Homogenous second order differential equation that seems not to be solvable(x2+y2)dx+(x2−xy)dy=0When I check for it to have a common partial derivative, I find that∂M∂y=(x2+y2)y=2yand the second ∂N∂x=(x2−xy)x=2x−yThus, it does not seem to have common partial derivative. What can be done to solve it?UPDATE:by Moo's suggestion we get:(1+u2)dx+(1−u)(xdu+udx)=0(1+u2)dx+(xdu+udx−uxdu−u2dx)=0(1+u)dx+x(1−u)du1+u1−udx=−xduthis gives:ln⁡x+c=−∫11+u+u1+uduln⁡x+c=−ln|1+u|+∫1du−∫11+uduln⁡x+c=u−2ln|u+1|Insert for u=yxln⁡x+c=−2ln|yx+1|+yx

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Homogenous second order differential equation that seems not to be solvable(x2+y2)dx+(x2−xy)dy=0When I check for it to have a common partial derivative, I find that∂M∂y=(x2+y2)y=2yand the second ∂N∂x=(x2−xy)x=2x−yThus, it does not seem to have common partial derivative. What can be done to solve it?UPDATE:by Moo&#039;s suggestion we get:(1+u2)dx+(1−u)(xdu+udx)=0(1+u2)dx+(xdu+udx−uxdu−u2dx)=0(1+u)dx+x(1−u)du1+u1−udx=−xduthis gives:ln⁡x+c=−∫11+u+u1+uduln⁡x+c=−ln|1+u|+∫1du−∫11+uduln⁡x+c=u−2ln|u+1|Insert for u=yxln⁡x+c=−2ln|yx+1|+yx

Mercedes Chang · Accepted Answer

Step 1Given:(1)&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;(x2+y2)dx+(x2−xy)dy=0It is not an Exact Equation, so we will try something else, rewriting&amp;nbsp;(1)(2)&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;(x2+y2)+(x2−xy)y&#039;=0Step 2Let (example of this approach)y=vx≈∼so≈∼y′=v+xv′Substituting into&amp;nbsp;(2)&amp;nbsp;and simplifyingx2(xv′+v−xv′v+1)=0Solving for&amp;nbsp;v&#039;v′=dv+1x(v−1)

Homogenous second order differential equation that seems not

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