Trying to solve a system of differential equations












1












$begingroup$


Let the following system $X'=AX+b(t)$ be a system of differential equations, where:



$A=begin{pmatrix}1&-4\1&1end{pmatrix}$ and $b=begin{pmatrix}cos2t\0end{pmatrix}$



Then When trying to solve, the roots of the characteristic polynomial of the matrix are complex, therefore I'm going to use:



$$D=begin{pmatrix}a&b\-b&aend{pmatrix}$$



since $lambda_{1,2}=1pm2i$ then $D=begin{pmatrix}1&2\-2&1end{pmatrix}$
and $P=begin{pmatrix}2&0\0&-1end{pmatrix}$, then $A=PDP^{-1}$



$X'=AX=PDP^{-1}Ximplies P^{-1}X'=DP^{-1}X$
if we note $P^{-1}X=begin{pmatrix}u\vend{pmatrix}$ then a matrix of fundamental solutions for the new system is: $M(t)=e^tbegin{pmatrix}cos(2t)&sin(2t)\-sin(2t)&cos(2t)end{pmatrix}$



Then $X=Pe^tbegin{pmatrix}cos(2t)&sin(2t)\sin(2t)&cos(2t)end{pmatrix}begin{pmatrix}c_1\c_2end{pmatrix}$



All in all we get: $$X=e^tbegin{pmatrix}2c_1cos(2t)+2c_2sin(2t)\-c_1sin(2t)-c_2sin(2t)end{pmatrix}$$



But that is not correct, what am I missing?






ordinary-differential-equations systems-of-equations







share|cite|improve this question











$endgroup$












  • $begingroup$
    It makes no sense to use a full matrix $D$ when $D$ should be diagonal. You need to use the eigendecomposition $A = left( begin{array}{cc} -2i & 2i\ 1 & 1 end{array} right) left( begin{array}{cc} 1-2i & 0\ 0 & 1+2i end{array} right) left( begin{array}{cc} -2i & 2i\ 1 & 1 end{array}right)^{-1}$ with a diagonal matrix in the middle.
    $endgroup$
    – Christoph
    Jan 16 at 7:19












  • $begingroup$
    You keep leaving a minus sign out of the fundamental matrix in your questions. As well, $exp(tA)=Pexp(tD)P^{-1}$. Note, too, that you’ve only solved the related homogeneous equation $X'=AX$. You still need to find a particular solution of the inhomogeneous equation.
    $endgroup$
    – amd
    Jan 17 at 0:39












  • $begingroup$
    @Christoph A $2times2$ real matrix with complex eigenvalues is similar to a matrix of the form $small{begin{bmatrix}a&-b\b&aend{bmatrix}}$. The OP’s decomposition is perfectly good.
    $endgroup$
    – amd
    Jan 17 at 0:40










  • $begingroup$
    @amd I see, you're right. I guess I'm sometimes ignorant of such things which work only in very special cases, like $2 times 2$.
    $endgroup$
    – Christoph
    Jan 17 at 5:16










  • $begingroup$
    @amd right, i'm sorry, And how do I search for the particular solution?
    $endgroup$
    – C. Cristi
    Jan 17 at 7:01
















1












$begingroup$


Let the following system $X'=AX+b(t)$ be a system of differential equations, where:



$A=begin{pmatrix}1&-4\1&1end{pmatrix}$ and $b=begin{pmatrix}cos2t\0end{pmatrix}$



Then When trying to solve, the roots of the characteristic polynomial of the matrix are complex, therefore I'm going to use:



$$D=begin{pmatrix}a&b\-b&aend{pmatrix}$$



since $lambda_{1,2}=1pm2i$ then $D=begin{pmatrix}1&2\-2&1end{pmatrix}$
and $P=begin{pmatrix}2&0\0&-1end{pmatrix}$, then $A=PDP^{-1}$



