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

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In summary, the correct roots for the characteristic equation are -\frac{1}{2}\pm\frac{1}{2}i and the correct homogeneous solution is y=c_1e^{-\frac{1}{2}t}\cos{\frac{1}{2}x} + c_2e^{-\frac{1}{2}t}\sin{\frac{1}{2}x}.
  • #1
bishy
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Homework Statement


[tex] 2y\prime\prime +2y\prime + y = 4 \sqrt{x}[/tex]

The Attempt at a Solution



charecteristic equation: [tex] x^2+x+\frac{1}{2}[/tex]
roots: [tex]\frac{1}{2}\pm\frac{1}{2}i[/tex]

homogenous solution: [tex] a \sin{\frac{1}{2}x} + b \cos{\frac{1}{2}x}[/tex]

Wronskian: [tex] \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}[/tex]

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.

[tex]y_{1}=\sin{\frac{1}{2}x} [/tex]

[tex]y_{2}=\cos{\frac{1}{2}x}[/tex]

[tex] u_{1} = \int{4\sqrt{x}\cos{\frac{1}{2}x}dx} [/tex]

[tex]u_{2} = \int{-4\sqrt{x}\sin{\frac{1}{2}x}dx}[/tex]

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

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

where the roots come from the form of [itex]\lambda \pm i\mu[/itex]. Your form was missing the exponential term. Hope that helps a bit.
 
  • #3
Yeah, jeffreydk is right. Your roots are incorrect and you neglected the exponential.
 

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

What is the concept of variations of parameters?

Variations of parameters is a method used to solve differential equations by finding a particular solution using a combination of known solutions to simpler equations.

How is the method of variations of parameters different from other methods of solving differential equations?

The method of variations of parameters is different from other methods because it allows for the use of known solutions to simpler equations, rather than relying solely on algebraic manipulation or integration.

What types of differential equations can be solved using variations of parameters?

Variations of parameters can be used to solve linear differential equations with constant coefficients, as well as some non-linear equations. It is most commonly used for second-order equations, but can also be applied to higher-order equations.

What are the steps involved in solving a differential equation using variations of parameters?

The first step is to find the complementary solution, which is the solution to the associated homogeneous equation. Then, a particular solution is found by assuming it can be written as a linear combination of known solutions to simpler equations. The coefficients in this combination are then determined using a system of equations. Finally, the general solution is found by combining the complementary solution and particular solution.

What are some applications of variations of parameters in real-world problems?

Variations of parameters can be used to model many physical systems, such as electrical circuits, mechanical systems, and chemical reactions. It is also commonly used in engineering and physics to analyze and predict the behavior of these systems.

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