign{ & y = \frac{1}{4} + \frac{{13}}{4}{e^{ - {x^4}}}{\text{ }}\left( {{\text{Particular solution 1}}} \right) \cr & y = \frac{1}{4} - \frac{3}{4}{e^{ - {x^4}}}{\text{ }}\left( {{\text{Particular solution 2}}} \right) \cr} $$

$$\eqalign{ & \frac{{dy}}{{dx}} + 4{x^3}y = {x^3} \cr & {\text{The differential equation has the form }}\frac{{dy}}{{dx}} + P\left( x \right)y = Q\left( x \right) \cr & {\text{With }}P\left( x \right) = 4{x^3}{\text{ and }}Q\left( x \right) = {x^3} \cr & {\text{Find the integrating factor }}I\left( x \right) = {e^{\int {P\left( x \right)} dx}} \cr & I\left( x \right) = {e^{\int {4{x^3}dx} }} = {e^{{x^4}}} \cr & {\text{Multiply the differential equation by the integrating factor}} \cr & {e^{{x^4}}}\left( {\frac{{dy}}{{dx}} + 4{x^3}y} \right) = {e^{{x^4}}}\left( {{x^3}} \right) \cr & {e^{{x^4}}}\frac{{dy}}{{dx}} + 4{x^3}{e^{{x^4}}}y = {x^3}{e^{{x^4}}} \cr & {\text{Write the left side in the form }}\frac{d}{{dx}}\left[ {I\left( x \right)y} \right] \cr & \frac{d}{{dx}}\left[ {y{e^{{x^4}}}} \right] = {x^3}{e^{{x^4}}} \cr & {\text{Integrate both sides}} \cr & y{e^{{x^4}}} = \int {{x^3}{e^{{x^4}}}} dx \cr & y{e^{{x^4}}} = \frac{1}{4}{e^{{x^4}}} + C \cr & {\text{Solve for }}y \cr & y = \frac{1}{4} + C{e^{ - {x^4}}} \cr & {\text{Use the initial condition }}\left( {0,\frac{7}{2}} \right) \cr & \frac{7}{2} = \frac{1}{4} + C{e^{ - {{\left( 0 \right)}^4}}} \cr & C = \frac{{13}}{4} \cr & y = \frac{1}{4} + \frac{{13}}{4}{e^{ - {x^4}}}{\text{ }}\left( {{\text{Particular solution 1}}} \right) \cr & {\text{Use the initial condition }}\left( {0, - \frac{1}{2}} \right) \cr & - \frac{1}{2} = \frac{1}{4} + C{e^{ - {{\left( 0 \right)}^4}}} \cr & C = - \frac{3}{4} \cr & y = \frac{1}{4} - \frac{3}{4}{e^{ - {x^4}}}{\text{ }}\left( {{\text{Particular solution 2}}} \right) \cr & \cr & {\text{Graph}} \cr} $$