What do I do with e^y in the differential equation dy/dt =(2t)/(e^y)?

What do I do with e^y in the differential equation dy/dt =(2t)/(e^y)?


[tex]What do I do with e^y in the differential equation dy/dt =(2t)/(e^y)?[/tex]

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This Post Has 3 Comments

  1. not only in mathematics, scientific proof also are required in the other field of science. also maths proofs are of particular structure, containing steps and logics. so proving is like doing science, it is science. that's why it is important to prove. 

    an inaccurate proof leads to errors, or stating results that are not accurate.  which wil mess the probelm up you dont want this

  2. 7! = 7×6×5×4×3×2

    3! = 3× 2

    as 7! is multiple of 3! because 7! contains 3! that is 3× 2 as factor

    so 7! /(3! ) = 7× 6× 5× 4

    = 42 × 20

    = 840

    its b) 840✌✌✌

    dhruv⭐

  3. As you said, the equation is separable:

    dy/dt = 2t / exp(y)

    exp(y) dy = 2t dt

    Integrate both sides:

    ∫ exp(y) dy = ∫ 2t dt

    exp(y) + C₁ = t ² + C₂

    Move the constant terms to one side. When you add them together, you get another constant, so you can ignore the subscript altogether:

    exp(y) = t ² + C

    Solve for y explicitly by taking the logarithm of both sides:

    ln(exp(y)) = ln(t ² + C )

    y = ln(t ² + C )

    C can be any number; if it happens to be 0, then you have

    y = ln(t ²) = 2 ln(t )

    so B is the correct choice.

    You can also approach this from the opposite angle: Assume y is one of the given solutions, then substitute it into the ODE. (Bit more trial-and-error involved, so not a good idea if you're short on time.)

    For example, if y = 2 exp(t ) as in choice A, you have dy/dt = 2 exp(t ), so the ODE would become

    2 exp(t ) = 2t / exp(2 exp(t ))

    which is clearly (I hope) not true.

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