Toronto Math Forum

MAT244-2018S => MAT244--Tests => Quiz-3 => Topic started by: Victor Ivrii on February 10, 2018, 05:17:23 PM

Title: Q3-T0401
Post by: Victor Ivrii on February 10, 2018, 05:17:23 PM
Find the solution of the given initial value problem
&y'' + 4y' + 3y = 0.\\
&y(0) = 2,\quad y'(0) = -1.
Title: Re: Q3-T0401
Post by: Darren Zhang on February 10, 2018, 06:08:19 PM
Substitution of the assumed solution $$y = e^{rt}$$ results in the characteristic equation $r^2+4r+3=0$, The roots of the equation are  r = -1/-3. Hence the general solution is $$y = c_{1}e^{-t}+c_{2}e^{-3t}$$. Its derivative is $$y' = -c_{1}e^{-t}+-3*c_{2}e^{-3t}$$
Based on the first equation y(0)=1, we can know that $$c_{1}+c_{2} = 2$$.
Based on the second equation y'(0)=-1, we can know that $$-c_{1}-3c_{2}=-1$$, therefore, $$c_{1}= \frac{5}{2}, c_{2} = -\frac{1}{2}$$
Therefore we can get the equation that  $$y = \frac{5}{2}e^{-t}-\frac{1}{2}e^{-3t}$$.
y ->0 when t -> \infty
Title: Re: Q3-T0401
Post by: Meng Wu on February 11, 2018, 09:30:55 AM
We assume that $y=e^{rt}$, and then it follows that $r$ must be a root of characteristic equation $$r^2+4r+3=(r+1)(r+3)=0$$
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
We also need $y'$ for the IVP,
To satisfy the first initial condition, we set $t=0$ and $y=2$, thus
To satisfy the second initial condition, set $t=0$ and $y'=-1$, thus
$$\cases{c_1+c_2=2\\-c_1-3c_2=-1} \implies \cases{c_1={5\over 2}\\c_2=-{1\over 2}}$$
Therefore, the solution of the initial value problem is
$$y={5\over 2}e^{-t}-{1\over 2}e^{-3t}$$
Note: $y \rightarrow 0$ as $t \rightarrow \infty$.