How do I solve this differential equation that are related?

Question

Suppose I have these $2$ equations:

$$\frac{dy}{dt} = -y + 3x$$ $$\frac{dx}{dt} = 4x - 2y.$$

Then I want to show that I can reduce them into a single ODE in $2$nd order expressed for $y(t)$. What I have done is that I found these by differentiating:

$$\frac{d^2y}{dt^2} = 5y - 9x$$ $$\frac{d^2x}{dt^2} = 10x -6y$$

and dividing these two equations together, I can get $y''(x)$ but not in $y(t)$ form. I tried to integrate from the very first ODE, but it gives me further algebraic expressions that I don't think is helpful.

Any hints will be appreciated.

Answer 1

According to the first equation, $3x= y+y'$, and deriving, $3x' = y' + y''$. Substitute both in the second equation and you will have a second order ODE in $y$.

Answer 2

from the second equation we get $$y=-\frac{1}{2}x'+2x$$ so we have $$y'=\frac{1}{2}x'+2x+3x$$ and we get $$y'=-\frac{1}{2}x''+2x'$$ and we get $$-\frac{1}{2}x''+2x'=\frac{1}{2}x'-2x+3x$$ can you solve this?

Answer 3

I think this would be better method for getting $y''(t)$: We rewrite the above system of equation into this form \begin{equation} \begin{cases} 3x-(D+1)y=0\\ (D-4)x+2y=0 ~~~~~~~~~~~~~~~~~where~~~ D\equiv \dfrac {d}{dt}. \end{cases} \end{equation} Then multiply first equation by $(D-4)$ and second equation by $3$ and subtracting we get $(D^2-3D+2)y=0$ which is equivalent to

$y''(t)-3y'(t)+2y(t)=0.$

Also from this ODE you can directly find $y(t)$.

Answer 4

$$ \frac{\mathrm{d}}{\mathrm{d}t}\begin{bmatrix}x\\y\end{bmatrix}= \begin{bmatrix}4&-2\\3&-1\end{bmatrix} \begin{bmatrix}x\\y\end{bmatrix} $$ Using the Jordan Decomposition, we get $$ \begin{align} \exp\left(t\begin{bmatrix}4&-2\\3&-1\end{bmatrix}\right) &=\begin{bmatrix}2&1\\3&1\end{bmatrix} \exp\left(t\begin{bmatrix}1&0\\0&2\end{bmatrix}\right) \begin{bmatrix}-1&1\\3&-2\end{bmatrix}\\ &=\begin{bmatrix}2&1\\3&1\end{bmatrix} \begin{bmatrix}e^t&0\\0&e^{2t}\end{bmatrix} \begin{bmatrix}-1&1\\3&-2\end{bmatrix}\\ &=\begin{bmatrix}2&1\\3&1\end{bmatrix} \begin{bmatrix}-e^t&e^t\\3e^{2t}&-2e^{2t}\end{bmatrix}\\ &=\begin{bmatrix}3e^{2t}-2e^t&2e^t-2e^{2t}\\3e^{2t}-3e^t&3e^t-2e^{2t}\end{bmatrix}\\ \end{align} $$ Therefore, $$ \begin{bmatrix}x\\y\end{bmatrix} =\begin{bmatrix}3e^{2t}-2e^t&2e^t-2e^{2t}\\3e^{2t}-3e^t&3e^t-2e^{2t}\end{bmatrix} \begin{bmatrix}x_0\\y_0\end{bmatrix} $$