Answered: Solve the initial-value problem x' =…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

Solve the initial-value problem x' = Bx, x(0) = x^I for the following B and x^I^.

Expert Solution

Step 1

Part A)-

Given:

Given matrix $B = (\begin{matrix} - 2 & - 1 \\ 1 & - 4 \end{matrix}), x' = (\begin{matrix} 3 \\ 0 \end{matrix})$

We want to find solution of $x' = B x$

Solution:

We want to find solution of the given initial value problem

Here matrix $B = (\begin{matrix} - 2 & - 1 \\ 1 & - 4 \end{matrix})$

consider,

$d e t (B - λ I) = 0 \Rightarrow |\begin{matrix} - 2 - λ & - 1 \\ 1 & - 4 - λ \end{matrix}| = 0 \Rightarrow (- 2 - λ) (- 4 - λ) - (- 1) = 0 \Rightarrow (2 + λ) (4 + λ) + 1 = 0 \Rightarrow 8 + 2 λ + 4 λ + λ^{2} + 1 = 0 \Rightarrow λ^{2} + 6 λ + 9 = 0 \Rightarrow {(λ + 3)}^{2} = 0 \Rightarrow λ = - 3, - 3$

Therefore here we have eigenvalue with multiplicity 2

Now we find corresponding eigenvector

For $λ = - 3$ consider,

$(\begin{matrix} - 2 - (- 3) & - 1 \\ 1 & - 4 - (- 3) \end{matrix}) = (\begin{matrix} 1 & - 1 \\ 1 & - 1 \end{matrix}) \Rightarrow r r e f (\begin{matrix} 1 & - 1 \\ 1 & - 1 \end{matrix}) \overset{R_{2} - R_{1}}{\to} (\begin{matrix} 1 & - 1 \\ 0 & 0 \end{matrix})$

Therefore to find eigenvector

$(\begin{matrix} 1 & - 1 \\ 0 & 0 \end{matrix}) (\begin{matrix} v_{1} \\ v_{2} \end{matrix}) = (\begin{matrix} 0 \\ 0 \end{matrix}) \Rightarrow v_{1} = v_{2}, v_{2} i s a r b r i t r a r y$

if $v_{2} = t \Rightarrow v_{1} = t$

Therefore eigenvector $V = (\begin{matrix} t \\ t \end{matrix}) = (\begin{matrix} 1 \\ 1 \end{matrix}) t$

Now we find generalized eigenvector as following

Consider,

$(A - λ I) w = V (\begin{matrix} - 2 - (- 3) & - 1 \\ 1 & - 4 - (- 3) \end{matrix}) (\begin{matrix} w_{1} \\ w_{2} \end{matrix}) = (\begin{matrix} 1 \\ 1 \end{matrix}) \Rightarrow (\begin{matrix} 1 & - 1 \\ 1 & - 1 \end{matrix}) (\begin{matrix} w_{1} \\ w_{2} \end{matrix}) = (\begin{matrix} 1 \\ 1 \end{matrix}) A u g m e n t e d m a t r i x, [\begin{matrix} 1 & - 1 \\ 1 & - 1 \end{matrix} \begin{matrix} 1 \\ 1 \end{matrix}] \overset{R_{2} - R_{1}}{\to} [\begin{matrix} 1 & - 1 \\ 0 & 0 \end{matrix} \begin{matrix} 1 \\ 0 \end{matrix}]$

So we get

$(\begin{matrix} 1 & - 1 \\ 0 & 0 \end{matrix}) (\begin{matrix} w_{1} \\ w_{2} \end{matrix}) = (\begin{matrix} 1 \\ 0 \end{matrix}) \Rightarrow w_{1} - w_{2} = 1 \Rightarrow w_{1} = 1 + w_{2}, w_{2} i s a r b r i t r a r y$

Therefore if $w_{2} = t \Rightarrow w_{1} = 1 + t$

Hence generalized eigenvector is $W = (\begin{matrix} 1 + t \\ t \end{matrix}) = (\begin{matrix} 1 \\ 0 \end{matrix}) + (\begin{matrix} 1 \\ 1 \end{matrix}) t$

Therefore we know that For repeated real eigenvalue $λ$ with multiplicity 2 and with eigenvector $V$ the general solution is of the form $X = C_{1} e^{λ t} V + C_{2} e^{λ t} (t W + V)$ where V is generalized eigenvector

Therefore solution of the given system is

$X = C_{1} e^{λ t} V + C_{2} e^{λ t} (t W + V) \Rightarrow X = C_{1} e^{- 3 t} (\begin{matrix} 1 \\ 1 \end{matrix}) + C_{2} e^{- 3 t} ((\begin{matrix} 1 \\ 1 \end{matrix}) t + (\begin{matrix} 1 \\ 0 \end{matrix}))$

Now given initial condition is $X (\begin{matrix} 0 \\ 0 \end{matrix}) = X' = (\begin{matrix} 3 \\ 0 \end{matrix})$ So applying this condition we get

$X (\begin{matrix} 0 \\ 0 \end{matrix}) = C_{1} e^{- 3 (0)} (\begin{matrix} 1 \\ 1 \end{matrix}) + C_{2} e^{- 3 (0)} ((\begin{matrix} 1 \\ 1 \end{matrix}) (0) + (\begin{matrix} 1 \\ 0 \end{matrix})) \Rightarrow (\begin{matrix} 3 \\ 0 \end{matrix}) = (\begin{matrix} C_{1} + C_{2} \\ C_{1} \end{matrix}) \Rightarrow C_{1} + C_{2} = 3, C_{1} = 0 \Rightarrow C_{2} = 3$

Hence the general solution of the given initial value problem is

$X = 3 e^{- 3 t} ((\begin{matrix} 1 \\ 1 \end{matrix}) t + (\begin{matrix} 1 \\ 0 \end{matrix}))$

Answer:

The general solution of the given initial value problem is

$X = 3 e^{- 3 t} ((\begin{matrix} 1 \\ 1 \end{matrix}) t + (\begin{matrix} 1 \\ 0 \end{matrix}))$

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