[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
