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Find all solutions of $$y'''-4y''+5y'-2y=-x^2+5x+2.$$

I know how to find the solutions of the corresponding homogenous differential equation $y'''-4y''+5y'-2y=0$. I've done that in the following way:

The characteristic equation is $$P(x) = x^3 - 4x^2 +5x-2=(x-1)^2(x-2).$$

So if we put $$f_1:x\mapsto e^x\\f_2:x\mapsto xe^x\\f_3:x\mapsto e^{2x}$$ then the solution basis is $$\{f_1, f_2, f_3\}$$ and all solutions are linear combinations of this basis.

However, I am not sure how to deal with the inhomogenous part of the differential equation.

Please share the general approach with me.

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  • $\begingroup$ I suppose a typo in the rhs. Please check and fix. $\endgroup$ – Claude Leibovici Aug 25 '14 at 6:55
  • $\begingroup$ Fixed it, my apologies :) $\endgroup$ – rehband Aug 25 '14 at 6:57
  • $\begingroup$ Go and learn some Laplace transform. It should be given by a convolution. $\endgroup$ – Troy Woo Aug 25 '14 at 6:57
  • $\begingroup$ $f_3$ is probably $e^{2x}$ $\endgroup$ – Claude Leibovici Aug 25 '14 at 7:00
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Use the method of undetermined coefficients. See second case here. You can try this out.

Essentially, you guess y looks like a polynomial e.g. $y = a_0+a_1x+a_2x^2+...+a_nx^n$ because that is what the RHS is. Then, plug in y.

So the question is what n to use?

Try 5. The highest order derivative is 3 and the degree of the RHS is 2. is y has $a_5x^5$ and is differentiated thrice, it will have an $x^2$. If it doesn't have a $x^5$, the coefficient will turn out $a_5=0$

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Hint

It may not be a very academic solution but since you found the solution for the homogenous part, let us set, because the rhs is a polynomial, that $$y=c_1 e^x+c_2 x e^{x}+c_3 e^{2x}+P(x)$$ Differentiate three times and substitute. You end with $$P'''(x)-4 P''(x)+5 P'(x)-2 P(x)=-x^2+5x+2$$

Try $P(x)=a+bx+x^2+dx^3$ and identify.

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  • $\begingroup$ You sure P(x) of degree 3 is the right guess? math.stackexchange.com/a/908413/140308 $\endgroup$ – BCLC Aug 25 '14 at 7:48
  • $\begingroup$ Not for sure ! It is clear that degree $2$ is clear. $\endgroup$ – Claude Leibovici Aug 25 '14 at 7:52
  • $\begingroup$ What? I mean it looks like the worst case scenario is that the third derivative is of degree 2 so why not try 5...? $\endgroup$ – BCLC Aug 25 '14 at 8:00
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    $\begingroup$ @BCLC. You could easily show that $P(x)$ must be a polynomial of the same degree of the highest power in the rhs because the lhs is a linear combination of the derivatives. $\endgroup$ – Claude Leibovici Aug 25 '14 at 8:03
  • $\begingroup$ Well Wikipedia seems to agree... en.wikipedia.org/wiki/… $\endgroup$ – BCLC Aug 25 '14 at 9:54
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For the particular solution, I usually just guess. For this particular problem, a quadratic polynomial $y(x)=ax^2+bx+c$ seems like a reasonable candidate (edit: because the RHS is also a quadratic polynomial). If you play around with the arithmetic, you will see that $$ y(x)=\frac{1}{2}x^2-3 $$ works.

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