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D^2y/Dx^2-y/Dx+y=2E^xsinx 求通解,过程具体谢谢

^y'' -2y'+y=2e^xsinx,设特解为y=(asinx +bcosx)e^x带入得到y' = (acosx - b sinx)e^x +(asinx +bcosx)e^x =e^x(a+b)cosx +(a-b)sinx e^xy'' = e^x[(a+b)cosx +(a-b)sinx] + e^x[-(a+b)sinx +(a-b)cosx]= e^x [2acosx -2b sinx]y'' -2y' +y = e^x[2acosx - 2b sinx -2(

y''+y=e^x特征方程为t^2+1=0,t=±i所以y1=C1sinx+C2cosx显然一个特解为y2=e^x/2所以通解为y=y1+y2=C1sinx+C2cosx+e^x/2

y'' - 8y' +16y = x+e^4x,特征方程s^2-8s +16=0, s1=s2=4y''-8y'+16y=0的通解为(a+bx)e^4xy''-8y'+16y=x的特解为c+dx带人得到-8d +16c +16dx =x ,d=1/16, c= 1/8y''-8y' +16y=e^4x特解为(m+nx)e^4x带人得到y' = (4m+4nx +n)e^4xy'' = (16m +16nx +4n + 4n)e^4x带入y''-8y' +16y=e^4x求出m,n然后就得到通解了

d^2y/(dx)^2=yy''-y=0特征方程:r^2-1=0r=±1故其通解是:y=C1e^x+C2e^(-x)

dy/dx=3x^2ydy=3x^2ydxdy/y=3x^2dx两边积分得到:∫dy/y=∫3x^2dxlny=x^3+cy=ce^(x^3).

y'' = E^(2x)积分得y' = 1/2 E^(2x) + C再积分得y = 1/4 E^(2x) + C1 x + C2

应该是y''-y'-2y=2e^x吧…… r^2-r-2=0 r=2或-1 特解是y*=-e^x 所以y=c1e^2x+c2e^(-x)-e^x

令p=dy/dx, 则d^2y/dx^2=pdp/dy代入方程:pdp/dy-e^yp=0dp/dy=e^ydp=e^ydy积分:p=e^y+cdy/dx=e^y+cdy/(e^y+c)=dxd(e^y)/[e^y(e^y+c)]=dxd(e^y)[1/e^y-1/(e^y+c)]=cdx积分:lne^y/(e^y+c)=cx+c1e^y/(e^y+c)=c1e^(cx)解得:y=ln{cc1e^(cx)/[1-c1e^(cx)]}

解:∵dy/dx=e^(2x) ==>dy/dx=e^(2x)/2+c1 (c1是积分常数) ==>y=e^(2x)/4+c1x+c2 (c2是积分常数) ∴原方程的通解是y=e^(2x)/4+c1x+c2 .

解:∵齐次方程d^2y/dx^2-2dy/dx+2y=0的特征方程是r^2-2r+2=0,则r=1±i (复数根) ∴此齐次方程的通解是y=(C1cosx+C2sinx)e^x (C1,C2是常数) ∵y=e^x+5/2是原方程的一个特解 ∴原方程的通解是y=(C1cosx+C2sinx)e^x+e^x+5/2.

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