Question

Find the particular solution dy/dx = -x+ycosx / 1+sinx

Answer 1

Answer:

Step-by-step explanation:

so -sinx + sinx = 0  

d2y/dx2 of ( cosx) = -cosx  

so -cosx +cosx = 0  

d2y/dx2 of (acosx+bsinx) = (-acosx-bsinx)  

so -acosx-bsinx+acosx+bsinx=0  

y=sin2x -----dy/dx---> 2cos2x -----d2y/dx2----> - 4sin2x  

y=cos2x -----dy/dx---> 2cos2x -----d2y/dx2----> - 4cos2x  

y=acos2x+bsin2x ---dy/dx -----> -a2sin2x+b2sin2x -----d2y/dx2-----> -a4cos2x-b4cos2x