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I want to solve the differential equation $$\dddot{y}-4\ddot{y}+5y-2=\sin t$$ but since now I only know how to solve first order differential equations using variation of parameters or separating variables.

What is the general procedure to solve such an equation?

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  • $\begingroup$ Isn't the variation of parameters a method for 2nd and higher order equations? Anyway, this is straightforward by the method of undetermined coefficients, such as the example here. $\endgroup$ Jan 29, 2018 at 22:47
  • $\begingroup$ Ah thank you very much I'll take a look. $\endgroup$ Jan 29, 2018 at 22:50
  • $\begingroup$ Also, you could look up systems of first order linear ODEs, because higher order ODEs can be reduced to that by introducing new functions $y'=z_1$, $y''=z_2$ and so on $\endgroup$
    – Yuriy S
    Jan 29, 2018 at 22:51

3 Answers 3

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In fact you must separate the answer by two: homogeneous and private and then add them up to obtain general answer since the ODE is linear. The homogeneous answer is the answer of the following ODE $$y'''-4y''+5y=0$$the characteristic equation is$$r^3-4r^2+5=0$$with roots $r_1,r_2,r_3$ and the homogeneous answer would be $$y_h=Ae^{r_1t}+Be^{r_2t}+Ce^{r_3t}$$for arbitrary constants $A$, $B$ and $C$. The private answer is that of the following ODE:$$y'''-4y''+5y=2+\sin t$$which according to sinusoid form of input should be of form:$$y_p=P\sin t+Q\cos t+R$$by substitution in the ODE we get:$$(9P+Q)\sin t+(9Q-P)\cos t+5R=2+\sin t$$which yields to $P=\dfrac{9}{82},Q=\dfrac{1}{82}$ and $R=\dfrac{2}{5}$. Therefore the private answer is $$y_p=\dfrac{9}{82}\sin t+\dfrac{1}{82}\cos t+\dfrac{2}{5}$$and the general answer would be obtained adding up the homogeneous and private answers as following:$$y_g=y_h+y_p=Ae^{r_1t}+Be^{r_2t}+Ce^{r_3t}+\dfrac{9}{82}\sin t+\dfrac{1}{82}\cos t+\dfrac{2}{5}$$

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    $\begingroup$ By "private answer" you mean "particular solution" I presume. The term you use might be confusing. $\endgroup$
    – Yuriy S
    Jan 29, 2018 at 23:01
  • $\begingroup$ Yes that's true... $\endgroup$ Jan 29, 2018 at 23:02
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    $\begingroup$ the characteristic equation should be $r^3-4r^2+5=0$. The "-2"-term should be part of the inhomogeneous part $\endgroup$
    – Kel
    Jan 29, 2018 at 23:09
  • $\begingroup$ Sorry my bad! I'm gonna fix it! $\endgroup$ Jan 29, 2018 at 23:10
  • $\begingroup$ Thanks for the comments :) $\endgroup$ Jan 29, 2018 at 23:41
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Simple Hint

$$y'''-4y''+5y-2=\sin t$$ $$y'''+y''-5y''-5y'+5y'+5y=2+\sin t$$ Substitute $g=y'+y$ to reduce the order $$g''-5g'+5g=2+\sin t$$ Then the characteristic equation becomes : $$R^2-5R+5=0$$ $$............$$

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Soving this equation with free CAS Maxima enter image description here

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