# Toronto Math Forum

## MAT244-2018S => MAT244--Tests => Quiz-3 => Topic started by: Victor Ivrii on February 10, 2018, 05:18:09 PM

Title: Q3-T0501
Post by: Victor Ivrii on February 10, 2018, 05:18:09 PM
Find the general solution of the given differential equation.
$$2y'' - 3y' + y = 0.$$
Title: Re: Q3-T0501
Post by: Darren Zhang on February 10, 2018, 05:57:47 PM
Substitution of the assumed solution $y=e^{rt}$ results in the characteristic equation $$2r^2-3r+1=0$$
The roots of the equation are $r = \frac{1}{2}, 1$. Hence the general solution is $y = c_{1}e^{\frac{t}{2}}+c_{2}e^{\frac{3t}{2}}$
Title: Re: Q3-T0501
Post by: Meng Wu on February 11, 2018, 09:29:22 AM
$$2y’’-3y’+y=0$$
We assume that $y=e^{rt}$, and then it follows that $r$ must be a root of characteristic equation $$2r^2-3r+1=(2r-1)(r-1)=0$$
Hence,
$$\cases{r_1={1\over2}\\r_2=1}$$
Since the general solution has the form of $$y=c_1e^{r_1t}+c_2e^{r_2t}$$
Therefore, the general solution of the given differential equation is
$$y=c_1e^{{1\over2}t}+c_2e^{t}$$