# Toronto Math Forum

## MAT244--2018F => MAT244--Tests => Quiz-3 => Topic started by: Victor Ivrii on October 12, 2018, 06:05:28 PM

Title: Q3 TUT 0501
Post by: Victor Ivrii on October 12, 2018, 06:05:28 PM
Find the general solution of the given differential equation:
$$2y''-3y'+y=0.$$
Title: Re: Q3 TUT 0501
Post by: Nick Callow on October 12, 2018, 06:16:17 PM
Find the solution of the given differential equation $2y''(t) - 3y'(t) + y(t) = 0$.

Since this a second order, linear, homogeneous differential equation, a good guess will be a solution of the form $y(t) = e^{rt}$. Consequently, $y'(t) = re^{rt}$ and $y''(t) = r^2e^{rt}$. Subbing this into the differential equation, we get:
$$2r^2e^{rt} - 3re^{rt} + e^{rt} = 0.$$
We can then factor out an $e^{rt}$ from the equation to produce
$$e^{rt}(2r^2 - 3r + 1)=0$$
We know that for any choice of $r$, $e^{rt} \neq 0$ therefore, we can discard it. We are left with the characteristic equation $2r^2 - 3r + 1$. Factoring this, we get $(r-1)(2r-1)$.

Since $r = 1$, $\frac{1}{2}$ we now have two particular solutions $y_1(t) = e^t$ and $y_2(t) = e^{\frac{t}{2}}$. The general solution of this will then be:
$$y(t) = c_1e^t + c_2e^{\frac{t}{2}}$$
where $c_1$, $c_2$ are constants.