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 October 22nd, 2019, 04:29 AM #1 Senior Member   Joined: Dec 2015 From: Earth Posts: 827 Thanks: 113 Math Focus: Elementary Math Non-order DE $\displaystyle y^{(4)} -y+e^{-x} =0$.
 October 22nd, 2019, 06:35 AM #2 Senior Member   Joined: Dec 2015 From: Earth Posts: 827 Thanks: 113 Math Focus: Elementary Math One solution can be found below : $\displaystyle (y'+y)^{(3)} -(y''-y')' -e^{x} (y''e^{-x} -e^{-x}y)=-e^{-x}$. $\displaystyle [e^{-x}(ye^{x})']^{(3)}-(y''-y')'-e^{x}\frac{dW(e^{-x} , y)}{dx}=-e^{-x}$. $\displaystyle \int d[e^{-x}(ye^{x})']''-\int d(y''-y')-\int e^{x}dW(e^{-x},y) =C+e^{-x}$. $\displaystyle [e^{-x}(ye^x )']''-y''+y'-[e^{x}W(e^{-x},y)-\int W(e^{-x},y)e^{x}dx]=C+e^{-x}$. $\displaystyle [e^{-x}(ye^x )']''-y''+\int ydx=C+e^{-x}$. $\displaystyle -e^{-x} +y'''-[e^{-x}(ye^x )']'''=y$. $\displaystyle -e^{-x} +y'''-[y'+y]'''=-e^{-x}-y'=y$. $\displaystyle e^{x}(y'+y)=\int dye^{x} =-\int dx=C-x$. $\displaystyle y=Ce^{-x}-xe^{-x}$. C-only one value. Thanks from topsquark
 October 22nd, 2019, 08:02 AM #3 Senior Member   Joined: Jun 2019 From: USA Posts: 380 Thanks: 205 If it's a 4th-order ODE, the general solution should have four constants to solve for. Why don't you just start with $y \sim e^{\lambda x}$? Thanks from topsquark and idontknow
 October 22nd, 2019, 08:17 AM #4 Senior Member   Joined: Dec 2015 From: Earth Posts: 827 Thanks: 113 Math Focus: Elementary Math $\displaystyle \lambda^{4} e^{\lambda x} -e^{\lambda x} +e^{-x} \approx 0$; How to continue this way ?
 October 22nd, 2019, 08:33 AM #5 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,697 Thanks: 2681 Math Focus: Mainly analysis and algebra 4th order ode with constant coefficients $$y''''-y=-e^{-x}$$ has characteristic equation (for the associated homogeneous equation) \begin{align} r^4-1 &= 0 \\ (r-1)(r+1)(r-i)(r+i) &= 0 \end{align} So the complementary solutions are $$y_c = Ae^x + Be^{-x} + C\sin{x} + D\cos{x}$$ Now you just need the particular solution which should be easy to determine by the method of undetermined coefficients. Thanks from topsquark and idontknow
October 22nd, 2019, 08:36 AM   #6
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Quote:
 Originally Posted by v8archie 4th order ode with constant coefficients $$y''''-y=-e^{-x}$$ has characteristic equation (for the associated homogeneous equation) \begin{align} r^4-1 &= 0 \\ (r-1)(r+1)(r-i)(r+i) &= 0 \end{align}
Seems like I'm nowhere at all; can you post the solution?

Last edited by skipjack; October 22nd, 2019 at 11:06 AM.

 October 23rd, 2019, 06:52 AM #7 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,697 Thanks: 2681 Math Focus: Mainly analysis and algebra The solution of any linear ODE with constant coefficients $$a_ny^{n} + a_{n-1}y^{n-1} + \ldots + a_1y' + a_0y =g(x)$$ is given by $$y = y_c + y_p$$ where $y_c$, the complimentary solution, is the general solution of the homogeneous equation $$a_ny^{n} + a_{n-1}y^{n-1} + \ldots + a_1y' + a_0y=0$$ and $y_p$, the particular solution, is any solution of the original equation. The solutions of the homogeneous equation are all of the form $e^{rx}$ (where $r$ is possibly complex) and by substituting $y=e^{rx}$ into the homogenous equation we find the characteristic polynomial which can be solved for values of $r$, the number of solutions determined by the degree of the characteristic polynomial which in turn is determined by the order of the original ODE. In this case we get \begin{align} r^4-1 &= 0 \\ (r-1)(r+1)(r-i)(r+i) &= 0 \end{align} It can be shown that real-valued solutions of the ODE with complex-conjugate values of $r = u \pm iv$ are equivalent to solutions $y = e^{ux}(A\sin vx + B\cos vx)$. Thus the complementary solution of $$y''''-y=-e^{-x}$$ is $$y_c = Ae^x + Be^{-x} + C\sin{x} + D\cos{x}$$ For the particular solution, using the method of undetermined coefficients (other methods are available) we guess the appropriate template for a solution, in this case $y_p = Exe^{-x}$ where $E$ is a constant to be determined. The $x$ in this term comes in because $Be^{-x}$ is already a solution (where $B$ is an arbitrary constant of integration). Thus we substitute $y_p=Exe^{-x}$ into the original ODE to get \begin{align} Exe^{-x} - 4Ee^{-x} - Exe^{-x} &= -e^{-x} \\ \implies E &= \tfrac14 \end{align} And so we have a solution \begin{align} y &= y_c + y_p \\ &= \tfrac14xe^{-x} + Ae^x + Be^{-x} + C\sin{x} + D\cos{x} \end{align} All of the above is a method standard for solving a second order ODE with constant coefficients extended to the fourth order equation you presented. This extension should be taught in the same course as the method for solving second order equations. Thanks from topsquark, idontknow and DarnItJimImAnEngineer Last edited by v8archie; October 23rd, 2019 at 07:00 AM.

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