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Agent_A

  • one year ago

Ordinary Differential Equations Question: (see photo)

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  1. Agent_A
    • one year ago
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  2. SolomonZelman
    • one year ago
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    For question one: y'+y=1 so you just have linear DE \(\Large e^{H(x)}= \LARGE e^{^{\LARGE \int dx}}\)

  3. SolomonZelman
    • one year ago
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    and then use the product rule backwards, as you should probably know to.

  4. SolomonZelman
    • one year ago
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    Ok, for question 2 \(\large\color{black}{ \displaystyle \frac{dz}{dt} =z^2+4 }\) integrate both sides with respect to z (and then you will have to solve for z)

  5. SolomonZelman
    • one year ago
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    3 is same as 2, but different variables.

  6. anonymous
    • one year ago
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    help!! me plzz

  7. SolomonZelman
    • one year ago
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    And then for 4. \(\large\color{black}{ \displaystyle \frac{dy}{dt}=\frac{ 1}{t^2y+t^2+y+1} }\) \(\large\color{black}{ \displaystyle \frac{dy}{dt}=\frac{ 1}{t^2(y+1)+y+1} }\) \(\large\color{black}{ \displaystyle \frac{dy}{dt}=\frac{ 1}{(t^2+1)(y+1)} }\) \(\large\color{black}{ \displaystyle (y+1)\frac{dy}{dt}=\frac{ 1}{t^2+1} }\) integrate both sides with respect to t.

  8. SolomonZelman
    • one year ago
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    ok, bye

  9. SolomonZelman
    • one year ago
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    Oh, actually, it would be better for question 2: 1/(4+z²) dz/dt=1 then integrate with respect to t on both sides, and then solve for z. ------------------ and bring the 1/(y²-9) over to dy/dt, and then integrate both sides with respect to t.

  10. SolomonZelman
    • one year ago
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    For question 1, if you want: \(\large\color{black}{ \displaystyle y'=y-1 }\) \(\large\color{black}{ \displaystyle \frac{ dy}{dx}=y-1 }\) \(\large\color{black}{ \displaystyle \frac{1}{y-1}\frac{ dy}{dx}=1 }\) integrate both sides with respect to x. (you will also then have to solve for y)

  11. Agent_A
    • one year ago
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    Thank You, @SolomonZelman!

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