Solve y''+2y' - 3y = 0, y(0) = 3, y'(0) = 11 Preview y(t) = |2e^(-3t)+5e^t Points possible: 1 This is attempt 3 of 3 Score on last attempt: 0. Score in gradebook: 0 License Submit

[tex]y''+2y'-3y=0 [/tex]Second order linear homogeneous differential equation with constant coefficients, ODE has a form of,[tex]ay''+by'+cy=0[/tex]From here we assume that for any equation of that form has a solution of the form, [tex]e^{yt}[/tex]Now the equation looks like this,[tex]((e^{yt}))''+2((e^{yt}))'-3e^{yt}=0[/tex]Now simplify to,[tex]e^{yt}(y^2+2y-3)=0[/tex]You can solve the simplified equation using quadratic equation since, [tex]e^{yt}(y^2+2y-3)=0\Longleftrightarrow y^2+2y-3=0[/tex]Using the QE we result with,[tex]\underline{y_1=1}, \underline{y_2=-3}[/tex]So,For two real roots [tex]y_1\neq y_2[/tex] the general solution takes the form of,[tex]y=c_1e^{y_1t}+c_2e^{y_2t}[/tex]Or simply,[tex]\boxed{y=c_1e^t+c_2e^{-3t}}[/tex]Hope this helps.r3t40