The solution space has multiple values, not just one. The solver has parameterized the solution space for you. There are multiple complex roots possible. E.g., a simpler example:
>> syms Y X n
>> eqv = Y==X^n;
>> solve(eqv,X)
Warning: Solutions are parameterized by the symbols: l. To include parameters and conditions in the
solution, specify the 'ReturnConditions' value as 'true'.
> In solve>warnIfParams (line 475)
In solve (line 357)
Warning: Solutions are valid under the following conditions: 0 <= real(n)*(pi - imag(log(Y)/n) +
2*l*pi*real(1/n)) & 0 < real(n) & 0 < real(n)*(pi + imag(log(Y)/n) - 2*l*pi*real(1/n)) & in(l,
'integer') | real(n)*(pi - imag(log(Y)/n) + 2*l*pi*real(1/n)) <= 0 & real(n) < 0 & real(n)*(pi +
imag(log(Y)/n) - 2*l*pi*real(1/n)) < 0 & in(l, 'integer'). To include parameters and conditions in the
solution, specify the 'ReturnConditions' value as 'true'.
> In solve>warnIfParams (line 482)
In solve (line 357)
ans =
Y^(1/n)*exp(-(pi*l*2i)/n)
The obvious solution by inspection is X = Y^(1/n). That happens to be the solution for the parameter choice l=0, but there are other solutions for other values of l also. Essentially it includes the n'th roots of 1 as a multipler in this case (the exp stuff).
E.g., to get rid of those complex n'th roots of 1:
>> solve(eqv,X,'Real',true)
Warning: Solutions are valid under the following conditions: 0 < Y. To include parameters and
conditions in the solution, specify the 'ReturnConditions' value as 'true'.
> In solve>warnIfParams (line 482)
In solve (line 357)
ans =
Y^(1/n)
And thus your code could be:
>> syms V1 V2 p1 p2 n
>> eqv = p1*V1^n==p2*V2^n;
>> solve(eqv,V2,'Real',true)
Warning: Solutions are valid under the following conditions: 0 < (V1^n*p1)/p2 &
in(((V1^n*p1)/p2)^(1/n), 'real'). To include parameters and conditions in the solution, specify the
'ReturnConditions' value as 'true'.
> In solve>warnIfParams (line 482)
In solve (line 357)
ans =
((V1^n*p1)/p2)^(1/n)
Best Answer