Chapter 10 - Differential Equations - 10.1 Solutions of Elementary and Separable Differential Equations - 10.1 Exercises - Page 535: 19

$$y = - 2x{e^{ - x}} - 2{e^{ - x}} + 44$$

$$\eqalign{ & 2\frac{{dy}}{{dx}} = 4x{e^{ - x}};\,\,\,\,\,\,\,y\left( 0 \right) = 42 \cr & {\text{Separating variables leads to}} \cr & dy = 2x{e^{ - x}}dx \cr & {\text{To solve this equation}}{\text{, determine the antiderivative of each side}} \cr & \int {dy} = \int {2x{e^{ - x}}dx} \cr & y = \int {2x{e^{ - x}}dx} \cr & {\text{solve the integral by parts }} \cr & {\text{setting }}\,\,\,\,\,\,u = 2x{\text{ then }}du = 2dx\,\,\,\,\,\,\,\,\,\, \cr & {\text{and}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,dv = {e^{ - x}}dx{\text{ then }}v = - {e^{ - x}} \cr & {\text{Substituting these values into the formula for integration by parts}} \cr & \int u dv = uv - \int {vdu} \cr & \int {2x{e^{ - x}}dx} = \left( {2x} \right)\left( { - {e^{ - x}}} \right) - \int {\left( { - {e^{ - x}}} \right)\left( 2 \right)} dx \cr & \int {2x{e^{ - x}}dx} = - 2x{e^{ - x}} + 2\int {{e^{ - x}}} dx \cr & \int {2x{e^{ - x}}dx} = - 2x{e^{ - x}} - 2{e^{ - x}} + C \cr & {\text{then}} \cr & y = - 2x{e^{ - x}} - 2{e^{ - x}} + C\,\,\,\,\,\,\left( {\bf{1}} \right) \cr & \cr & {\text{We can find the constant }}C{\text{ using the initial value problem}}\,\,y\left( 0 \right) = 42 \cr & y\left( 0 \right) = 42{\text{ implies that }}y = 42{\text{ when }}x = 0 \cr & {\text{Substituting these values into }}\left( {\bf{1}} \right) \cr & 42 = - 2\left( 0 \right){e^{ - 0}} - 2{e^{ - 0}} + C \cr & 42 = - 2 + C \cr & C = 44 \cr & {\text{Substitute }}C = 44{\text{ into }}\left( {\bf{1}} \right){\text{ to find the particular solution}} \cr & y = - 2x{e^{ - x}} - 2{e^{ - x}} + 44 \cr} $$