Toronto Math Forum
MAT2442018F => MAT244Tests => Quiz3 => Topic started by: Victor Ivrii on October 12, 2018, 06:05:28 PM

Find the general solution of the given differential equation:
$$2y''3y'+y=0.$$

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 $(r1)(2r1)$.
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.