## Calculus with Applications (10th Edition)

$$y = \frac{x}{3}{e^{3x}} - \frac{1}{9}{e^{3x}} + 1$$
\eqalign{ & x\frac{{dy}}{{dx}} = {x^2}{e^{3x}};\,\,\,\,\,\,\,y\left( 0 \right) = \frac{8}{9} \cr & {\text{Separating variables leads to}} \cr & dy = x{e^{3x}}dx \cr & {\text{To solve this equation}}{\text{, determine the antiderivative of each side}} \cr & \int {dy} = \int {x{e^{3x}}dx} \cr & y = \int {x{e^{3x}}} \cr & {\text{solve the integral by parts }} \cr & {\text{setting }}\,\,\,\,\,\,u = x{\text{ then }}du = dx\,\,\,\,\,\,\,\,\,\, \cr & {\text{and}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,dv = {e^{3x}}dx{\text{ then }}v = \frac{1}{3}{e^{3x}} \cr & {\text{Substituting these values into the formula for integration by parts}} \cr & \int u dv = uv - \int {vdu} \cr & \int {x{e^{3x}}} = \left( x \right)\left( {\frac{1}{3}{e^{3x}}} \right) - \int {\left( {\frac{1}{3}{e^{3x}}} \right)dx} \cr & \int {x{e^{3x}}} = \frac{x}{3}{e^{3x}} - \frac{1}{3}\int {{e^{3x}}dx} \cr & \int {x{e^{3x}}} = \frac{x}{3}{e^{3x}} - \frac{1}{3}\left( {\frac{1}{3}{e^{3x}}} \right) + C \cr & \int {x{e^{3x}}} = \frac{x}{3}{e^{3x}} - \frac{1}{9}{e^{3x}} + C \cr & {\text{then}} \cr & y = \frac{x}{3}{e^{3x}} - \frac{1}{9}{e^{3x}} + C\,\,\,\,\,\,\left( {\bf{1}} \right) \cr & \cr & {\text{we can find the constant }}C{\text{ using the initial value problem}}\,\,y\left( 0 \right) = \frac{8}{9} \cr & y\left( 0 \right) = \frac{8}{9}{\text{ implies that }}y = \frac{8}{9}{\text{ when }}x = 0 \cr & {\text{substituting these values into }}\left( {\bf{1}} \right) \cr & \frac{8}{9} = \frac{0}{3}{e^{3\left( 0 \right)}} - \frac{1}{9}{e^{3\left( 0 \right)}} + C \cr & \frac{8}{9} = - \frac{1}{9} + C \cr & C = 1 \cr & {\text{substitute }}C = 1{\text{ into }}\left( {\bf{1}} \right){\text{ to find the particular solution}} \cr & y = \frac{x}{3}{e^{3x}} - \frac{1}{9}{e^{3x}} + 1 \cr}