y"-2y'+2y=x*(e^x)*cosx通解很容易求,Y(x)=(C1+C2x)e^(3x),但是特解好像很难求,我算了N多遍都和答案不一样,比答案多一项.高人帮我算一算,到底等于多少啊?就算特解就行.那个通解我抄错了,应该是Y(x)=e^
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y"-2y'+2y=x*(e^x)*cosx通解很容易求,Y(x)=(C1+C2x)e^(3x),但是特解好像很难求,我算了N多遍都和答案不一样,比答案多一项.高人帮我算一算,到底等于多少啊?就算特解就行.那个通解我抄错了,应该是Y(x)=e^
y"-2y'+2y=x*(e^x)*cosx
通解很容易求,Y(x)=(C1+C2x)e^(3x),但是特解好像很难求,我算了N多遍都和答案不一样,比答案多一项.高人帮我算一算,到底等于多少啊?就算特解就行.
那个通解我抄错了,应该是Y(x)=e^x(C1cosx+C2sinx),抱歉~
y*=x*e^x*(ax+b)cos(x)----这个是不对的,特解形式不是这样的.
y"-2y'+2y=x*(e^x)*cosx通解很容易求,Y(x)=(C1+C2x)e^(3x),但是特解好像很难求,我算了N多遍都和答案不一样,比答案多一项.高人帮我算一算,到底等于多少啊?就算特解就行.那个通解我抄错了,应该是Y(x)=e^
算起来好像很复杂,我算出的是:
(x/4)(xsinx+cosx)e^x-(e^x*sinx)/8
不知道对不对.
y"-2y'+2y=x*(e^x)*cosx 对应的齐次方程为
r²-2r+2=0
r=1±i
λ=1 ω=1 ,所以λ+iω=1+i是特征方程的根。设方程的特解为
y*=x*e^x*(ax+b)cos(x)
y``-2y`+2y
=2ae^x*cosx-2*exp(x)*(a*x+b)*sin(x)-2*a*x*exp(x)*sin(x)
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