anonymous
  • anonymous
solve the DE: 2xy^3dx=(1-x^2)dy
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
Notice that you can manipulate this DE into a separable DE by dividing both sides by (1-x^2) and y^3 which gives you the following: \[(2x/(1-x^{2}))dx=(1/y^{3})dy\]this DE now has the separable form \[M(x)dx=N(y)dy\] Now integrate the left side of the DE with respect to x and the right side with respect to y.\[\int\limits_{}^{}(2x/(1-x^{2}))dx=\int\limits_{}^{}y^{-3}dy\]Using u=1-x^2 and performing u-substitution, the integral on the left hand side is \[-\ln(1-x^{2})+k\]and the right hand side integral is \[-1/2y^{-2}+c\] So,\[-\ln(1-x^{2}) + k = -1/2y^{-2} + c\]The c absorbs the -k when it is subtracted over to the right hand side of the equation giving you: \[-\ln(1-x^{2})=-1/2y^{-2}+c\]Now since you didn't provide an initial condition, there is no way to find an explicit solution. Only a general solution can be found because there is no initial condition to solve for the constant c. So, we simplify to get y alone. \[1/(2y^{2})=\ln(1-x^{2})+c\]\[1/2(\ln(1-x^{2})+c)=y^{2}\]\[\pm \sqrt{1/(\ln((1-x^{2})^{2})+2c)}=y\] There is no way to tell which sign on the square root is correct since no initial condition was given to determine interval of validity.

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