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     y'(x) -2z =0, z'(x) +y-2z=0

    Now I solved this, by differentiating the former equation to get  y=e^x(Acosx+Bsinx) and using  z= \frac{1}{2}y'(x), z= \frac{1}{2}e^x[(A+B)cosx +(B-A)sinx]

    But if I solve differentiating the equation in  z(x) first, I get

    z=e^x(Acosx+Bsinx) and using the second ODE above, y=  e^x[(A+B)sinx + (B-A)cosx]

    why are the answers different, are both acceptable ?
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    (Original post by NotNotBatman)
     y'(x) -2z =0, z'(x) +y-2z=0

    Now I solved this, by differentiating the former equation to get  y=e^x(Acosx+Bsinx) and using  z= \frac{1}{2}y'(x), z= \frac{1}{2}e^x[(A+B)cosx +(B-A)sinx]

    But if I solve differentiating the equation in  z(x) first, I get

    z=e^x(Acosx+Bsinx) and using the second ODE above, y=  e^x[(A+B)sinx + (B-A)cosx]

    why are the answers different, are both acceptable ?
    Has to do with arbitrary constant A and B and the usual idea of stuff getting absorbed into them. Either answer is fine.
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    (Original post by RDKGames)
    Has to do with arbitrary constant A and B and the usual idea of stuff getting absorbed into them. Either answer is fine.
    That's what I was thinking, but couldn't confirm it. Thanks.
 
 
 
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