Question

@SithsAndGiggles techniques in answering reduction of order non homo?

Answer 1

Hint:
we can rewrite the left side of your first equation, as follows:
\[\left( {{D^2} + D} \right)y = \left( {{D^2} - 4D + 4D} \right)y = \left( {{D^2} - 4D} \right)y + 4Dy\]
so your first equation becomes:
\[\left( {{D^2} - 4D} \right)y = \sin x - 4Dy\]

Answer 2

and after substitutionh into your second equation, we can write:
\[\sin x - 4Dy + 4y = \exp \left( x \right)\]

Answer 3

why is it 4?

Answer 4

since the left side of your second equation contains the term 4y

Answer 5

oh i forget one step and i messed up the whole problem

Answer 6

@Michele_Laino, I don't think this is a system of ODEs, but rather two separate ODEs to be solved via reduction of order.

Answer 7

@Michele_Laino, also, you seem to made a slight mistake. Writing \(-4D+4D\) would gives \(0D\), thus removing the first derivative. I think you meant to \(-4D+5D\) ?

Answer 8

Assuming these ODEs are actually independent of each other... First, solve the respective associated homogeneous ODEs:
\[\begin{array}{c|c}
\text{ODE}&\text{auxiliary eq.}&\text{roots to aux. eq.}&\text{homog. sol.}\\
\hline
(D^2+D)y=0&r^2+1=0&r=\pm i&C_1\cos x+C_2\sin x\\
\hline
(D^2-4D+4)y=0&r^2-4r+4=0&r=2_{(2)}&(C_1+C_2x)e^{2x}
\end{array}\]
where \(2_{(2)}\) denotes that \(r=2\) is a root of multiplicity \(2\).

Answer 9

You are right! @SithsAndGiggles I have made a big error. Here is the right step:
\[\begin{gathered}
\left( {{D^2} + D} \right)y = \left( {{D^2} - 4D + 4D + D} \right)y = \hfill \\
= \left( {{D^2} - 4D} \right)y + 5Dy = \sin x \hfill \\
\end{gathered} \]

Answer 10

so we can write:
\[\sin x - 5Dy + 4y = \exp \left( x \right)\]
I'm very sorry @silverxx

Answer 11

thanks for your comment @SithsAndGiggles

Answer 12

Thank you somuchguys =)