微分方程y''=(y')平方+1的通解为
令p=y' ==> y''=dp/dx=dp/dy* dy/dx=p*dp/dy
dp/dy=1/p* (p^2 +1) ==> dp* p/(p^2+1)=dy ===> 1/2 *ln(p^2+1)=y+C*
1+p^2=e^(2y+C) ===> p=+ - √[e^(2y+C)-1]
===> dy/dx = =+ - √[e^(2y+C)-1]
====>dy/√[e^(2y+C)-1]=+ -dx
两边求不定积分:
∫1/e^(2y+C) d√[e^(2y+C)-1]=+ -x+C'
左边,令u=√[e^(2y+C)-1]
∫1/(u^2+1) du =+ - x+ C'
ar...全部
令p=y' ==> y''=dp/dx=dp/dy* dy/dx=p*dp/dy
dp/dy=1/p* (p^2 +1) ==> dp* p/(p^2+1)=dy ===> 1/2 *ln(p^2+1)=y+C*
1+p^2=e^(2y+C) ===> p=+ - √[e^(2y+C)-1]
===> dy/dx = =+ - √[e^(2y+C)-1]
====>dy/√[e^(2y+C)-1]=+ -dx
两边求不定积分:
∫1/e^(2y+C) d√[e^(2y+C)-1]=+ -x+C'
左边,令u=√[e^(2y+C)-1]
∫1/(u^2+1) du =+ - x+ C'
arctan令p=y' ==> y''=dp/dx=dp/dy* dy/dx=p*dp/dy
dp/dy=1/p* (p^2 +1) ==> dp* p/(p^2+1)=dy ===> 1/2 *ln(p^2+1)=y+C*
1+p^2=e^(2y+C) ===> p=+ - √[e^(2y+C)-1]
===> dy/dx = =+ - √[e^(2y+C)-1]
====>dy/√[e^(2y+C)-1]=+ -dx
两边求不定积分:
∫1/e^(2y+C) d√[e^(2y+C)-1]=+ -x+C'
左边,令u=√[e^(2y+C)-1]
∫1/(u^2+1) du =+ - x+ C'
====>arctan√[e^(2y+C)-1]+C''==+ - x+ C'
相当烦琐的一道题,总算搞完了。
方法是对的。结果就不管了。头疼
。收起
