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I was asked to solve a coupled linear system like this :

$$\frac{dx}{dt} = y $$ $$\frac{dy}{dt} = x $$

I tried to solve that by making $$x dx = y dy$$ Integrating I obtain $$x^2 - y^2 = C $$ where $C$ is a constant

I think I made some mistake, can any one help me with this?

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    Because both of the equations are the same except for x and y "reversed", $\frac{dx}{dt}=x$. – 2017-02-19

3 Answers 3

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Differentiating the first identity yields $$x''=y'$$ And substituting in the second equation $$x''=x$$ Can you take it from here?

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we have $dx/dt=y$ and $dy/dt=x$ so $d²x/dt²=x$ so $x=exp(t+C)$ where C is a constant

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    @ MEKTOUBI: your "solution" is not complete ! The general solution of the equation $x''=x$ is given by: $x(t)=c_1e^t+c_2e^{-t}$. – 2017-02-17
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Well, you can use Laplace transform. So we have that:

$$ \begin{cases} \text{x}'\left(t\right)=\text{y}\left(t\right)\\ \\ \text{y}'\left(t\right)=\text{x}\left(t\right) \end{cases}\tag1 $$

Take the Laplace transform of both sides:

$$ \begin{cases} \text{s}\cdot\text{X}\left(\text{s}\right)-\text{x}\left(0\right)=\text{Y}\left(\text{s}\right)\\ \\ \text{s}\cdot\text{Y}\left(\text{s}\right)-\text{y}\left(0\right)=\text{X}\left(\text{s}\right) \end{cases}\tag2 $$

So, we also get that:

$$ \begin{cases} \text{X}\left(\text{s}\right)=\frac{\text{Y}\left(\text{s}\right)+\text{x}\left(0\right)}{\text{s}}\\ \\ \text{Y}\left(\text{s}\right)=\frac{\text{X}\left(\text{s}\right)+\text{y}\left(0\right)}{\text{s}} \end{cases}\tag3 $$

So, substitute them into each other:

$$ \begin{cases} \text{X}\left(\text{s}\right)=\frac{\frac{\text{X}\left(\text{s}\right)+\text{y}\left(0\right)}{\text{s}}+\text{x}\left(0\right)}{\text{s}}\\ \\ \text{Y}\left(\text{s}\right)=\frac{\frac{\text{Y}\left(\text{s}\right)+\text{x}\left(0\right)}{\text{s}}+\text{y}\left(0\right)}{\text{s}} \end{cases}\space\space\space\Longleftrightarrow\space\space\space\begin{cases} \text{X}\left(\text{s}\right)=\frac{\text{s}\cdot\text{x}\left(0\right)+\text{y}\left(0\right)}{\text{s}^2-1}\\ \\ \text{Y}\left(\text{s}\right)=\frac{\text{s}\cdot\text{y}\left(0\right)+\text{x}\left(0\right)}{\text{s}^2-1} \end{cases}\tag4 $$

So, using inverse Laplace transform we get that:

$$ \begin{cases} \text{x}\left(t\right)=\text{x}\left(0\right)\cdot\cosh\left(t\right)+\text{y}\left(0\right)\cdot\sinh\left(t\right)\\ \\ \text{y}\left(t\right)=\text{y}\left(0\right)\cdot\cosh\left(t\right)+\text{x}\left(0\right)\cdot\sinh\left(t\right) \end{cases}\tag5 $$