System of Differential Equations discrepancy

sol = DSolve[ {x'[t] == -1/10 x[t] + 3/40 y[t], y'[t] == 1/10 x[t] - 1/5 y[t] }, {x[t], y[t]}, t] 

\begin{align*} x(t) =& \frac{1}{4} c_1 e^{-t/4} \left(3 e^{t/5}+1\right)+\frac{3}{8} c_2 e^{-t/4} \left(e^{t/5}-1\right)\\ y(t) =& \frac{1}{2} c_1 e^{-t/4} \left(e^{t/5}-1\right)+\frac{1}{4} c_2 e^{-t/4} \left(e^{t/5}+3\right) \end{align*}

is the same as your solution.

Proof:

\begin{align*} x\left( t\right) & =\frac{1}{4}e^{-\frac{t}{4}}\left( 1+3e^{\frac{t}{5} }\right) c_{1}+\frac{3}{8}e^{-\frac{t}{4}}\left( -1+e^{\frac{t}{5}}\right) c_{2}\\ y\left( t\right) & =\frac{1}{2}e^{-\frac{t}{4}}\left( -1+e^{\frac{t}{5} }\right) c_{1}+\frac{1}{4}e^{-\frac{t}{4}}\left( 3+e^{\frac{t}{5}}\right) c_{2} \end{align*} Hence \begin{align*} x\left( t\right) & =\frac{1}{4}\left( e^{-\frac{t}{4}}+3e^{\frac{-t}{20} }\right) c_{1}+\frac{3}{8}\left( -e^{-\frac{t}{4}}+e^{\frac{-t}{20}}\right) c_{2}\\ y\left( t\right) & =\frac{1}{2}\left( -e^{-\frac{t}{4}}+e^{\frac{-t}{20} }\right) c_{1}+\frac{1}{4}\left( 3e^{-\frac{t}{4}}+e^{\frac{-t}{20}}\right) c_{2} \end{align*} Or \begin{align*} x\left( t\right) & =e^{-\frac{t}{4}}\left( \frac{1}{4}c_{1}-\frac{3} {8}c_{2}\right) +e^{\frac{-t}{20}}\left( \frac{3}{4}c_{1}+\frac{3}{8} c_{2}\right) \\ y\left( t\right) & =e^{-\frac{t}{4}}\left( -\frac{1}{2}c_{1}+\frac{3} {4}c_{2}\right) +e^{\frac{-t}{20}}\left( \frac{1}{2}c_{1}+\frac{1}{4} c_{2}\right) \end{align*} Let $\frac{1}{4}c_{1}-\frac{3}{8}c_{2}=-A$ and let $\frac{3}{4}c_{1}+\frac {3}{8}c_{2}=3B$ where $A,B$ are new constants. Hence $-\frac{1}{2}c_{1} +\frac{3}{4}c_{2}=2A$ and $\frac{1}{4}c_{1}+\frac{1}{8}c_{2}=B$ or $\frac {1}{2}c_{1}+\frac{1}{4}c_{2}=2B$ Hence the above becomes \begin{align*} x\left( t\right) & =-Ae^{-\frac{t}{4}}+3Be^{\frac{-t}{20}}\\ y\left( t\right) & =2Ae^{-\frac{t}{4}}+2Be^{\frac{-t}{20}} \end{align*} But $A,B$ are constants. We can call them anything we want. Let $A=c_{1}$ and $B=c_{2}$. The above becomes \begin{align*} x\left( t\right) & =-c_{1}e^{-\frac{t}{4}}+3c_{2}e^{\frac{-t}{20}}\\ y\left( t\right) & =2c_{1}e^{-\frac{t}{4}}+2c_{2}e^{\frac{-t}{20}} \end{align*}

Which is your solution