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If I have differential equation of fourth order with four solutions $\alpha _1,\alpha _2,-\alpha _1,-\alpha _2 $ where we have appearing the solutions in the form $e^{\text{$i\alpha $}_1 x},e^{\text{$i\alpha $}_2 x},e^{-\text{$i\alpha $}_1 x},e^{-\text{$i\alpha $}_2 x}$. Is it possible to force Mathematica to reduce order of this equation on 2 but to keep two positive solutions $ \alpha _1,\alpha _2 $ which should be solutions of this equation

 DSolve[Y''''[x] + a1 Y''[x] + a2*Y[x] == 0, Y[x], x]

output

$$ Y(x)\to c_1 e^{\frac{x \sqrt{-\sqrt{\text{a1}^2-4 \text{a2}}-\text{a1}}}{\sqrt{2}}}+c_2 e^{-\frac{x \sqrt{-\sqrt{\text{a1}^2-4 \text{a2}}-\text{a1}}}{\sqrt{2}}}+c_3 e^{\frac{x \sqrt{\sqrt{\text{a1}^2-4 \text{a2}}-\text{a1}}}{\sqrt{2}}}+c_4 e^{-\frac{x \sqrt{\sqrt{\text{a1}^2-4 \text{a2}}-\text{a1}}}{\sqrt{2}}} $$

So solutions are now in the form $e^{\text{$i\beta $}_1 x},e^{-\text{$i\beta $}_1 x}$

Now I want $ \alpha _1,\alpha _2 $ to be equal with two solutions of reduced equation ${\text{$\beta $}_1 },{-\text{$\beta $}_1 }$ on second order which has solutions $\sqrt{-\text{b1}-\text{b2}}$ and $-\sqrt{-\text{b1}-\text{b2}}$. Or to make other possible combination but to neglect two of them, maybe other combination $ ({+/-\alpha _i,+/-\alpha _j})$ where $ i,j=1,2. $ Reduced equation

DSolve[Y''[x] + b1 Y[x] + b2*Y[x] == 0, Y[x], x]

$$ Y(x)\to c_1 e^{x \sqrt{-\text{b1}-\text{b2}}}+c_2 e^{x \left(-\sqrt{-\text{b1}-\text{b2}}\right)} $$

Maybe idea can be first to extract solutions of first equation and then make conditions for second. It is good if I can keep two positive solutions from first equation. Maybe Mathematica has some functions for that or any help can be good.

Atention: Constants C1,... can be different.

share|improve this question
    
Because the $\alpha_j$ are multiplied by $i = \sqrt{-1}$, there is no consistent sense in which you can determine which one of each pair of $\pm \alpha_j$ are "positive"--pick either one. Moreover, the solution you show does not match the form you describe at the beginning. Given that, I cannot figure out what you're trying to accomplish. –  whuber Jan 30 '13 at 19:30
    
@whuber, I want to reduced order of equation 1 on order 2 and to keep 2 solutions from first equation, i is printing mistake, I changed –  Pipe Jan 30 '13 at 20:03
    
You are lacking suitable criteria for selecting the solutions. One good one might be the behavior at $\infty$, for instance. Without such criteria, the question is not resolvable, because a homogeneous linear fourth order ODE does not have just four solutions: it has an entire four-dimensional space of solutions. You need to tell us how to identify the two-dimensional subspace you might be looking for. –  whuber Jan 30 '13 at 20:34
    
@whuber Dr whuber when I am speaking about solutions, I think about coefficient with x in e^alpha*x, so if coefficients are the same in first and second reduced equation my problem is solved –  Pipe Jan 30 '13 at 20:48
    
@ whuber you made me confuse, values under the root are already negative naturally and solutions are in the form e^(+/-)i*alpha*x and e^(+/-)i*beta*x.. alpha and beta should be the same –  Pipe Jan 30 '13 at 21:13
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