Chapter 8 - Mathematical Modeling With Differential Equations - Chapter 8 Review Exercises - Page 595: 21

$$y = - 1 + 4{e^{\frac{{{x^2}}}{2}}}$$

$$\eqalign{ & y' - xy = x,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{initial condition }}y\left( 0 \right) = 3 \cr & {\text{write }}y'{\text{ as }}\frac{{dy}}{{dx}} \cr & \frac{{dy}}{{dx}} - xy = x \cr & {\text{The equation is in the form }}\frac{{dy}}{{dp}} + p\left( x \right)y = q\left( x \right) \cr & {\text{with }}p\left( x \right) = - x{\text{ and }}q\left( x \right) = x \cr & \cr & {\text{Use the method of the integrating factors}}{\text{, calculate }}\mu = {e^{\int {p\left( x \right)dx} }} \cr & \mu = {e^{\int {p\left( x \right)dx} }} = {e^{\int { - x} dx}} \cr & \mu = {e^{ - \frac{{{x^2}}}{2}}} \cr & \cr & {\text{Multiply both sides of the equation by }}\mu \cr & {e^{ - \frac{{{x^2}}}{2}}}\frac{{dy}}{{dx}} - x{e^{ - \frac{{{x^2}}}{2}}}y = x{e^{ - \frac{{{x^2}}}{2}}} \cr & {\text{Express the result as }}\frac{d}{{dx}}\left[ {\mu y} \right] = {e^x}{\text{. Then}}{\text{,}} \cr & \frac{d}{{dx}}\left[ {{e^{ - \frac{{{x^2}}}{2}}}y} \right] = x{e^{ - \frac{{{x^2}}}{2}}} \cr & {\text{Integrate both sides}} \cr & {e^{ - \frac{{{x^2}}}{2}}}y = \int {x{e^{ - \frac{{{x^2}}}{2}}}} dx \cr & {e^{ - \frac{{{x^2}}}{2}}}y = - {e^{ - \frac{{{x^2}}}{2}}} + C \cr & {\text{Solve for }}y \cr & y = - 1 + C{e^{\frac{{{x^2}}}{2}}} \cr & \cr & {\text{Use the initial condition }}y\left( 0 \right) = 3 \cr & 3 = - 1 + C{e^{\frac{{{{\left( 0 \right)}^2}}}{2}}} \cr & C = 4 \cr & \cr & {\text{Then}}{\text{, the particular solution of the differential equation is}} \cr & y = - 1 + 4{e^{\frac{{{x^2}}}{2}}} \cr} $$