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I want to solve the differential equation $y'''(x)-2y''(x)+y'(x)=e^x$.

I have done the following:

Consider the homogeneous equation $y'''(x)-2y''(x)+y'(x)$.

$k^3-2k^2+k=0 \Rightarrow k=0 \text{ single root } , k=1 \text{ double root } $

So, the solution of the homogeneous problem is $y_h(x)=c_1+c_2e^x+c_3xe^x$.

Since $1$ is a root of the characteristic polynomial of multiplicity $2$, we consider that the partial solution is of the form $y_p(x)=Ax^2e^x$, or not?

Then we have the following:

$$y_p(x)=Ax^2e^x \\ y_p'(x)=2Axe^x+Ax^2e^x \\ y_p''(x)=2Ae^x+4Axe^x+Ax^2e^x \\ y_p'''(x)=6Ae^x+6Axe^x+Ax^2e^x$$ right?

Replacing these at the differential equation we get $$2Ax^2e^x-2Axe^t+2Ae^x=e^x$$ or have I done something wrong?

How could we continue?

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EDIT:

Considering $y_p=(Ax^2+Bx+C)e^{x}$.

Finding the derivatives $y'_p,y''_p,y'''_p$ and replacing it at the problem I find $A=\frac{1}{2}$.

Is this correct?

What values do $B$ and $C$ get?

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The equation is $$ (D-1)^2Dy=e^x\tag{1} $$ and applying $D-1$ again, we get $$ (D-1)^3Dy=0\tag{2} $$ Equation $(2)$ has a general solution of the form $$ y=\left(Ax^2+Bx+C\right)e^x+F\tag{3} $$ and if we plug $(3)$ into $(1)$ we find that $A=\frac12$.

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  • $\begingroup$ I see... Thank you very much!! :-) $\endgroup$ – Mary Star Apr 1 '16 at 16:18
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make the ansatz $$y_p=(Ax^2+Bx+C)e^{x}$$

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  • $\begingroup$ Finding the derivatives $y_p', y_p'', y_p'''$ and replacing it at the problem I find $A=\frac{1}{2}$. Is this correct? What values do $B$ and $C$ get? $\endgroup$ – Mary Star Apr 1 '16 at 15:27

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