Solution To System Of ODEs

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 ?

(edited 6 years ago)

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.

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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