MAT244--2019F > Term Test 1

Problem 3 (main)

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Rui Xiang:
Here is my solution.

BJM:
Here is the solution.

Yuying Chen:
$\text{(a)}\\$
$r^2-2r-3=0\\$
$(r+1)(r-3)=0\\$
$r=-1, r=3\\$
$\text{Homogeneous Equation:$y_c(x)=c_1e^{-x}+c_2e^{3x}$}\\ \\$

$\text{Since we know$\cosh{x}=\frac{e^{x}+e^{-x}}{2} = \frac{1}{2}e^x+\frac{1}{2}e^{-x}$}\\$
$\text{Therefore,}\\$
$y^{\prime\prime}-2y^{\prime}-3y=8e^x+8e^{-x}\\$
$Y(x)=Ae^x+Bxe^{-x}\\$
$Y^{\prime}(x)=Ae^x+Be^{-x}-Bxe^{-x}\\$
$Y^{\prime\prime}(x)=Ae^x-Be^{-x}-Be^{-x}+Bxe^{-x}\\$
$Ae^x-2Be^{-x}+Bxe^{-x}-2Ae^x-2Be^{-x}+2Bxe^{-x}-3Ae^x-3Bxe^{-x} = 8e^x+8e^{-x}\\$
\begin{cases}
(A-2A-3A)e^{x}=8e^{x}\\\
(-2B-2B)e^{-x}=8e^{-x}
\end{cases}
\begin{cases}
A=-2\\
B=-2
\end{cases}
$Y(x)=-2e^x-2xe^{-x}\\$
$\text{General Solution:}\\$
$y(t)=c_1e^{-x}+c_2e^{3x}-2e^x-2xe^{-x}\\$

$\text{(b)}\\$
$y^{\prime}=-c_1e^{-x}+3c_2e^{3x}-2e^{x}-2e^{-x}+2xe^{-x}\\$
$y(0)=0\Rightarrow c_1+c_2-2=0\\$
$y^{\prime}(0)=0\Rightarrow -c_1+3c_2-2-2=0\\$
\begin{cases}
c_1=\frac{1}{2}\\
c_2=\frac{3}{2}
\end{cases}

$\text{Thus,}$
$y(t)=\frac{1}{2}e^{-x}+\frac{3}{2}e^{3x}-2e^x-2xe^{-x}$

Xinqiao Li:
a) Find the general solution for equation $y'' - 2y' -3y = 16coshx$
b) Find solution, satisfying $y(0) = 0, y'(0) = 0$

a)
Consider $y'' - 2y' -3y = 0$

Assume $y = e^{rx}$, then the characteristic polynomial of the equation is given by $r^2 -2r -3 = 0$

This simplifies to $(r - 3)(r +1) = 0$ which gives us two roots $r_1 = 3$ and $r_2 = -1$

Then the complementary solution is $y_c = c_1e^{3x} + c_2e^{-x}$

Since $y'' - 2y' -3y = 16cosh = 8e^x + 8e^{-x}$

We guess the particular solution is of the form

$y = Ae^x + Bxe^{-x}$
$y' = Ae^x + B(e^{-x} - xe^{-x}) = Ae^x + Be^{-x} - Bxe^{-x}$
$y'' = Ae^x - Be^{-x} - B(e^{-x} - xe^{-x}) = Ae^x - 2Be^{-x} + Bxe^{-x}$

Substituting back to the original equation:

$y'' - 2y' -3y = Ae^x - 2Be^{-x} + Bxe^{-x} - 2Ae^x - 2Be^{-x} + 2Bxe^{-x} - 3Ae^x - 3Bxe^{-x} = -4Ae^x - 4Be^{-x} = 8e^x + 8e^{-x}$

Therefore, $A = -2$ and $B = -2$
The particular solution is $y_p = - 2e^x - 2xe^{-x}$
The general solution is :
$$y = c_1e^{3x} + c_2e^{-x} - 2e^x - 2xe^{-x}$$

b)
$$y = c_1e^{3x} + c_2e^{-x} - 2e^x - 2xe^{-x}$$
$$y' = 3c_1e^{3x} - c_2e^{-x} - 2e^x + 2xe^{-x} - 2e^{-x}$$

Plug in the initial conditions $y(0) = 0, y'(0) = 0$, we have,
$0 = c_1 + c_2 - 2$
$0 = 3c_1 - c_2 - 2 - 2$

so, $c_1 = \frac{3}{2}$ and $c_2 = \frac{1}{2}$
The particular solution is
$$y = \frac{3}{2}e^{3x} + \frac{1}{2}e^{-x} - 2e^x - 2xe^{-x}$$

yueyangyu:
a)
First solve the homogenous part:
\begin{aligned}
r^{2}-2r-3 &=0\\
r^{2}-2r+1&=4\\
(r-1)^{2} &=4\\
r_1 &=3 \\r_2 &=-1
\end{aligned}
So the solution to homogenous part is:
\begin{aligned}
y_c(x) =c_1e^{3x}+c_2e^{-x}
\end{aligned}
Next we solve
\begin{aligned}
y^{\prime\prime}-2y^{\prime}-3y &=16cosh(x)\\
&=16*\frac{e^{x}+e^{-x}}{2} \\
&=8e^{x}+8e^{-x}
\end{aligned}
Let
\begin{aligned}
y_p(x)&=Ae^{x}+Be^{-x}*x\\
y_p^{\prime}(x) &= Ae^{x}+Be^{-x}-Bxe^{-x}\\
y_p^{\prime\prime}(x) &= Ae^{x}-Be^{-x}-Be^{-x}+Bxe^{-x}\\
&= Ae^{x}-2Be^{-x}+Bxe^{-x}\\
\end{aligned}
Thus we can have
\begin{aligned}
Ae^{x}-2Be^{-x}+Bxe^{-x}-2(Ae^{x}+Be^{-x}-Bxe^{-x})-3(Ae^{x}+Be^{-x}*x) &=8e^{x}+8e^{-x}\\
-4Ae^{x}-4Be^{x} &=8e^{x}+8e^{-x}\\
A &=-2\\
B &=-2
\end{aligned}
Therefore, we can have:
\begin{aligned}
y_p(x) &=-2e^{x}-2xe^{-x}
\end{aligned}
From the above, we get
\begin{aligned}
y&=y_c(x)+y_p(x)\\
&= c_1e^{3x}+c_2e^{-x}-2e^{x}-2xe^{-x}
\end{aligned}

b)
\begin{aligned}
y &= c_1e^{3x}+c_2e^{-x}-2e^{x}-2xe^{-x}\\
y_p^{\prime} &= 3c_1e^{3x}-c_2e^{-x}-2e^x+2xe^{-x}-2e^{-x}\\
\end{aligned}
since
\begin{aligned}
y(0) &=0\\
y^{\prime} &= 0
\end{aligned}
we can have
\begin{aligned}
c_1 &=3/2\\
c_2 &=1/2
\end{aligned}
Thus, the solution is
\begin{aligned}
y &= 3/2e^{3x}+1/2e^{-x}-2e^{x}-2xe^{-x}
\end{aligned}