$X'=AX=PDP^{-1}Ximplies P^{-1}X'=DP^{-1}X$
if we note $P^{-1}X=begin{pmatrix}u\vend{pmatrix}$ then a matrix of fundamental solutions for the new system is: $M(t)=e^tbegin{pmatrix}cos(2t)&sin(2t)\-sin(2t)&cos(2t)end{pmatrix}$



Then $X=Pe^tbegin{pmatrix}cos(2t)&sin(2t)\sin(2t)&cos(2t)end{pmatrix}begin{pmatrix}c_1\c_2end{pmatrix}$



All in all we get: $$X=e^tbegin{pmatrix}2c_1cos(2t)+2c_2sin(2t)\-c_1sin(2t)-c_2sin(2t)end{pmatrix}$$



But that is not correct, what am I missing?






ordinary-differential-equations systems-of-equations







share|cite|improve this question











$endgroup$












  • $begingroup$
    It makes no sense to use a full matrix $D$ when $D$ should be diagonal. You need to use the eigendecomposition $A = left( begin{array}{cc} -2i & 2i\ 1 & 1 end{array} right) left( begin{array}{cc} 1-2i & 0\ 0 & 1+2i end{array} right) left( begin{array}{cc} -2i & 2i\ 1 & 1 end{array}right)^{-1}$ with a diagonal matrix in the middle.
    $endgroup$
    – Christoph
    Jan 16 at 7:19












  • $begingroup$
    You keep leaving a minus sign out of the fundamental matrix in your questions. As well, $exp(tA)=Pexp(tD)P^{-1}$. Note, too, that you’ve only solved the related homogeneous equation $X'=AX$. You still need to find a particular solution of the inhomogeneous equation.
    $endgroup$
    – amd
    Jan 17 at 0:39












  • $begingroup$
    @Christoph A $2times2$ real matrix with complex eigenvalues is similar to a matrix of the form $small{begin{bmatrix}a&-b\b&aend{bmatrix}}$. The OP’s decomposition is perfectly good.
    $endgroup$
    – amd
    Jan 17 at 0:40










  • $begingroup$
    @amd I see, you're right. I guess I'm sometimes ignorant of such things which work only in very special cases, like $2 times 2$.
    $endgroup$
    – Christoph
    Jan 17 at 5:16










  • $begingroup$
    @amd right, i'm sorry, And how do I search for the particular solution?
    $endgroup$
    – C. Cristi
    Jan 17 at 7:01














1












1








1


1



$begingroup$


Let the following system $X'=AX+b(t)$ be a system of differential equations, where:



$A=begin{pmatrix}1&-4\1&1end{pmatrix}$ and $b=begin{pmatrix}cos2t\0end{pmatrix}$



Then When trying to solve, the roots of the characteristic polynomial of the matrix are complex, therefore I'm going to use:



$$D=begin{pmatrix}a&b\-b&aend{pmatrix}$$



since $lambda_{1,2}=1pm2i$ then $D=begin{pmatrix}1&2\-2&1end{pmatrix}$
and $P=begin{pmatrix}2&0\0&-1end{pmatrix}$, then $A=PDP^{-1}$



$X'=AX=PDP^{-1}Ximplies P^{-1}X'=DP^{-1}X$
if we note $P^{-1}X=begin{pmatrix}u\vend{pmatrix}$ then a matrix of fundamental solutions for the new system is: $M(t)=e^tbegin{pmatrix}cos(2t)&sin(2t)\-sin(2t)&cos(2t)end{pmatrix}$



Then $X=Pe^tbegin{pmatrix}cos(2t)&sin(2t)\sin(2t)&cos(2t)end{pmatrix}begin{pmatrix}c_1\c_2end{pmatrix}$



All in all we get: $$X=e^tbegin{pmatrix}2c_1cos(2t)+2c_2sin(2t)\-c_1sin(2t)-c_2sin(2t)end{pmatrix}$$



But that is not correct, what am I missing?






