## Calculus with Applications (10th Edition)

$$y = 5{e^{3x - {x^2}}}$$
\eqalign{ & \frac{{dy}}{{dx}} = \left( {3 - 2x} \right)y;\,\,\,\,\,\,\,\,\,y\left( 0 \right) = 5 \cr & {\text{Separating variables leads to}} \cr & \frac{{dy}}{y} = \left( {3 - 2x} \right)dx \cr & {\text{To solve this equation}}{\text{, determine the antiderivative of each side}} \cr & \int {\frac{1}{y}dy} = \int {\left( {3 - 2x} \right)dx} \cr & {\text{integrating}} \cr & \ln \left| y \right| = 3x - {x^2} + C\,\,\,\,\,\left( {\bf{1}} \right) \cr & \cr & {\text{we can find the constant }}C{\text{ using the initial value problem}}\,\,y\left( 0 \right) = 5 \cr & y\left( 0 \right) = 5{\text{ implies that }}y = 5{\text{ when }}x = 0 \cr & {\text{substituting these values into }}\left( {\bf{1}} \right) \cr & \ln \left| 5 \right| = 3\left( 0 \right) - {\left( 0 \right)^2} + C \cr & C = \ln 5 \cr & {\text{substitute }}C = \ln 5{\text{ into }}\left( {\bf{1}} \right){\text{ to find the particular solution}} \cr & \ln \left| y \right| = 3x - {x^2} + \ln 5 \cr & {\text{solve for }}y \cr & {e^{\ln \left| y \right|}} = {e^{3x - {x^2}}}{e^{\ln 5}} \cr & y = 5{e^{3x - {x^2}}} \cr}