Separation Of Variables And Integrating Factors Exercise #2

CLAREMONT MCKENNA COLLEGE Separation of Variables and Integrating Factors Solve for y=2+2xy1+�2 Solution: Rewrite the equation in the form, ��=(�)+(�)� ��(�)�=(�) where (�)�=2��1+�2 , ��=�, and (�)=2 Then, �2��1+�2=2 �2�1+�2(�)=2 Next, we solve for the exponent of the integrating factor, -(�)�=2�1+�2�� Let =�2+1,�=2���, and ��=12��� (�)�=2��1+�2=1� =ln||=ln |�2+1| Thus, further simplifying, (�)=�ln|�2+1|1=(1+�2)1 (�)=11+�2 Next, we substitute the equation in the form, �=1(�)(�)(�)�� �=111+�211+�2(2)��Simplify the integral further, �=1+�221+�2�� Recall the property of reverse trig functions for tan, then d/dx becomes: ���tan1�=1�2+1 �=1+�221+�2��=1+�2(2tan1(�)+�) Distribute x raised to 2, �=1+ 2tan1(�)+2�2tan1(�)+2�+2��2 Letting 2c become �1 while 2��2 become �2, we will have, �=1+ 2tan1(�)+2�2tan1(�)+�1+�2�2