Find the general solution by varied parameter y"+y=cosx-sinx
Answer (1)
27th Aug, 2020
Hello aspirant.
By converting D operator, the equation will be
(D^2+1)y=cosx-sinx
The complementary equation:
D^2+1=0
D=+j,-j
y(cf)=Aexp(jx)+Bexp(-jx)
y(cf)=Csin(x+a),c and a are constant
y(pI)=1/(D^2+1)*(cosx-sinx)
y(pI)=x*1/2D*(cosx-sinx)
y(pI)=x/2*integration(cosx-sinx)dx
y(pI)=x(sinx+cosx)/2
Complete solution:
y=y(cf)+y(PI)
y=Csin(x+a)+x(sinx+cosx)/2
Hope this will help you.
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