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Find a linear homogeneous constant-coefficient differential equation with the general solution given by \[ y=A e^{-3 x}+B x e^{-3 x}+C \]

User Czioutas
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Answer:

To find the linear homogeneous constant-coefficient differential equation with the given general solution we begin by taking the derivative of y with respect to x.

\[ \frac{dy}{dx} = -3A e^{-3x} + Be^{-3x} - 3Bxe^{-3x} \]

Now we can set up the differential equation by equating the derivative of y to a linear combination of y itself:

\[ y' + ay = 0 \]

where a is a constant coefficient. Substituting the expressions for y and y' into the differential equation we have:

\[ -3A e^{-3x} + Be^{-3x} - 3Bxe^{-3x} + a(A e^{-3x}+B x e^{-3x}+C) = 0 \]

Simplifying this equation we get:

\[ (-3A + aA)e^{-3x} + (B - 3Bx + aBx)e^{-3x} + aC = 0 \]

For this equation to hold for all values of x each term must be equal to zero. Therefore we have the following system of equations:

\[ -3A + aA = 0 \]

\[ B - 3Bx + aBx = 0 \]

\[ aC = 0 \]

Solving these equations we find that aA = 3A B - 3Bx + aBx = 0 and aC = 0. Since a can be any constant it is clear that A = 0 C = 0 and B can be any constant.

Hence the linear homogeneous constant-coefficient differential equation with the general solution given by \[ y=A e^{-3 x}+B x e^{-3 x}+C \] is:

\[ y' + 3y = 0 \]

User Snowangel
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