How to solve this (simple) non-linear system of 2 variables I am trying to solve the following system:
$ x(\sin{t} - \lambda \cos{t}) - y(\lambda \sin{t} + \cos{t}) + \lambda K e^{\lambda t} = 0$
$ x(\cos{t} + \lambda \sin{t}) - y(\lambda \cos{t} - \sin{t}) + \lambda^2 K e^{\lambda t} = 0$
I need to get a solution $x=x(t), y=y(t)$.
I have used Mathematica to get the solution $x=-e^{\lambda t}K\sin{t}, y= e^{\lambda t}K\cos{t}$ but I don't know how to get there by hand
Edit: As noticed in the answer, the valid solution is $x=\lambda e^{\lambda t}K\sin{t}, y= \lambda e^{\lambda t}K\cos{t}$
 A: Nota Bene: It appears to me that the solutions for $x(t)$, $y(t)$ posited by the OP in the question as stated are off by a factor of $\lambda$, as is (hopefully) made clear below.  Having stated this caveat:
According to the above remarks, we may not be able to get theer by hand; however, traveling by foot we make the following journey:
First of all note that, though the title advertises a "non-linear system of $2$ variables", the presented set of equations is really linear in the unknowns $x$ and $y$,  viz.,
$x(\sin{t} - \lambda \cos{t}) - y(\lambda \sin{t} + \cos{t}) + \lambda K e^{\lambda t} = 0, \tag{1}$
$x(\cos{t} + \lambda \sin{t}) - y(\lambda \cos{t} - \sin{t}) + \lambda^2 K e^{\lambda t} = 0, \tag{2}$
exhibits no power of $x$ and/or $y$, jointly and/or severally, of degree higher than the first.  Indeed, we may write (1), (2) as a matrix-vector equation in $x$ and $y$, having coefficients which are (admittedly non-linear) functions of $t$:  let $A(t)$ be the $2 \times 2$ matrix
$A = \begin{bmatrix} (\sin t - \lambda \cos t) & -(\lambda \sin t + \cos t) \\ (\cos t + \lambda \sin t) & -(\lambda \cos t - \sin t) \end{bmatrix} \tag{3}$
and $\mathbf r$ be the vector
$\mathbf r = \begin{pmatrix}x \\ y \end{pmatrix}; \tag{4}$
then (1), (2) may be written
$A(t) \mathbf r = \begin{pmatrix} -\lambda K e^{\lambda t} \\ -\lambda^2 K e^{\lambda t} \end{pmatrix} = -\lambda K \begin{pmatrix} e^{\lambda t} \\ \lambda e^{\lambda t} \end{pmatrix}. \tag{5}$
To reiterate:  though the coefficients and non-homogeneous term are decidedly non-linear functions of $\lambda$ and $t$, this system, being linear in $\mathbf r$, may be solved by elementary methods.  Indeed, for any $2 \times 2$ matrix $B$:
$B = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \tag{6}$
with $\det(B) = ad - bc \ne 0$ we have the well-known formula for $B^{-1}$:
$B = (\det(B))^{-1} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}; \tag{7}$
we apply this formula to $A(t)$ in concert with the well-known plug and grind methodology.  First, we compute $\det(A(t))$:
$\det(A(t)) = -(\sin t - \lambda \cos t)(\lambda \cos t - \sin t) + (\lambda \sin t + \cos t)(\cos t + \lambda \sin t)$
$=  (\lambda \cos t - \sin t)^2 + (\lambda \sin t + \cos t)^2 = \lambda^2 + 1, \tag{8}$
whence, in accord with (7),
$A^{-1}(t) = (\lambda^2 + 1)^{-1}\begin{bmatrix} (\sin t - \lambda \cos t) & (\cos t + \lambda \sin t) \\ -(\cos t + \lambda \sin t) &  (\sin t - \lambda \cos t) \end{bmatrix}; \tag{9}$
finally, from (5) we have
$\begin{pmatrix} x \\ y \end{pmatrix} = \mathbf r = -\lambda K A^{-1}(t) \begin{pmatrix} e^{\lambda t} \\ \lambda e^{\lambda t} \end{pmatrix}; \tag{10}$
combining (10) with (9) and turning the crank on the algebra machine yields
$x(t) = -\lambda K e^{\lambda t}\ \sin t, \tag{11}$
$y(t) =  \lambda K e^{\lambda t} \cos t, \tag{12}$
which I do believe can be readily checked by direct substitution into (5).  Now, having checked and re-checked this work rather carefully for a couple hours, I think I've got it right; I hope I've got it right.  I would really appreciate the efforts of anyone who wants to cross-check my calculations.  The only way I can explain the discrepancy is that a factor of $\lambda$ was dropped from each component of the inhomogeneous term on the right-hand side of (5), but such speculation on my part is exactly that.  Meanwhile, the mystery stands, though it appears to me that my solution is correct.
Hope this helps.  Cheers,
and as always,
Fiat Lux!!!
