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I recently came across a question in which we had to to solve the set of differential equations:

$10dx/dt+x+y/2=0 $ and $6d(x-y)/dt= y$

I tried a lot to solve these equations but I was unable to do so. I tried adding them eliminating $t$, but I couldn't even solve the resulting differential equation. Substituting $y=mx$ resulted in a messy calculation. Would someone please help me to solve this question?

shsh23
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1 Answers1

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Hint:

This is a linear system of differential equations. Rewrite it in standard form: \begin{cases} x'= -\dfrac 1{10}x-\dfrac 1{20}y \\[1ex] y'=-\dfrac 1{10}x -\dfrac{13}{60}y \end{cases} (subtracting the 2nd equation from the 1st), which you can write in matrix form: $$\begin{bmatrix}x'\\y'\end{bmatrix}=A\begin{bmatrix}x\\ y\end{bmatrix}, \qquad A=\begin{bmatrix}-\frac 1{10}-\frac 1{20}\\-\frac 1{10}-\frac{13}{60}\end{bmatrix}$$ The solution is given by $$\begin{bmatrix}x(t)\\ y(t)\end{bmatrix}=\exp(At)\begin{bmatrix}x(0)\\ y(0)\end{bmatrix}$$ so all you have to do is calculating the exponential of $At$, which requires to determine a basis of eigenvectors.

Bernard
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