ordinary-differential-equations systems-of-equations







share|cite|improve this question











$endgroup$




Let the following system $X'=AX+b(t)$ be a system of differential equations, where:



$A=begin{pmatrix}1&-4\1&1end{pmatrix}$ and $b=begin{pmatrix}cos2t\0end{pmatrix}$



Then When trying to solve, the roots of the characteristic polynomial of the matrix are complex, therefore I'm going to use:



$$D=begin{pmatrix}a&b\-b&aend{pmatrix}$$



since $lambda_{1,2}=1pm2i$ then $D=begin{pmatrix}1&2\-2&1end{pmatrix}$
and $P=begin{pmatrix}2&0\0&-1end{pmatrix}$, then $A=PDP^{-1}$



$X'=AX=PDP^{-1}Ximplies P^{-1}X'=DP^{-1}X$
if we note $P^{-1}X=begin{pmatrix}u\vend{pmatrix}$ then a matrix of fundamental solutions for the new system is: $M(t)=e^tbegin{pmatrix}cos(2t)&sin(2t)\-sin(2t)&cos(2t)end{pmatrix}$



Then $X=Pe^tbegin{pmatrix}cos(2t)&sin(2t)\sin(2t)&cos(2t)end{pmatrix}begin{pmatrix}c_1\c_2end{pmatrix}$



All in all we get: $$X=e^tbegin{pmatrix}2c_1cos(2t)+2c_2sin(2t)\-c_1sin(2t)-c_2sin(2t)end{pmatrix}$$



But that is not correct, what am I missing?






ordinary-differential-equations systems-of-equations




ordinary-differential-equations systems-of-equations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 17 at 7:02







C. Cristi

















asked Jan 16 at 7:04









C. CristiC. Cristi

1,634318




1,634318












  • $begingroup$
    It makes no sense to use a full matrix $D$ when $D$ should be diagonal. You need to use the eigendecomposition $A = left( begin{array}{cc} -2i & 2i\ 1 & 1 end{array} right) left( begin{array}{cc} 1-2i & 0\ 0 & 1+2i end{array} right) left( begin{array}{cc} -2i & 2i\ 1 & 1 end{array}right)^{-1}$ with a diagonal matrix in the middle.
    $endgroup$
    – Christoph
    Jan 16 at 7:19












  • $begingroup$
    You keep leaving a minus sign out of the fundamental matrix in your questions. As well, $exp(tA)=Pexp(tD)P^{-1}$. Note, too, that you’ve only solved the related homogeneous equation $X'=AX$. You still need to find a particular solution of the inhomogeneous equation.
    $endgroup$
    – amd
    Jan 17 at 0:39












  • $begingroup$
    @Christoph A $2times2$ real matrix with complex eigenvalues is similar to a matrix of the form $small{begin{bmatrix}a&-b\b&aend{bmatrix}}$. The OP’s decomposition is perfectly good.
    $endgroup$
    – amd
    Jan 17 at 0:40










  • $begingroup$
    @amd I see, you're right. I guess I'm sometimes ignorant of such things which work only in very special cases, like $2 times 2$.
    $endgroup$
    – Christoph
    Jan 17 at 5:16










  • $begingroup$
    @amd right, i'm sorry, And how do I search for the particular solution?
    $endgroup$
    – C. Cristi
    Jan 17 at 7:01


















  • $begingroup$
    It makes no sense to use a full matrix $D$ when $D$ should be diagonal. You need to use the eigendecomposition $A = left( begin{array}{cc} -2i & 2i\ 1 & 1 end{array} right) left( begin{array}{cc} 1-2i & 0\ 0 & 1+2i end{array} right) left( begin{array}{cc} -2i & 2i\ 1 & 1 end{array}right)^{-1}$ with a diagonal matrix in the middle.
    $endgroup$
    – Christoph
    Jan 16 at 7:19












