Can You Help Me Solve This Non-Elementary Differential Equation?

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SUMMARY

The discussion centers on solving the non-elementary differential equation 2y'' + 2y' + y = 4√x. The characteristic equation derived is x² + x + ½, leading to complex roots of -½ ± ½i. The homogeneous solution is correctly identified as a sin/cos combination, but the participants note an omission of the exponential term in the general solution. The method of variation of parameters is suggested for finding the particular solution, with integrals u₁ and u₂ defined for further computation.

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Homework Statement


2y\prime\prime +2y\prime + y = 4 \sqrt{x}

The Attempt at a Solution



charecteristic equation: x^2+x+\frac{1}{2}
roots: \frac{1}{2}\pm\frac{1}{2}i

homogenous solution: a \sin{\frac{1}{2}x} + b \cos{\frac{1}{2}x}

Wronskian: \left(\begin{array}{cc}\sin{\frac{1}{2}x}&\cos{\frac{1}{2}x}\\\frac{1}{2}\cos{\frac{1}{2}x}&-\frac{1}{2}\sin{\frac{1}{2}x}\end{array}\right) = -\frac{1}{2}

It would be nice to know if up to here, everyone else gets the same answer. After this I get into non elementary functions which is no where near the level of difficulty included within the course I'm taking. I haven't attempted to solve what comes next, frankly because I have no clue where to even begin. I think I probably made a mistake above, if someone can point me in the right direction that would be awesome. The method used is variation of parameters.

y_{1}=\sin{\frac{1}{2}x}

y_{2}=\cos{\frac{1}{2}x}

u_{1} = \int{4\sqrt{x}\cos{\frac{1}{2}x}dx}

u_{2} = \int{-4\sqrt{x}\sin{\frac{1}{2}x}dx}

ick
 
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With x^2+x+\frac{1}{2} as your characteristic equation, I think your roots should be -\frac{1}{2}\pm\frac{1}{2}i. And then with the complex roots, the homogeneous solution should be in the form of

y=c_1e^{\lambda t}\cos(\mu t)+c_2e^{\lambda t}\sin(\mu t)

where the roots come from the form of \lambda \pm i\mu. Your form was missing the exponential term. Hope that helps a bit.
 
Yeah, jeffreydk is right. Your roots are incorrect and you neglected the exponential.
 

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