text: # Verify solution of system of first order ODE
  
  I am given a solution $v_1$(t)=$-e^{t/2}cost$ for a system of first order ODE and I am asked to verify this solution can solves the linear second order ODE in terms of $y_1$(t)
  
  Here is the matrix ODE [![enter image description] After get $y_2$(t), I copy the output of y2[t] and use D to take its derivative
  
  D[(-2 y1[t] + y1[t] - 2 Derivative[1][y1][t])/(
  2 + 3 Cos[t] Sin[t]), t]
  
  
  Next, I equate $y_2'$(t) to the second component of the coefficient matrix (second row of the matrix), and finally, i obtained second ODE in terms of $y_1$(t), $y_1'$(t) and $y_1''$(t) (*)
  
  {-(((3 Cos[t]^2 - 3 Sin[t]^2) (-2 y1[t] + 3 Cos[t]^2 y1[t] -
  2 Derivative[1][y1][t]))/(2 + 3 Cos[t] Sin[t])^2) + (-6 Cos[
  t] Sin[t] y1[t] - 2 Derivative[1][y1][t] +
  3 Cos[t]^2 Derivative[1][y1][t] - 2 (Prime])[t]}]
  
  
  Summary
  
  Sorry if it is hard to follow, it is quite messy to right details but basically i have the matrix ode and i used mathematica to rearrange the equation and then substitution so that finally i get a single second ODE with only $y_1$(t) $y_1’$(t) and $y_1’’$(t). Then i plug in the solution i have into this equation and verify that it is 0. But i can’t get it and i am not sure if it is due to systax error or any possible mistakes i could get during computing with mathematica.
  
  Is there any syntax errors or any mistakes I have made throughout the process? Any help please :):). Thank you very much
  
  • I am given a solution for a system of first order ODE Where is the solution you are given? I do not see it? is it the v1 there? Your question is a little hard to follow. Apr 26, 2018 at 6:29
  • @Nasser Sorry for that i have edited the post little bit. Could you try to help me with that? Thanks so much :):)
  – user57821
  Apr 26, 2018 at 6:40
  • I don't think $y_1(t)=e^{-t/2} \cos (t)$ is a solution; and Mathematica is telling you so.
  – ulvi
  Apr 26, 2018 at 6:53
  • @ulvi it is $-e^{t/2}cos(t)$.The question i had is verify that $v_1$(t) can solve the second ODE i have just found in terms of $y_1$(t).... so it should be 0 when i plug the solution in :(
  – user57821
  Apr 26, 2018 at 6:59
  • @tabitha96 It is not nice, towards someone who answered, to delete a topic without notice.
  – Kuba
  Apr 26, 2018 at 7:57
  
  The solution you gives does not satisfy the ODE?
  
  \begin{align*} \begin{pmatrix} y_{1}^{\prime}\\ y_{2}^{\prime} \end{pmatrix} = \begin{pmatrix} a_{11} & a_{12}\\ a_{21} & a_{22} \end{pmatrix} \begin{pmatrix} y_{1}\\ y_{2} \end{pmatrix} \end{align*}
  
  Hence \begin{align*} y_{1}^{\prime} & =a_{11}y_{1}+a_{12}y_{2}\\ y_{2}^{\prime} & =a_{21}y_{1}+a_{22}y_{2} \end{align*}
  
  But $y_{1}=y$ and $y_{2}=y^{\prime}$, therefore the above becomes
  
  \begin{align*} y_{1}^{\prime} & =a_{11}y+a_{12}y^{\prime}\\ y_{2}^{\prime} & =a_{21}y+a_{22}y^{\prime} \end{align*}
  
  And also $y_{1}^{\prime}=y^{\prime},y_{2}^{\prime}=y^{\prime\prime}$, therefore the above becomes
  
  \begin{align*} y^{\prime} & =a_{11}y+a_{12}y^{\prime}\\ y^{\prime\prime} & =a_{21}y+a_{22}y^{\prime} \end{align*}
  
  Solving for $y^{\prime}$ from first equation above gives $y^{\prime} =\frac{a_{11}y}{1-a_{12}}$. Substituting this in the second equation gives
  
  \begin{align*} y^{\prime\prime}=a_{21}y+a_{22}\frac{a_{11}y}{\left( 1-a_{12}\right) } \end{align*}
  
  Or
  
  \begin{align*} y^{\prime\prime}=\left( a_{21}+a_{22}\frac{a_{11}}{1-a_{12}}\right) y \end{align*}
  
  To verify that $v\left( t\right) =-e^{\frac{t}{2}}\cos\left( t\right)$ is a solution, plugin into the above and see if it satisfies it.
  
  ClearAll[y,t,v]
  a={{-1+3/2Cos[t]^2,1-3/2 Sin[t]Cos[t]},
  {-1-3/2 Sin[t]Cos[t],-1+3/2 Sin[t]^2}};
  
  
  ode=y''[t]==(a[[2,1]] +a[[2,2]] a[[1,1]] /(1-a[[1,2]]))y[t];
  ode=Simplify[ode]
  
  
  v[t_]:=-Exp[t/2] Cos[t];
  ode/.{y[t]->v[t],y''[t]->v''[t]}//Simplify
  
  
  It does not satisfy it. If it did, we'll get True at the end.
  
  Are you sure -Exp[t/2] Cos[t]` is supposed to be a solution?