  • $begingroup$
    You keep leaving a minus sign out of the fundamental matrix in your questions. As well, $exp(tA)=Pexp(tD)P^{-1}$. Note, too, that you’ve only solved the related homogeneous equation $X'=AX$. You still need to find a particular solution of the inhomogeneous equation.
    $endgroup$
    – amd
    Jan 17 at 0:39












  • $begingroup$
    @Christoph A $2times2$ real matrix with complex eigenvalues is similar to a matrix of the form $small{begin{bmatrix}a&-b\b&aend{bmatrix}}$. The OP’s decomposition is perfectly good.
    $endgroup$
    – amd
    Jan 17 at 0:40










  • $begingroup$
    @amd I see, you're right. I guess I'm sometimes ignorant of such things which work only in very special cases, like $2 times 2$.
    $endgroup$
    – Christoph
    Jan 17 at 5:16










  • $begingroup$
    @amd right, i'm sorry, And how do I search for the particular solution?
    $endgroup$
    – C. Cristi
    Jan 17 at 7:01
















$begingroup$
It makes no sense to use a full matrix $D$ when $D$ should be diagonal. You need to use the eigendecomposition $A = left( begin{array}{cc} -2i & 2i\ 1 & 1 end{array} right) left( begin{array}{cc} 1-2i & 0\ 0 & 1+2i end{array} right) left( begin{array}{cc} -2i & 2i\ 1 & 1 end{array}right)^{-1}$ with a diagonal matrix in the middle.
$endgroup$
– Christoph
Jan 16 at 7:19






$begingroup$
It makes no sense to use a full matrix $D$ when $D$ should be diagonal. You need to use the eigendecomposition $A = left( begin{array}{cc} -2i & 2i\ 1 & 1 end{array} right) left( begin{array}{cc} 1-2i & 0\ 0 & 1+2i end{array} right) left( begin{array}{cc} -2i & 2i\ 1 & 1 end{array}right)^{-1}$ with a diagonal matrix in the middle.
$endgroup$
– Christoph
Jan 16 at 7:19














$begingroup$
You keep leaving a minus sign out of the fundamental matrix in your questions. As well, $exp(tA)=Pexp(tD)P^{-1}$. Note, too, that you’ve only solved the related homogeneous equation $X'=AX$. You still need to find a particular solution of the inhomogeneous equation.
$endgroup$
– amd
Jan 17 at 0:39






$begingroup$
You keep leaving a minus sign out of the fundamental matrix in your questions. As well, $exp(tA)=Pexp(tD)P^{-1}$. Note, too, that you’ve only solved the related homogeneous equation $X'=AX$. You still need to find a particular solution of the inhomogeneous equation.
$endgroup$
– amd
Jan 17 at 0:39














$begingroup$
@Christoph A $2times2$ real matrix with complex eigenvalues is similar to a matrix of the form $small{begin{bmatrix}a&-b\b&aend{bmatrix}}$. The OP’s decomposition is perfectly good.
$endgroup$
– amd
Jan 17 at 0:40




$begingroup$
@Christoph A $2times2$ real matrix with complex eigenvalues is similar to a matrix of the form $small{begin{bmatrix}a&-b\b&aend{bmatrix}}$. The OP’s decomposition is perfectly good.
$endgroup$
– amd
Jan 17 at 0:40












$begingroup$
@amd I see, you're right. I guess I'm sometimes ignorant of such things which work only in very special cases, like $2 times 2$.
$endgroup$
– Christoph
Jan 17 at 5:16




$begingroup$
@amd I see, you're right. I guess I'm sometimes ignorant of such things which work only in very special cases, like $2 times 2$.
$endgroup$
– Christoph
Jan 17 at 5:16












$begingroup$
@amd right, i'm sorry, And how do I search for the particular solution?
$endgroup$
– C. Cristi
Jan 17 at 7:01




$begingroup$
@amd right, i'm sorry, And how do I search for the particular solution?
$endgroup$
– C. Cristi
Jan 17 at 7:01










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