Answer:
[tex]y = \frac{1}{3}e^{2x} + \frac{5}{3}e^{-x}[/tex].
Step-by-step explanation:
Let's find a particular solution [tex]y_p = ae^{bx}[/tex]:
[tex]y' = abe^{bx}[/tex], so
[tex]abe^{bx} + ae^{bx} = e^{2x}[/tex]
[tex]ae^{bx}(b + 1) = e^{2x}[/tex], then, b=2 and 3a = 1, so a= 1/3.
Our particular solution is [tex]y_p = \frac{1}{3}e^{2x}[/tex]. Now, we are going to find the solution of the homogeneus equation with constants coefficients.
Let y = [tex]e^{\lambda x}[/tex]
[tex]y' = \lambda e^{\lambda x}[/tex], so
[tex]\lambda e^{\lambda x}+ e^{\lambda x} = 0[/tex]
[tex]e^{\lambda x}(\lambda + 1) = 0[/tex]
[tex]\lambda= -1[/tex]. Then [tex]y_h = Ce^{-x}[/tex] and the solution is
[tex]y = y_p + y_h = \frac{1}{3}e^{2x} + Ce^{-x}[/tex]. Now, we use the initial condition to find C:
[tex]y(0) = \frac{1}{3}e^{0} + Ce^{0} = \frac{1}{3} + C = 2[/tex]
[tex]C = 2- \frac{1}{3} = \frac{5}{3}.[/tex] The final result is
[tex]y = \frac{1}{3}e^{2x} + \frac{5}{3}e^{-x}[/tex].
