Dgdrxrt

I was analyzing stability for the following system of differential equations:
$$z_1'=z_1+(6+e^-t)z_2$$
$$z_2'=-z_1-4tanh(t)z_2$$
In an effort to check my answer, I attempted to solve the system, but I'm not sure if this can even be accomplished. The following 2nd-order DE is what resulted when I differentiated the first equation above and made the appropriate substitutions:
$$z_1''+left(-1+frace^-t6+e^-t+frac4(6+e^-t)tanh(t)6+e^-tright)z_1'+left(frac-e^-t6+e^-t+6+e^-t-4tanh(t)right)z_1=0.$$
Is there a technique for solving this diff equ? Perhaps my more general question is do we have a strategy for solving 2nd-order linear DEs of the form $z_1''+a(t)z_1'+b(t)z_1=0$ ?

Thanks for your help.

http://eqworld.ipmnet.ru/en/solutions/sysode/sode-toc1.htm mentions the solving method of some special cases and general case of systems of linear first-order ODEs with functional coefficients. Note that the special cases mentioned in http://eqworld.ipmnet.ru/en/solutions/sysode/sode-toc1.htm are all having the simpler solving method than that the general case.

According to this question, unfortunately it is obviously only belongs to the general case (http://eqworld.ipmnet.ru/en/solutions/sysode/sode0107.pdf). So you should be unavoidable to handle some second-order linear ODEs with complicated functional coefficients.

$z_1''=z_1'+(6+e^-t)z_2'-e^-tz_2$

$z_1''=z_1'+(6+e^-t)(-z_1-4tanh(t)z_2)-e^-tz_2$

$z_1''=z_1'-(6+e^-t)z_1-(e^-t+4(6+e^-t)tanh(t))z_2$

$z_1''=z_1'-(6+e^-t)z_1-(e^-t+4(6+e^-t)tanh(t))dfracz_1'-z_16+e^-t$

$z_1''=-left(dfrace^-t6+e^-t-1+4tanh(t)right)z_1'+left(dfrace^-t6+e^-t-6-e^-t+4tanh(t)right)z_1$

$z_1''+left(dfrace^-t6+e^-t-1+4tanh(t)right)z_1'+left(6+e^-t-dfrace^-t6+e^-t-4tanh(t)right)z_1=0$

$dfracd^2z_1dt^2+left(-dfrac66+e^-t+dfrac4(1-e^-2t)1+e^-2tright)dfracdz_1dt+left(dfrac36+11e^-t+e^-2t6+e^-t-dfrac4(1-e^-2t)1+e^-2tright)z_1=0$

$dfracd^2z_1dt^2-left(dfrac4e^-2t-4e^-2t+1+dfrac6e^-t+6right)dfracdz_1dt+left(dfrac4e^-2t-4e^-2t+1+dfrace^-2t+11e^-t+36e^-t+6right)z_1=0$

Let $u=e^-t$ ,

Then $dfracdz_1dt=dfracdz_1dudfracdudt=-e^-tdfracdz_1du=-udfracdz_1du$

$dfracd^2z_1dt^2=dfracddtleft(-udfracdz_1duright)=dfracdduleft(-udfracdz_1duright)dfracdudt=left(-udfracd^2z_1du^2-dfracdz_1duright)(-u)=u^2dfracd^2z_1du^2+udfracdz_1du$

$therefore u^2dfracd^2z_1du^2+udfracdz_1du+left(dfrac4u^2-4u^2+1+dfrac6u+6right)udfracdz_1du+left(dfrac4u^2-4u^2+1+dfracu^2+11u+36u+6right)z_1=0$

$dfracd^2z_1du^2+dfrac1uleft(5-dfrac8u^2+1+dfrac6u+6right)dfracdz_1du+dfrac1u^2left(4-dfrac8u^2+1+dfracu^2+11u+36u+6right)z_1=0$

$dfracd^2z_1du^2+left(dfrac5u-dfrac8u(u^2+1)+dfrac6u(u+6)right)dfracdz_1du+left(dfrac4u^2-dfrac8u^2(u^2+1)+dfracu^2+11u+36u^2(u+6)right)z_1=0$

$dfracd^2z_1du^2+left(dfrac8uu^2+1-dfrac1u+6-dfrac2uright)dfracdz_1du+left(dfrac8u^2+1+dfrac16(u+6)+dfrac56u+dfrac2u^2right)z_1=0$

Let $z_1=uz$ ,

Then $dfracdz_1du=udfracdzdu+z$

$dfracd^2z_1du^2=udfracd^2zdu^2+dfracdzdu+dfracdzdu=udfracd^2zdu^2+2dfracdzdu$

$therefore udfracd^2zdu^2+2dfracdzdu+left(dfrac8uu^2+1-dfrac1u+6-dfrac2uright)left(udfracdzdu+zright)+left(dfrac8u^2+1+dfrac16(u+6)+dfrac56u+dfrac2u^2right)uz=0$

$udfracd^2zdu^2+2dfracdzdu+left(dfrac8u^2u^2+1-dfracuu+6-2right)dfracdzdu+left(dfrac8uu^2+1-dfrac1u+6-dfrac2uright)z+left(dfrac8uu^2+1+dfracu6(u+6)+dfrac56+dfrac2uright)z=0$

$udfracd^2zdu^2+left(dfrac8u^2u^2+1-dfracuu+6right)dfracdzdu+left(dfrac16uu^2+1-dfrac2u+6+1right)z=0$

$dfracd^2zdu^2+left(dfrac8uu^2+1-dfrac1u+6right)dfracdzdu+left(dfrac16u^2+1-dfrac2u(u+6)+dfrac1uright)z=0$

$dfracd^2zdu^2+left(dfrac8uu^2+1-dfrac1u+6right)dfracdzdu+left(dfrac16u^2+1+dfrac13(u+6)+dfrac23uright)z=0$