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Jason21

  • 2 years ago

Please Explain about the basis for the space of functions that satify (a) y' - 2y =0, (b) y' - y/x=0? its question number 33 in section 3.5

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  1. MichaelT
    • 2 years ago
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    (a) the exponential function \[y=e^{2x}\] happens to solve \[y'-2y=0\] and since it's a first order differential equation, all other solutions are multiples of it, i.e. \[c e^{2x} \]. (b) the simple function \[y=x\] solves this equation. and similarly, all solutions are its multiples.

  2. uwipostgrad2b
    • 2 years ago
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    y'-2y =0 To sole this equation one can write the characteristic equation to get the eigenvalues. Then using each specific eigenvalue, we get unique eigenvectors for each specific one. Now these eigenvalues, together with their vectors form a basis or SOLUTION SET for the equation meaning all other solutions of the equation are simply multiples of the answers found. Suffice to say, the eigenvalues, together with their eigenvectors are linearly independent and so span R^2. Note that the nature of the eigenvalues have a say in the general solution: whether complex, real, and also if we have repeated roots, etc... *** I hope this is sufficient enough to make you want to solve the problem as it is in the doing that we learn. To learn mathematics, one must write and write and write... indeed learning math is simply hard work and we must desire to get our hands dirty.

  3. uwipostgrad2b
    • 2 years ago
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    actually the solution is as above... i just gave you the method to solve those equations of 2nd order... i felt compelled to tell you.

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