Hmm... do you know how to get the correct solution?
It depends on L A simple multiplication of two operators L=L1⋅L2 as you wrote in the first equation or one linear operator meaning L=(dtd+2)(dt2d2+3)=dt3d3+2dt2d2+3dtd+6 as you wrote with characteristic polinomial I think to this Then your method is correct with that m2=3⋅i m3=−3⋅i For simple multiplication you will get same base solutions but then the real and complex solutions will be separate solutions so you can not get the linear combination of them (only for the two complex conjugate to get real functions) because the initial equation is not linear. 1. x1(t)=A⋅e−2t 2. THe base solutions
so for real x2a=2B(u1+u2)=B⋅cosh(i⋅3t)=B⋅cos(3t) x2b=2iC(u1−u2)=iC⋅sinh(i⋅3t)=C⋅sin(3t) So another solution for x(t) x2(t)=B⋅cos(3t)+C⋅sin(3t)
It depends on L A simple multiplication of two operators L=L1⋅L2 as you wrote in the first equation or one linear operator meaning L=(dtd+2)(dt2d2+3)=dt3d3+2dt2d2+3dtd+6 as you wrote with characteristic polinomial I think to this Then your method is correct with that m2=3⋅i m3=−3⋅i For simple multiplication you will get same base solutions but then the real and complex solutions will be separate solutions so you can not get the linear combination of them (only for the two complex conjugate to get real functions) because the initial equation is not linear. 1. x1(t)=A⋅e−2t 2. THe base solutions
so for real x2a=2B(u1+u2)=B⋅cosh(i⋅3t)=B⋅cos(3t) x2b=2iC(u1−u2)=iC⋅sinh(i⋅3t)=C⋅sin(3t) So another solution for x(t) x2(t)=B⋅cos(3t)+C⋅sin(3t)
Thanks for the answer.
Although, I don't fully understand why do the real and complex solutions have to be separated if the initial equation is not linear. Could explain this a bit further or give me a link to your reference. Cheers
Although, I don't fully understand why do the real and complex solutions have to be separated if the initial equation is not linear. Could explain this a bit further or give me a link to your reference. Cheers
In first half of my answer I assume L is one linear operator of L=(dtd+2)(dt2d2+3)=dtd(dt2d2)+2dt2d2+3dtd+6 so L[x(t)]=0 does not mean the following product as you wrote (dtdx+2x)(dt2d2x+3x)=0 but dt3d3x+2dt2d2x+3dtdx+6x=0 equation I wrore that you have the correct solution for this except two m values; m1=−2 m2=3⋅i m3=−3⋅i So the base solutions are e−2t,e−3it,e3it from this the general solution in Complex is the linear combination of base solutions (only if the equation is linear and it is) x(t)=Ae−2t+Be3it+Ce3it or functions with real variables (with further linear combinations as i wrote) x(t)=Ae−2t+Bcos(3t)+Csin(3t)
At the second half of my answer I assumed that L=L1⋅L2 where L1=(dtd+2) and L2=(dt2d2+3) so L[x(t)]=0 which means (dtdx+2x)(dt2d2x+3x)=0 That is (dtdx+2x)=0 OR (dt2d2x+3x)=0 Which means to solve the two equations separately and not simoultaneously because L1[(x(t)]=0 involves L(x)=0 and L2[x(t)]=0 involves that too. So there is two solutions for x(t) from the 1st equation x1(t)=A⋅e−2t from the 2nd x2(t)=B⋅e3it+C⋅e−3it or for real functions x2(t)=B⋅cos(3t)+C⋅sin(3t)
In first half of my answer I assume L is one linear operator of L=(dtd+2)(dt2d2+3)=dtd(dt2d2)+2dt2d2+3dtd+6 so L[x(t)]=0 does not mean the following product as you wrote (dtdx+2x)(dt2d2x+3x)=0 but dt3d3x+2dt2d2x+3dtdx+6x=0 equation I wrore that you have the correct solution for this except two m values; m1=−2 m2=3⋅i m3=−3⋅i So the base solutions are e−2t,e−3it,e3it from this the general solution in Complex is the linear combination of base solutions (only if the equation is linear and it is) x(t)=Ae−2t+Be3it+Ce3it or in real (with further linear combinations as i wrote) x(t)=Ae−2t+Bcos(3t)+Csin(3t)
At the second half of my answer I assumed that L=L1⋅L2 where L1=(dtd+2) and L2=(dt2d2+3) so L[x(t)]=0 which means (dtdx+2x)(dt2d2x+3x)=0 That is (dtdx+2x)=0 OR (dt2d2x+3x)=0 Which means to solve the two equations separately and not simoultaneously because L1[(x(t)]=0 involves L(x)=0 and L2[x(t)]=0 involves that too. So there is two solutions for x(t) from the 1st equation x1(t)=A⋅e−2t from the 2nd x2(t)=B⋅e3it+C⋅e−3it or for real functions x2(t)=B⋅cos(3t)+C⋅sin(3t)
I have no queation or answer books or links just my knowledge and skills Substituting back the solution into the original equation it works, and it is enough to prove that this was a good solution.