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show: if $y_1$ is a solution to 3rd order linear homogeneous equation $y'''+p_2(x)y''+p_1(x)y'+p_0(x)y = 0 $ (1) and $y_2 = u(x)y_1(x)$ then $v =u'$ satisfies 2nd order linear homogeneous DE

So far...

  • I took $y_2 ' $, $y_2''$, and $y_2'''$ and substitues into the above formula (1)

  • how do we end up showing $v =u'$ satisfies 2nd order linear homogeneous DE? Any hints?

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  • $\begingroup$ After substituting in the derivatives of $y_{2}$, factoring $u, u', u''$ and $u'''$ and noting that the ODE with $u$ factored is the given ODE and hence $0$, you end up with $$u''' (y_{1}) + u''(3y_{1}' + p_{2}y_{1}) + u'(3y_{1}'' + 2p_{2}y_{1}' + p_{1}y_{1}) = 0$$ Making the substitution $v = u'$ gives you a second order ODE in $v$. $\endgroup$ – Mattos Dec 1 '16 at 6:10
  • $\begingroup$ after substitution i get: $u′′′(y_1)+u′′(3y_1'+p_2y_1)+u′(3y_1''+2p_2y_1'+p_1y_1)+u(y_1'''+p_2y_1''+p_1y_1'+p_0y_1)=0$ $\endgroup$ – ISuckAtMathPleaseHELPME Dec 1 '16 at 6:32
  • $\begingroup$ $y_{1}''' + p_{2}y_{1}'' + p_{1}y_{1}' + p_{0}y_{1} = 0$ $\endgroup$ – Mattos Dec 1 '16 at 6:33
  • $\begingroup$ what happens to $p_0y_1$ $\endgroup$ – ISuckAtMathPleaseHELPME Dec 1 '16 at 6:34
  • $\begingroup$ You'll have to be more specific, I don't know what you are asking. $\endgroup$ – Mattos Dec 1 '16 at 6:39

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