# System of Differential Equations discrepancy

Using the following code in Mathematica:

{X, Y} = {x, y} /.
DSolve[{x'[t] == -1/10 x[t] + 3/40 y[t],
y'[t] == 1/10 x[t] - 1/5 y[t]}, {x, y}, t] // FullSimplify //
First; {X[t], Y[t]} // Expand


I get the following solution:

{1/4 E^(-t/4) C[1] + 3/4 E^(-t/20) C[1] - 3/8 E^(-t/4) C[2] +
3/8 E^(-t/20) C[2], -(1/2) E^(-t/4) C[1] + 1/2 E^(-t/20) C[1] +
3/4 E^(-t/4) C[2] + 1/4 E^(-t/20) C[2]}


But if I do it by hand, I get: $$\begin{bmatrix}x\\y\end{bmatrix}=c_1 e^{-\dfrac{t}{4}} \begin{bmatrix}-1\\2\end{bmatrix}+c_2 e^{-\dfrac{t}{20}} \begin{bmatrix}3\\2\end{bmatrix}$$

So, $$x(t)=-c_1e^{-\dfrac{t}{4}}+3c_2e^{-\dfrac{t}{20}} \text{ and } y(t)=2c_1e^{-\dfrac{t}{4}}+2c_2e^{-\dfrac{t}{20}}$$

Why is there a discrepancy? How do I simplify Mathematica answer to mine?

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*}

• Thanks! Nasser. Appreciate the explanation. Commented Apr 30, 2020 at 17:55

How to find the transformation from Mathematica's solution to the OP's soluytion:

mmaSol =  (* DSolve[]'s solution *)
{1/4 E^(-t/4) C[1] + 3/4 E^(-t/20) C[1] -
3/8 E^(-t/4) C[2] + 3/8 E^(-t/20) C[2], -(1/2) E^(-t/4) C[1] +
1/2 E^(-t/20) C[1] + 3/4 E^(-t/4) C[2] + 1/4 E^(-t/20) C[2]};
opSol =  (* OP's solution *)
{-C[3] E^(-t/4) + 3 C[4] E^(-t/20),
2 C[3] E^(-t/4) + 2 C[4] E^(-t/20)};

mma2op = First@Solve[mmaSol == opSol, {C[1], C[2]}]
(*  {C[1] -> -C[3] + 3 C[4], C[2] -> 2 (C[3] + C[4])}  *)


The transformation is applied with /. (ReplaceAll):

mmaSol /. mma2op


Check:

(mmaSol /. mma2op) == opSol // Simplify
(*  True  *)