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Given that $y=\frac{1}{2} e^{2 x}+\left(x-\frac{1}{3}\right) e^{x}$ is a special solution of the differential equation $y^{\prime \prime}+a y^{\prime}+b y=c e^{x}$ (a, b, c are constants), I want to find the value of a, b, c.

y[x_] := 1/2 E^(2 x) + (x - 1/3) E^x
SolveAlways[(D[#, {x, 2}] + a*D[#, x] + b*# == c*E^x) &@y[x], x]

But the above code can't solve this problem (the answer is a=-3,b=2,c=-1). How can I use function SolveAlways to solve this problem?

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1 Answer 1

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From the documentation for SolveAlways:

SolveAlways works primarily with linear and polynomial equations.

So let's make it linear! There are three linearly independent functions in here, the constant function, $x$, and $e^x$, so let's set the exponential function to y:

y[x_] = 1/2 E^(2 x) + (x - 1/3) E^x;
(D[#, {x, 2}] + a*D[#, x] + b*# == c*E^x) &@y[x] // Simplify;
SolveAlways[% /. Exp[x] -> y, {x, y}];
(* {{a -> -3, b -> 2, c -> -1}} *)
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