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Solve the following ODE using the method of undetermined coefficients:

$$y''-4y'+4y=(x+1)e^{2x}$$

My work. I have the homogeneous solution: $$y_h=C_1e^{2t}+C_2te^{2t}$$ I use this function to get the particular solution of the ODE: $$y=(Ax+B)xe^{2x}$$ And after substitution on the ODE, I get: $$2A=(x+1)e^{2x}$$ What about $B$? I don't know what I am missing. Could someone give me a hint? Thank you.

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  • $\begingroup$ What are the boundary conditions? $\endgroup$ – Joe Goldiamond Jun 10 '18 at 16:58
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Hint. Since the characteristic polynomial is $(z-2)^2$ then the multiplicity $m$ of the root $2$ is $2$. It follows that for $f(x)=(x+1)e^{2x}$ you should try a particular solution of the form $$y(x)=(Ax+B)x^me^{2x}=(Ax+B)x^2e^{2x}.$$

P.S. If you use the form $y(x)=(Ax+B)2e^{2x}$ then after the substitution you should get $2Ae^{2x}=xe^{2x}+e^{2x}$ which is impossible to solve.

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  • $\begingroup$ Thank you Robert! I finally solved it $\endgroup$ – Guille Jun 10 '18 at 17:26
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For the particular solution, Robert's suggestion is the best $$y_p=(ax^3+bx^2)e^{2x}$$

Since you already have terms in $e^{2x},xe^{2x}$

Another solution

$$y''-4y'+4y=(x+1)e^{2x}$$ $$y''e^{-2x}-4y'e^{-2x}+4ye^{-2x}=(x+1)$$ $$y''e^{-2x}-2y'e^{-2x}-2y'e^{-2x}+4ye^{-2x}=(x+1)$$ $$(y'e^{-2x})'-2(ye^{-2x})'=(x+1)$$ Simply integrate $$y'e^{-2x}-2ye^{-2x}=\frac {x^2}2+x+K_1$$ $$(ye^{-2x})'=\frac {x^2}2+x+K_1$$ Integrate again $$ye^{-2x}=\frac {x^3}6+\frac {x^2}2+K_1x+K_2$$ Therefore $$\boxed{y(x)=e^{2x}\left(\frac {x^3}6+\frac {x^2}2+K_1x+K_2\right )}$$

You can see that the particular solution $y_p$ is $$y_p=e^{2x}(\frac {x^3}6+\frac {x^2}2)$$

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    $\begingroup$ Thanks Isham! I get to that solution $\endgroup$ – Guille Jun 10 '18 at 17:32
  • $\begingroup$ @Guille that's great.... $\endgroup$ – Isham Jun 10 '18 at 17:33
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You've probably made a mistake substituting it back in. After simplifying, you should arrive at $$ 6Axe^{2x}+2Be^{2x}=xe^{2x}+e^{2x} $$

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  • $\begingroup$ After the substitution you should get $2Ae^{2x}=xe^{2x}+e^{2x}$. The problem is the form of the particular solution! $\endgroup$ – Robert Z Jun 10 '18 at 17:23
  • $\begingroup$ @RobertZ Ah yes, I didn't see that. $\endgroup$ – Infiaria Jun 10 '18 at 17:50

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