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second order differential
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本帖最后由 noob1988 于 14-1-2010 04:27 PM 编辑
second order differential equation...
y"-4y'+4y=4e^(2x)
我找到
yc=e^2x(a+bx)
现在要找yp
用y= 4e^(2x)
然后我要let y=?
我每会到这边我就不会了...
有谁可以教我要怎样很确定的知道要用什么?
还有1st order...
x dy/dx = 1/(y-3) , y(1)=4
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发表于 14-1-2010 05:26 PM
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e^2x 用了,xe^2x 用了,所以就用(x^2)e^2x 咯。
1st order 的,就是x 搬一边,y 搬一边,然后integrate 而已吧? |
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楼主 |
发表于 14-1-2010 06:11 PM
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e^2x 用了,xe^2x 用了,所以就用(x^2)e^2x 咯。
1st order 的,就是x 搬一边,y 搬一边,然后integrate ...
antimatter 发表于 14-1-2010 05:26 PM ![](http://chinese4.cari.com.my/myforum/images/common/back.gif)
就这样看吧了?
像这一题呢?
y"+6y'+9y=-3x+5
b^2-4ac=0
yc=e^(3x) (A+Bx)
y=-3x+5
yp=? 是不是ax+b?还是?
还有
y"-5y'+6y=36x
b^2-4ac>0
yc=Ae^3x+B^2x
y=36x
yp=ax?还是yp=ax+b?
几时要加b?几时不用?
1st order的那个...
我用了2张纸就是做不到答案
不懂是我的答案错还是朋友给我的答案错... |
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楼主 |
发表于 14-1-2010 06:12 PM
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e^2x 用了,xe^2x 用了,所以就用(x^2)e^2x 咯。
1st order 的,就是x 搬一边,y 搬一边,然后integrate ...
antimatter 发表于 14-1-2010 05:26 PM ![](http://chinese4.cari.com.my/myforum/images/common/back.gif)
就这样看吧了?
像这一题呢?
y"+6y'+9y=-3x+5
b^2-4ac=0
yc=e^(3x) (A+Bx)
y=-3x+5
yp=? 是不是ax+b?还是?
还有
y"-5y'+6y=36x
b^2-4ac>0
yc=Ae^3x+B^2x
y=36x
yp=ax?还是yp=ax+b?
几时要加b?几时不用?
1st order的那个...
我用了2张纸就是做不到答案
不懂是我的答案错还是朋友给我的答案错... |
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发表于 14-1-2010 06:24 PM
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你只要记住把yp 放进去differential equation 里面,一定能得到等式的右边的。你就用ax 来做,然后看得到什么答案再放进去differential equation 里面看对不对。不对的话就用ax+b,在不对的话就用ax^2+bx+c,再不对就用kuasa 3 的……就这样而已。
把你的答案(和步骤)放上来啦,不然我哪里懂你在做什么? |
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发表于 16-1-2010 09:41 PM
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antimatter,can i have ur msn ?cuz i will sit for further math this year. |
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发表于 17-1-2010 01:22 AM
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Haha, why not you just put your questions here? Then it is good for ppl to discuss here also. ^^ |
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发表于 17-1-2010 07:24 PM
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本帖最后由 數學神童 于 17-1-2010 07:25 PM 编辑
using the substitution x=e^z,show that the differential equation
x^2(d^2y/dx^2)+px(dy/dx)+qy=0 where p and q are constants,can be transformed into the differential equation
(d^2y/dz^2)+r(dy/dz)+sy=0 where r and s are constants to be determined in terms of p and q. |
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发表于 17-1-2010 10:31 PM
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x=e^z, dx/dz=e^z.
d2y/dx2=d/dx(dy/dx)=d/dx(dy/dz x dz/dx)=d/dx(dy/dz x exp(-z))
=d/dz(dy/dz x exp(-z)) x dz/dx =(exp(-z) d2y/dz2 - exp(-z) dy/dz) x exp(-z)
=exp(-2z) d2y/dz2 - exp(-2z) dy/dz
Try to do the same thing for dy/dx and then substitute both the d2y/dx2 and dy/dx inside the differential equation. Collect the d2y/dz2 terms and dy/dz terms repectively and you will see the new form. |
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