Toronto Math Forum
MAT2442018S => MAT244Tests => Quiz3 => Topic started by: Victor Ivrii on February 10, 2018, 05:17:23 PM

Find the solution of the given initial value problem
\begin{align*}
&y'' + 4y' + 3y = 0.\\
&y(0) = 2,\quad y'(0) = 1.
\end{align*}

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

$$y''+4y'+3y=0,y(0)=2,y'(0)=1$$
We assume that $y=e^{rt}$, and then it follows that $r$ must be a root of characteristic equation $$r^2+4r+3=(r+1)(r+3)=0$$
Hence,
$$\cases{r_1=1\\r_2=3}$$
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
$$y=c_1e^{t}+c_2e^{3t}$$
We also need $y'$ for the IVP,
$$y'=c_1e^{t}3c_2e^{3t}$$
To satisfy the first initial condition, we set $t=0$ and $y=2$, thus
$$c_1+c_2=2$$
To satisfy the second initial condition, set $t=0$ and $y'=1$, thus
$$c_13c_2=1$$
Hence,
$$\cases{c_1+c_2=2\\c_13c_2=1} \implies \cases{c_1={5\over 2}\\c_2={1\over 2}}$$
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 $t \rightarrow \infty$.