→ sapphireBOB :ln(2e^u +u) 對u微分 08/18 16:30
→ Crazycraze :真是謝謝 自己基礎不好^^ 08/18 16:35
x/y x/y
solve [1+2e ]dx + 2e [1-x/y]dy=0
我的做法是
let u=x/y dx=ydu+udy
u u
=> [1+2e ] [ydu+udy] + 2e [1-u]dy=0
u u
=> [1+2e ] [ydu+udy] = 2e [u-1]dy
u u
=> 2e [u-1]/[1+2e ] = [ydu+udy]/dy = y(du/dy)+u
u u u u
=> [2ue -2e -u-2ue ]/[1+2e ] = y(du/dy)
u u
=> -[2e +u]/[1+2e ] = y(du/dy)
u u
=> [(1+2e )/(2e +u)]du + dy/y = 0
u u
=>∫[(1+2e )/(2e +u)]du +∫dy/y = C
^^^^^^^^^^^^^^^^^^^^^
請問此積分如何求呢?
或是小弟頭昏，前面的運算有出錯呢?
重算很多次了，麻煩各位幫我看看，謝謝~
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