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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?

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2 Answers 2

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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

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  • $\begingroup$ Thanks! Nasser. Appreciate the explanation. $\endgroup$
    – Ashish
    Commented Apr 30, 2020 at 17:55
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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  *)
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