Differential Equations with set of equations

Question

The functions $f(x)$ and $g(x)$ satisfy $f(0) = 3,$ $g(0) = -4,$ and \begin{align*} f'(x) &= 7f(x) + 2g(x), \\ g'(x) &= -4f(x) + g(x). \end{align*}Find $f(x).$

I've tried substituting values for f(x) and g(x) but haven't gotten anywhere. Is there another method?

Edit: I haven't worked with eigenvectors yet

Answer 1

You can write it as a linear system

$$\begin{pmatrix}f\\g\end{pmatrix}' = \begin{pmatrix}7 & 2\\-4 & 1\end{pmatrix}\begin{pmatrix}f\\g\end{pmatrix}$$

To me this would suggest diagonalising the coeffient matrix and write it as some $ADA^{-1}$ where $D$ is diagonal. The diagonal system is easy: $f'=af$ you can do with given $f(0)$, etc. and then compute back what the solutions are as linear combinations of those.

Answer 2

Yes! Write it in matrix form:

$$ \tag{1} \label{DE} \begin{bmatrix} f \\ g \end{bmatrix}' = \begin{bmatrix} 7 & 2 \\ -4 & 1 \end{bmatrix} \begin{bmatrix} f \\ g \end{bmatrix} $$ the general solution is given by $$ \begin{bmatrix} f(x) \\ g(x) \end{bmatrix} = \exp(tA) \begin{bmatrix} f(x_0) \\ g(x_0) \end{bmatrix} $$ where $A$ is 2x2 the matrix in \eqref{DE}. In order to evaluate $\exp(tA)$ you need to calculate the eigenvectors and eigevalues of $A$. The charateristic polynomial is $$ p_A(\lambda) = (\lambda-7)(\lambda-1) + 8 = \lambda^2 -8\lambda +15 = (\lambda-5)(\lambda-3), $$ hence $\lambda_1 = 3, \lambda_2 = 5$. Next we need to find the corresponding eigenvectors. For $\lambda_1 = 3$, we have $v_1 = (1,-2)^T $, for $\lambda = 5$, we have $v_2 = (1,-1)^T$. And so the solution is given by $$ \begin{bmatrix} f(x) \\ g(x) \end{bmatrix} = c_1e^{3x}v_1 + c_2e^{5x}v_2. $$ You can find $c_1$ and $c_2$ using the initial conditions \begin{align} 3 & = c_1+c_2 \\ -4 &= -2c_1 -c_2 \end{align}

Answer 3

You can also insert $2g= f'-7f$ into the second equation to find a scalar equation of order 2 for $f$, $$ f''-7f'=2g'=-8f+2g=-8f+(f'-7f),\\ f''-8f'+15f=0. $$ which can be easily solved via its characteristic polynomial.