# Solving higher order ODEs

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?

• 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. Jan 29, 2018 at 22:47
• Ah thank you very much I'll take a look. Jan 29, 2018 at 22:50
• 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 Jan 29, 2018 at 22:51

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:$$y_g=y_h+y_p=Ae^{r_1t}+Be^{r_2t}+Ce^{r_3t}+\dfrac{9}{82}\sin t+\dfrac{1}{82}\cos t+\dfrac{2}{5}$$

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

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 : $$R^2-5R+5=0$$ $$............$$

Soving this equation with free CAS Maxima