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So the question is to solve this differential equation: $y'' - 4y' + 5y = 1 + t$.

I've got the general solution to $\ C_1e^{2t}\cos(2t) + C_2e^{2t}\sin(2t) $

However, I don't really know how to proceed with the particular solution.

I know that t is a polynomial of the first degree and should be written as $At + B$. After that I'm pretty much stuck. One particular question I have though, should $1$ be a constant $D$ which would make the equation for the particular solution: $y = At + B + D$? Or is that redundant since we already have the constant $B$?

As you see, my main concern is how handle the "set-up" of the particular solution.

Follow up question: Is it some general rule to follow when setting up these particular solutions? I'm feeling pretty lost and want to really learn the concept.

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  • $\begingroup$ Yes having extra constants like that is redundant as it can be rewritten as another constant $\endgroup$ – Triatticus Feb 3 '18 at 21:55
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    $\begingroup$ B + D, or maybe just call it C. Anyway, you have the form, why not put it in for y? $\endgroup$ – Kaynex Feb 3 '18 at 21:57
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    $\begingroup$ What you wrote is not the general solution, it is the homogenous solution. $\endgroup$ – Joel Feb 3 '18 at 22:03
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The method of undetermined coefficients has been called 'educated guesswork' for finding particular solutions. One attempts trial solutions based on the form of the right-hand-side (RHS), bearing in mind the form of the complementary function. In this case, the RHS is a polynomial of degree 1, and as it does not appear in your complementary function, you pick a trial solution as a polynomial of degree 1 i.e. $y_P=At+B$. Just one additive constant. Note there are other, more systematic, methods for solving these problems such as variation of parameters.

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  • $\begingroup$ For arguments sake (or more for me to understand the concept better), what would one do if it did appear in my complementary function? $\endgroup$ – gbgult Feb 3 '18 at 22:07
  • $\begingroup$ The problem is if your ODE is Ly=f and f is part of the CF, then Lf=0 so trying yp=f is going to fail. You then typically try to find particular solutions by multiplying by (powers) of t i.e try yp= A t f say, where A is a const. $\endgroup$ – PM. Feb 3 '18 at 22:10
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make for the particular solution the ansatz $$y_p=At^2+Bt+C$$

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First, your homogeneous solution contains a mistake. It should be:

$$y_h = C_1e^{2t}\cos(t) + C_2e^{2t}\sin(t)$$


Now, this equation allows for a simple particular solution, but a more general way to find a particular solution is making one of the constants depend on $t$ and substituting this into the equation:

$$y_p=C(t)e^{2t}\cos(t)=C y_0$$

Where:

$$y_0''-4y_0'+5y_0=0$$


$$y_p'= C'y_0+Cy_0'$$

$$y_p''= C''y_0+2C'y_0'+Cy_0''$$

Now substitute into the original equation:

$$C''y_0+2C'y_0'+\color{blue}{Cy_0''}-4(C'y_0+\color{blue}{Cy_0'})+\color{blue}{5C y_0}=t+1$$

The sum of the blue parts is equal to $0$ by definition so our equation becomes:

$$y_0 C''+2(y_0'-2y_0)C'-t-1=0$$

or:

$$e^{2t} \cos t~ C''-2 e^{2t} \sin t~ C'-t-1=0$$

Making a substitution:

$$C'(t)=f(t)$$

$$e^{2t} \cos t~ f'-2 e^{2t} \sin t~ f-t-1=0$$

And we obtained a 1st order ODE, which can be solved and in the end get us $C(t)$ and consequently, $y_p(t)$.

The solution is simple in this case, as the homogeneous equation becomes:

$$f_h'-2 \tan t~ f_h=0$$

$$f_h(t)=\frac{A}{\cos^2 t}$$

Now again, we make $A$ depend on $t$:

$$f_p(t)=\frac{A(t)}{\cos^2 t}=A f_0$$

So we find the general solution for $f(t)$, which we then integrate to find $C(t)$ and our particular solution $y_p(t)$ is known.


Again, for this case it doesn't make sense to use this complicated procedure, but for the more general case of inhomogeneous equations this is a sure way to obtain a solution.

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  • $\begingroup$ Thanks for the great comment. First of all, where is the mistake? Isn't the roots: $2 +_- 2i$ which makes my homogenous solution correct? Or have I mistaken how to handle imaginary numbers? Secondly, I'm gonna try to understand how to use your way of finding a $Y_p (t)$. Do you have any tips of videos/online sources where I could see some more examples? That's the best way for me to learn. $\endgroup$ – gbgult Feb 4 '18 at 9:33
  • $\begingroup$ No, the roots are $2 \pm i$, hence your mistake. If this is an assignment, make sure to fix your solution. $$ $$ As for the method, it's called variation of parameters (of constants) and is pretty standard. See Wikipedia for example en.wikipedia.org/wiki/Variation_of_parameters $\endgroup$ – Yuriy S Feb 4 '18 at 9:38
  • $\begingroup$ Oh, sorry about that, spotted my, silly, mistake now. Thanks, I'll look into that! $\endgroup$ – gbgult Feb 4 '18 at 9:49

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