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how do we solve the initial-value problem? −7y''−7y=sin(x); y(0)=2y(0)=−4
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for the homogenous solution, use -7(r^2+1)=0 or r^2= -1 This gives exp(-ix) and exp(ix), or Asin(x) + Bcos(x), A and B to be determined. for the particular, assume yp= x(Csin(x) + Dcos(x)) we multiply by x to be orthogonal to the homogenous solution. sub yp into the equation, simplify and match coeffs. not sure what the initial conditions are. y(0)= 2 ? y'(0)= -4?
yea they are
omar sorry it isnt right
and phi could you please show me a step my step solution?
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