# Mathematica can't solve this ODE system

I have a differential equations.But mathematica dosen't work.


eq1 = a'[x] - 1/2*a[x] - 1/2*Exp[x]*b[x];

eq2 = b'[x] + 1/2*b[x] - 1/2*Exp[x]*a[x];

DSolve[{eq1 == 0, eq2 == 0}, {a[x], b[x]}, x]



But it return itself.


Out[1] = DSolve[{-(a[x]/2) - 1/2 E^x b[x] + Derivative[1][a][x] ==
0, -(1/2) E^x a[x] + b[x]/2 + Derivative[1][b][x] == 0}, {a[x],
b[x]}, x]

$$$$

• That simply means that Mathematica can't find an analytical solution. Add initial conditions and use NDSolve instead. Commented Nov 11, 2022 at 5:05
• Maple 2022.2 answers $$\left\{a\! \left(x\right) {=} {\mathrm e}^{\frac{x}{2}} \left(\textit{_}\mathit{C1} \mathrm{BesselI}\! \left(0,\frac{{\mathrm e}^{x}}{2}\right)+\textit{_}\mathit{C2} \mathrm{BesselK}\! \left(0,\frac{{\mathrm e}^{x}}{2}\right)\right), \\ b\! \left(x\right){=}{\mathrm e}^{\frac{x}{2}} \left(\mathrm{BesselI}\! \left(1,\frac{{\mathrm e}^{x}}{2}\right) \textit{_}\mathit{C1}-\mathrm{BesselK}\! \left(1,\frac{{\mathrm e}^{x}}{2}\right) \textit{_}\mathit{C2}\right)\right\} .$$ Commented Nov 11, 2022 at 6:42
• Manually elimination of b[x] gives a solvable ode Commented Nov 11, 2022 at 7:28

Remove[eq1, eq2, solA, A, solB, B, a, b]
eq1 = a'[x] - 1/2*a[x] - 1/2*Exp[x]*b[x];
eq2 = b'[x] + 1/2*b[x] - 1/2*Exp[x]*a[x];


You can do this. Note that each of the expressions has a parameter, but not its derivative. From the second expression get $$a[x]$$:

solA = Solve[eq2 == 0, a[x]]


Then substitute this into the first ODE and solve it like one ordinary differential equation for the opposite variable (in this case for b[x]):

A = a[x] /. solA // First
ReplaceAll[eq1, {a[x] -> A, a'[x] -> D[A, x]}]
solB = DSolve[% == 0, b[x], x] // FullSimplify


Then substitute into the first equation and get a couple of solutions $$A,B$$:

B = ReplaceAll[A, {solB[[1]] // First, D[solB[[1]] // First, x]}] // FullSimplify


The same is done and vice versa.

Remove[eq1, eq2, solA, A, solB, B, a, b]
eq1 = a'[x] - 1/2*a[x] - 1/2*Exp[x]*b[x];
eq2 = b'[x] + 1/2*b[x] - 1/2*Exp[x]*a[x];


From the first expression get $$b[x]$$:

solB = Solve[eq1 == 0, b[x]]


hen substitute this into the second ODE and solve it like one ordinary differential equation for the opposite variable (in this case for a[x]):

B = b[x] /. solB // First

ReplaceAll[eq2, {b[x] -> B, b'[x] -> D[B, x]}]

solA = DSolve[% == 0, a[x], x] // FullSimplify


Then substitute into the second equation and get a couple of solutions $$A,B$$:

A = ReplaceAll[B, {solA[[1]] // First, D[solA[[1]] // First, x]}] // FullSimplify
`

The most interesting thing is that such a naive approach gives, perhaps, two pairs of solutions separately. In the figure below, the solution plots for each of the cases are symmetrical about the x-axis.