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Question Number 112740 by Mr.D.N. last updated on 09/Sep/20
Solvedifferentialequationthemethodofvariationofparameters:d2ydx2+4y=4cosec2x
Answered by mathmax by abdo last updated on 09/Sep/20
y″+4y=4sin(2x)h→r2+4=0⇒r=+−2i⇒yh=ae2ix+be−2ix=αcos(2x)+βsin(2x)=αu1+βu2W(u1,u2)=|cos(2x)sin(2x)−2sin2x2cos(2x)|=2(cos22x+sin2(2x))=2≠0W1=|0sin(2x)4sin(2x)2cos(2x)|=−4W2=|cos(2x)0−2sin(2x)4sin(2x)|=4cotan(2x)V1=∫W1Wdx=∫−42dx=−2xV2=∫W2Wdx=∫4cotan(2x)2dx=2∫cos(2x)sin(2x)dx=ln(∣sin(2x)∣)⇒yp=u1v1+u2v2=cos(2x)(−2x)+sin(2x)ln∣sin(2x)∣thegeneralsolutionisy=yh+ypy=αcos(2x)+βsin(2x)−2xcos(2x)+sin(2x)ln∣sin(2x)∣
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