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Write the linear system of differential equations in matrix form then solve the system.

dx/dt = x + y

dy/dt = 4x + y

x(0) = 1, y(0) = 2

User Jpecht
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1 Answer

5 votes

In matrix form, the system is


(\mathrm d)/(\mathrm dt)\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}1&1\\4&1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}

First find the eigenvalues of the coefficient matrix (call it
\mathbf A).


\det(\mathbf A-\lambda\mathbf I)=\begin{vmatrix}1-\lambda&1\\4&1-\lambda\end{vmatrix}=(1-\lambda)^2-4=0\implies\lambda^2-2\lambda-3=0


\implies\lambda_1=-1,\lambda_=3

Find the corresponding eigenvector for each eigenvalue:


\lambda_1=-1\implies(\mathbf A+\mathbf I)\vec\eta_1=\vec0\implies\begin{bmatrix}2&1\\4&2\end{bmatrix}\begin{bmatrix}\eta_(1,1)\\\eta_(1,2)\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}


\lambda_2=3\implies(\mathbf A-3\mathbf I)\vec\eta_2=\vec0\implies\begin{bmatrix}-2&1\\4&-2\end{bmatrix}\begin{bmatrix}\eta_(2,1)\\\eta_(2,2)\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}


\implies\vec\eta_1=\begin{bmatrix}1\\-2\end{bmatrix},\vec\eta_2=\begin{bmatrix}1\\2\end{bmatrix}

Then the system has general solution


\begin{bmatrix}x\\y\end{bmatrix}=C_1\vec\eta_1e^(\lambda_1t)+C_2\vec\eta_2e^(\lambda_2t)

or


\begin{cases}x(t)=C_1e^(-t)+C_2e^(3t)\\y(t)=-2C_1e^(-t)+2C_2e^(3t)\end{cases}

Given that
x(0)=1 and
y(0)=2, we have


\begin{cases}1=C_1+C_2\\2=-2C_1+2C_2\end{cases}\implies C_1=0,C_2=2

so that the system has particular solution


\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}e^(3t)\\2e^(3t)\end{bmatrix}

User Amin Abbaspour
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