Toronto Math Forum

MAT244-2018S => MAT244--Tests => Quiz-3 => Topic started by: Victor Ivrii on February 10, 2018, 05:20:01 PM

Title: Q3-T0801
Post by: Victor Ivrii on February 10, 2018, 05:20:01 PM

Find the solution of the given initial value problem.
&2y'' + y' - 4y = 0,\\
&y(0) = 0,\qquad y'(0) = 1.
Title: Re: Q3-T0801
Post by: Darren Zhang on February 10, 2018, 06:24:35 PM
The characteristic equation is $$2r^2+r-4r=0$$ with roots $$r = -\frac{1}{4}+\frac{\sqrt{33}}{4}, -\frac{1}{4}-\frac{\sqrt{33}}{4}$$
Therefore, the general equation is $$y = c_{1}exp(-1-\sqrt{33})t/4+c_{2}exp(-1+\sqrt{33})t/4$$
$$y = (-\frac{1}{4}+\frac{\sqrt{33}}{4})c_{1}exp(-1-\sqrt{33})t/4+ (-\frac{1}{4}-\frac{\sqrt{33}}{4})c_{2}exp(-1+\sqrt{33})t/4$$
In order to satisfy the initial conditions, we require that $$c_{1}+c_{2}=0$$
Therefore, we can get $$c_{1}=\frac{-2}{\sqrt{33}},  c_{2}=\frac{2}{\sqrt{33}}$$.
Therefore, the final solution is $$y = \frac{-2}{\sqrt{33}}exp(-1-\sqrt{33})t/4+\frac{2}{\sqrt{33}}exp(-1+\sqrt{33})t/4$$,
y->\infty as t -> \infty
Title: Re: Q3-T0801
Post by: Meng Wu on February 11, 2018, 09:30:21 AM
We assume that $y=e^{rt}$, and then it follows that $r$ must be a root of characteristic equation $$2r^2+r-4=0$$
We use the quadratic formula which is
$$r={-b\pm \sqrt{b^2-4ac}\over 2a}$$
$$\cases{r_1={-1+\sqrt{33}\over 4}\\r_2={-1-\sqrt{33}\over 4}}$$
Since the general solution has the form of $$y=c_1e^{r_1t}+c_2e^{r_2t}$$
Then the general solution of the given differential equation is
$$y=c_1e^{{-1+\sqrt{33}\over 4}t}+c_2e^{{-1-\sqrt{33}\over 4}t}$$
We also need $y'$ for the IVP,
$$y'={-1+\sqrt{33}\over 4}c_1e^{{-1+\sqrt{33}\over 4}t}+{-1-\sqrt{33}\over 4}c_2e^{{-1-\sqrt{33}\over 4}t}$$
To satisfy the first initial condition, we set $t=0$ and $y=0$, thus
To satisfy the second initial condition, set $t=0$ and $y'=1$, thus
$${-1+\sqrt{33}\over 4}c_1+{-1-\sqrt{33}\over 4}c_2=1$$
$$\cases{c_1+c_2=0\\{-1+\sqrt{33}\over 4}c_1+{-1-\sqrt{33}\over 4}c_2=1} \implies \cases{c_1={2\over \sqrt{33}}\\c_2=-{2\over \sqrt{33}}}$$
Therefore, the solution of the initial value problem is
$$y={2\over \sqrt{33}}e^{{-1+\sqrt{33}\over 4}t}-{2\over \sqrt{33}}e^{{-1-\sqrt{33}\over 4}t}$$
Note: $y \rightarrow \infty$ as $t \rightarrow \infty$.