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S

  • 2 years ago

By inspection, find a one parameter family of solutions of the differential equation xy' = 4y. Verify that each member of its family also satisfies the initial condition y(0) = 0

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  1. imranmeah91
    • 2 years ago
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    x y' -4y=0 y' - 4/x y=0 now we can better inspect

  2. S
    • 2 years ago
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    Ok thanks i get it so far, but I'm confused by the wording.. one parameter family.. i'm not sure where to go from there

  3. imranmeah91
    • 2 years ago
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    since it say inspection, I don't think we have to try ti solve it

  4. Jemurray3
    • 2 years ago
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    When they say by inspection, they mean just try to see through the problem to the solution without any serious work. It can be done. Note: \[ x y' - 4y = 0 \rightarrow xy' = 4y\] So taking the derivative of the function and multiplying by x is the same as multiplying the function by 4. To me, this suggests something of the form \[ y = A x^4\] where a is an arbitrary constant.

  5. Jemurray3
    • 2 years ago
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    A*

  6. S
    • 2 years ago
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    OOo i think i get it now thank you!

  7. Jemurray3
    • 2 years ago
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    The full solution, in contrast, is the following: \[ xy' - 4y = 0\] \[ y' - \frac{4}{x} y = 0\] Introduce an integrating factor \[e^{\mu(x)} \] so \[[ye^{\mu(x)} ]' = y'e^\mu + \mu'ye^\mu\] Identifying \[ \mu'e^\mu = -\frac{4}{x}e^\mu \rightarrow \mu = e^{-4 \ln(x)} = e^{\ln(x^{-4})} = x^{-4} \] So multiplying through and grouping together, \[ [x^{-4}y]' = 0\] so \[ x^{-4}y = A\] and finally \[y =Ax^4\] That would be showing all the rigorous work... what we did is called by inspection, or just looking at it ;)

  8. S
    • 2 years ago
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    Alright that explains it, thank you!

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