lissethsegura90 lissethsegura90
  • 11-06-2017
  • Mathematics
contestada

y^-1 * dy+y e^cosx * sinx * dx=0

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LammettHash
LammettHash LammettHash
  • 11-06-2017
[tex]\dfrac{\mathrm dy}y+ye^{\cos x}\sin x\,\mathrm dx=0[/tex]
[tex]\implies\dfrac{\mathrm dy}{y^2}=-e^{\cos x}\sin x\,\mathrm dx[/tex]

[tex]\displaystyle\int\frac{\mathrm dy}{y^2}=-\int e^{\cos x}\sin x\,\mathrm dx[/tex]
[tex]-\dfrac1y=e^{\cos x}+C[/tex]
[tex]y=-\dfrac1{e^{\cos x}+C}[/tex]
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