How am I supposed to find the solution to the non-homogeneous ode ?Y→'=(−422−1)Y→+(x−12x−1+4)

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hestyllvql

Answered question

2022-04-22

How am I supposed to find the solution to the non-homogeneous ode ?
${\vec{Y}}^{'} = (\begin{matrix} - 4 & 2 \\ 2 & - 1 \end{matrix}) \vec{Y} + (\begin{matrix} x^{- 1} \\ 2 x^{- 1} + 4 \end{matrix})$

Answer & Explanation

Ciara Melton

Beginner2022-04-23Added 12 answers

Step 1
The answer is
The Nonhomogeneous system is given by
Complementary solution:
$Y' (x) = AY (x)$
The eigenvalues are give by
$λ_{1} = - 5, λ_{2} = 0$
and the eigenvectors are given by
Since $λ_{1} \neq λ_{2}$ so the complementary solution is of the form
$Y_{c} (x) = c_{1} v_{1} e^{λ_{1} x} + c_{2} v_{2} e^{λ_{2} x}$
Hence
Particular solution:
$Y' (x) = AY (x) + F (x)$
Using variation of parameters:
The fundamental matrix is of the form
The particular solution is of the form
$Y_{p} (x) = Φ (x) \int Φ^{- 1} (x) F (x) d x$
Hence, we have
$Y_{p} (x) = Φ (x) \int Φ^{- 1} (x) F (x) d x$
$= [\begin{matrix} - 2 e^{- 5 x} & 1 \\ e^{- 5 x} & 2 \end{matrix}] \int [\begin{matrix} - \frac{2}{5} e^{5 x} & \frac{1}{5} e^{5 x} \\ \frac{1}{5} & \frac{2}{5} \end{matrix}] [\begin{matrix} \frac{1}{x} \\ \frac{2}{x} + 4 \end{matrix}] d x$
$= \frac{1}{5} [\begin{matrix} - 2 e^{- 5 x} & 1 \\ e^{- 5 x} & 2 \end{matrix}] \int [\begin{matrix} - 2 e^{5 x} & e^{5 x} \\ 1 & 2 \end{matrix}] [\begin{matrix} \frac{1}{x} \\ \frac{2}{x} + 4 \end{matrix}] d x$
$= \frac{1}{5} [\begin{matrix} - 2 e^{- 5 x} & 1 \\ e^{- 5 x} & 2 \end{matrix}] \int [\begin{matrix} - \frac{2}{x} e^{5 x} + (\frac{2}{x} + 4) e^{5 x} \\ \frac{1}{x} + 2 (\frac{2}{x} + 4) \end{matrix}] d x$
$= \frac{1}{5} [\begin{matrix} - 2 e^{- 5 x} & 1 \\ e^{- 5 x} & 2 \end{matrix}] [\begin{matrix} \frac{4}{5} e^{5 x} \\ 8 x + 5 \log x \end{matrix}]$
$= \frac{1}{5} [\begin{matrix} - \frac{8}{5} + 8 x + \log x \\ \frac{4}{5} + 16 x + 2 \log x \end{matrix}]$
General solution:
$Y (x) = Y_{c} (x) + Y_{p} (x)$
Therefore the answer is
$Y (x) = c_{1} [\begin{matrix} - 2 \\ 1 \end{matrix}] e^{- 5 x} + c_{2} [\begin{matrix} 1 \\ 2 \end{matrix}] e^{0 \cdot x} + \frac{1}{5} [\begin{matrix} - \frac{8}{5} + 8 x + \log x \\ \frac{4}{5} + 16 x + 2 \log x \end{matrix}]$

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