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gazc99

  • 2 years ago

Diff eqn dy/dx= (ycos(x))/(1+2y^2) int condt y(0)=1

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  1. hartnn
    • 2 years ago
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    this is easily separable. bring it in the form f(x)dx = f(y)dy then integrate both sides.

  2. hartnn
    • 2 years ago
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    can you ? or need help with that ?

  3. Mimi_x3
    • 2 years ago
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    because im bored...\[\frac{dy}{dx} = \frac{ycos(x)}{1+2y^{2}} \] \[=>\frac{dy}{dx} = \frac{y}{1+2y^{2}} *\cos(x) => dy = \frac{y}{1+2y^{2}} *\cos(x)dx => \frac{dy(1+2y^{2})}{y} = cos(x) \] \[=> \int\limits\frac{1+2y^{2}}{y} dy = \int\limits \cos(x) \] \[\int\limits\frac{1}{y} dy + \frac{2y^{2}}{y} dy = \ln(y) + \frac{y^{2}}{2} \] \[=> \ln(y)+\frac{y^{2}}{2} = sinx +c\] very easy DE..

  4. sirm3d
    • 2 years ago
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    hmm, it's \[ \ln y+y^2=\sin x + C\] now put \(x=0\) and \(y=1\) to find \(C\)

  5. Mimi_x3
    • 2 years ago
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    wait there's a typo..\[ \int\limits\frac{2y^{2}}{y} dy => \int\limits2y dy => y^2\]

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