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I am looking to solve for the above nonhomogeneous ODE. I know how to find the general solution for the reduced equation of the homogeneous form, that is, $$y'' - 4y' - 5y = 0.$$

The characteristic equation is $r^{2} - 4r - 5 = 0$, which gives two real and distinct roots $r=-1,5$.

So the complementary solution is $y_{c}=c_{1}e^{5x} + c_{2}e^{-x}$.

Now I am looking to guess the particular on the right-hand side but I am not sure about how to do that in order to find the general solution of the above nonhomogenous ODE.

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  • 2
    $\begingroup$ If you cannot see the one below, you could try polynomials, i.e. assume $y = Ax^2 + Bx = C$ satisfies the nonhomogeneous ODE, then plug it into the equation to solve for $A,B,C$ you might be able to get a same answer. $\endgroup$ – xbh Nov 28 '18 at 9:30
  • $\begingroup$ Define $z=5y+2$ and see what happens. $\endgroup$ – Claude Leibovici Nov 28 '18 at 9:44
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$$y''-4y'-5y=2$$

If $y=-\frac25$ everywhere, then $y'=y''=0$.

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Hint:

Let $y=z+a+bx+cx^2+d\cdot x^3$

$y'=z'+b+2cx+3d\cdot x^2$

$y''=z''+2c+6\cdot dx$

$$2=y''-4y'-5y=z''-4z'-5z+x^3(d)+x^2(c-12d)+x(b-8c-30d)-5a-4b+2c$$

Set $0=d=c-12d=b-8c-30d\implies b=c=d=0;$

$2=2c-4b-5a\iff a=?$

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