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### Messages - Klaire

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1
##### Term Test 1 / Re: Problem 4 (noon)
« on: October 23, 2019, 07:43:53 AM »
a,First solve the homogenous part:
\begin{aligned}
r^{2}-8r+25 &=0\\
r^{2}-8r+16&=-9\\
(r-4)^{2} &=9i^2\\
r_1 &=3i+4 \\r_2 &=-3i+4
\end{aligned}
So the solution to homogenous part is:
\begin{aligned}
y_c(x) =c_1e^{4x}cos(3x)+c_2e^{4x}sin(3x)
\end{aligned}
Next we solve
\begin{aligned}
y^{\prime\prime}-8y^{\prime}+25y &=18e^{4x}\\
\end{aligned}
Let
\begin{aligned}
y_p(x)&=Ae^{4x}\\
y_p^{\prime}(x) &= 4Ae^{4x}\\
y_p^{\prime\prime}(x) &= 16Ae^{4x}\\

\end{aligned}
Thus we can have
\begin{aligned}
16Ae^{4x}-32Ae^{4x}+25Ae^{4x} &=18e^{4x}\\
A &= 2\\
\end{aligned}
Therefore, we can have:
\begin{aligned}
y_p(x) &=2e^{4x}\\
\end{aligned}

\begin{aligned}
y^{\prime\prime}-8y^{\prime}+25y &=104Cos(3x)\\
y_p(x) &= Acos(3x)+Bsin(3x)\\
y_p^{\prime}(x) &= -3Asin(3x)+3Bcos(3x)\\
y_p^{\prime\prime}(x) &= -9Acos(3x)-9Bsin(3x)\\
\end{aligned}
Thus we can have
\begin{aligned}
-9Acos(3x)-9Bsin(3x)-8(-3Asin(3x)+3Bcos(3x))+25Acos(3x)+25Bsin(3x) &=104cos(3x)\\
(16A-24B)cos(3x)+(16B+24A)sin(3x) &= 104cos(3x)\\
A &= 2\\
B &=-3\\
y_p(x) &= 2cos(3x)-3sin(3x)
\end{aligned}
From the above, we get
\begin{aligned}
y &= y_c(x)+y_p(x)=c_1e^{4x}cos(3x)+c_2e^{4x}sin(3x)+2e^{4x}+2cos(3x)-3sin(3x)
\end{aligned}

b,
Since
\begin{aligned}
y(0) &= 0\\
y^{\prime}(0) &= 0\\
\end{aligned}
we can get
\begin{aligned}
c_1 &= -4\\
c_2 &= 1/3
\end{aligned}
So the solution is
\begin{aligned}
y &= -4e^{4x}cos(3x)+1/3e^{4x}sin(3x)+2e^{4x}+2cos(3x)-3sin(3x)
\end{aligned}

2
##### Term Test 1 / Re: Problem 3 (main)
« on: October 23, 2019, 06:51:26 AM »
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 Never use * for multiplication
\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}

OK. V.I.

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