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May 26th, 2018, 05:16 AM   #1
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Differential Equations-HELP

Hey guys, I'm new to here. Attached with this are two problems in a tutorial of our university maths department. Hope you could help me to solve them.
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Last edited by skipjack; May 26th, 2018 at 07:18 AM.
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May 26th, 2018, 07:32 AM   #2
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(1) (D² + 2D + 1)y = $x \cos x$

Solution: y = (A + B$x$)$e^{-x}$ + ($x\sin x - \sin x + \cos x$)/2, where A and B are constants.

There are various ways of obtaining that. For example, you can start by using $e^x\!$ as an integrating factor. What method(s) would you be expected to know?

Can you confirm that the second question gives the equations below?
(2D² - D + 9)y - (D² + D + 3)z = 0
(2D² + D + 7)y - (D² - D + 5)z = 0
They're a bit hard to read.
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Last edited by skipjack; May 26th, 2018 at 10:36 PM.
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May 26th, 2018, 06:08 PM   #3
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For the second, you can take the sum of the two equations and the difference of the two equations to get the equations
\begin{align*}2(D-1)(y+z) &= 0 \\
2(D^2 + 4 )(2y-z) &= 0 \end{align*}

You can solve these for the two functions $f(x)=y(x)+z(x)$ and $g(x)=2y(x)-z(x)$ respectively, and then find $y(x)$ and $z(x)$ as functions of $f(x)$ and $g(x)$.

I get \begin{align*}
y(x) &= c_1 e^x + c_2 \sin{(2x)} + c_3 \cos{(2x)} \\
z(x) &= 2c_1 e^x - c_2 \sin{(2x)} - c_3 \cos{(2x)}
\end{align*}
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Last edited by v8archie; May 26th, 2018 at 06:40 PM.
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May 26th, 2018, 08:26 PM   #4
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thanks alot guys.....
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May 27th, 2018, 05:41 AM   #5
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yep.That's the second part.
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