MAT244-2018S > Quiz-3

Q3-T0401

(1/1)

**Victor Ivrii**:

Find the solution of the given initial value problem

\begin{align*}

&y'' + 4y' + 3y = 0.\\

&y(0) = 2,\quad y'(0) = -1.

\end{align*}

**Darren Zhang**:

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

**Meng Wu**:

$$y''+4y'+3y=0,y(0)=2,y'(0)=-1$$

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

$$-c_1-3c_2=-1$$

Hence,

$$\cases{c_1+c_2=2\\-c_1-3c_2=-1} \implies \cases{c_1={5\over 2}\\c_2=-{1\over 2}}$$

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 $t \rightarrow \infty$.

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