Answer to Question #94172 in Differential Equations for Sarafa Ibrahim

Question

Solve the differential equations y"-3y'+2y=0

Answer 1

It's a second-order linear ordinary differential equation.

An equation $y''+py'+qy=0$ has solution $y=C_1e^{\lambda_1x}+C_2e^{\lambda_3x}$ where $\lambda_1,\lambda_2$ are the roots of $x^2+px+q=0$ if $\lambda_1\ne\lambda_2$ and $y=(C_1+C_2x)e^\lambda$ if $\lambda_1=\lambda_2\ (=\lambda)$ .

https://en.wikipedia.org/wiki/Linear_differential_equation#Second-order_case

So, $x^2-3x+2=0$ has the roots $1$ and $2$ , according to converse Vieta's theorem (as $-(1+2)=-3$ and $1\cdot 2=2$ ).

So, the solution is $y=C_1e^x+C_2e^{2x}$ .

You can see here https://www.wolframalpha.com/input/?i=y%22-3y%27%2B2y%3D0 , it's correct.