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I have these sets of equations,

$x'_1(t)=x_2(t)$

$x'_2(t)=-x_2(t)-\lambda _2(t)$

$\lambda '_1(t)=-x_1(t)$

$\lambda '_2(t)=-\lambda _1(t)+\lambda _2(t)$

where $x^T(0)=\left[2\quad 5\right]$ and $\lambda _i^T(t=tf)=0$ where $tf=5$

I know this is simply a form of;

$x'(t)=Ax$

Edit: Added matrix A

where $A=\left[ \begin{array}\\ 0&1&0&0\\ 0&-1&0&-1\\ -1&0&0&0\\ 0&0&-1&1 \end{array}\right]$

But couldn't solve.

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    @Moo I have added the matrix A – 2017-01-14
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    I couldn't make the mathematica simplify the result. Thus can't conclude the solution :/ – 2017-01-14
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    Same command sucks in Mathematica whcih gives something 1/2 RootSum[ 1 - #1^2 + #1^4 &, (-E^(t #1) + E^(t #1) #1^2)/(-1 + 2 #1^2) &] – 2017-01-14
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    Now it is something like this -> http://www.wolframalpha.com/input/?i=E%5E((-0.866025+-++++++0.5+I)+t)+((0.25+%2B+++++++0.144338+I)+%2B+(0.25+-+0.144338+I)+E%5E((0.+%2B+1.+I)+t)+%2B+(0.25+-++++++++0.144338+I)+E%5E(+++++1.73205+t)+%2B+(0.25+%2B+0.144338+I)+E%5E((1.73205+%2B+1.+I)+t)) – 2017-01-14
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    Thanks for your patient responds. – 2017-01-14

1 Answers 1

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The solution to a matrix differential equation $x'(t)=Ax(t)$ is known to be

$$x(t)=e^{tA}x_0,$$

where $e^{X}$ denotes the matrix exponential and $x_0$ is the initial value.

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    I have calculated it in mathematica whcih resulted 4x4 =16 equations with complex numbers. Whcih doesn't seem to be right. – 2017-01-14
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    I don't know what you mean. In your case the matrix $A$ clearly does not contain complex numbers (though some values of $x(t)$ possibly do, not sure about that). You can evaluate the matrix exponential e.g. using the matlab command expm. – 2017-01-14
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    eigenvalues are complex and distinct so the matrix exponential. – 2017-01-14