Solve for $x$ and $y$ where $x^{\prime} = x + y + t$ and $y^{\prime} = 2x - t$. Given $x = 0$, $y = 1$ at $t = 1$.

Question

Given the equations \begin{align} x' &= x + y + t \\ y' &= 2 x - t \end{align} with $x(1) = 0$ and $y(1) = 1$, how can solutions be obtained?

I tried solving it but I did not get any idea.

Answer 1

There are two standard approaches. One is to differentiate the first and plug the second into it. That gives a second order equation in one variable. Do you know how to handle those?

The second is to formulate it as a vector equation. Here it is $$\begin {pmatrix}x\\y \end {pmatrix}'=\begin {pmatrix}1&1\\2&0 \end {pmatrix}\begin {pmatrix}x\\y \end {pmatrix}+\begin {pmatrix}t\\-t \end {pmatrix} $$ which is like an inhomogeneous first order equation, but working on vector quanties. You can diagonalize the $2 \times 2$ matrix to make exponentiating it easy.

Answer 2

For $x^{\prime} = x + y + t$ and $y^{\prime} = 2x - t$, Given $x(1) = 0$, $y(1) = 1$, then \begin{align} x'' &= x' + y' +1 = x' + 2 x - t + 1. \end{align} Now, $x'' - x' - 2 x = 1-t$ has solution $$x(t) = c_1 e^{-t} + c_2 e^{2 t} + \frac{2 t -3}{4}.$$ This also leads to \begin{align} y(t) &= 2 \int x(u) \, du - \frac{t^{2}}{2} + d_{0} \\ &= - 2 c_{1} \, e^{-t} + c_{2} \, e^{2 t} - \frac{t^{2} - 3 t}{2} + d_{0}. \end{align} From here one can use the conditions $x(1) = 0$ and $y(1) = 1$ to find the constants. This yields \begin{align} 0 &= c_{1} e^{-1} + c_{2} e^{2} - \frac{1}{4} \\ 1 &= -2 c_{1} e^{-1} + c_{2} e^{2} + 1 + d_{0} \end{align} and \begin{align} x(t) &= \frac{1}{12} \, \left[ (8 d_{0} + 1) \, e^{- (t-1)} + 2(2d_{0}+1) \, e^{2(t-1)} + 6 t - 9 \right] \\ y(t) &= \frac{1}{6} \, \left[ - (8 d_{0} +1) \, e^{-(t-1)} + (2 d_{0} +1) \, e^{2(t-1)} - 3 t^{2} + 9 t + 6 d_{0} \right]. \end{align}

Since the constant of integration, $d_{0}$, either needs another boundary condition, or can be arbitrary, then consider the case when $d_{0} = 0$ for which the solutions are \begin{align} x(t) &= \frac{1}{12} \, \left[ e^{- (t-1)} + 2\, e^{2(t-1)} + 6 t - 9 \right] \\ y(t) &= \frac{1}{6} \, \left[ - e^{-(t-1)} + e^{2(t-1)} - 3 t^{2} + 9 t \right]. \end{align}