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arrow_back Find the general solution for $x>3$. For $x>3$, we have $$ \frac{d y_{2}}{d x}+2 y_{2}=-2, \quad y_{2}(3)=y_{1}(3) $$

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Find the general solution for $x>3$. For $x>3$, we have
$$
\frac{d y_{2}}{d x}+2 y_{2}=-2, \quad y_{2}(3)=y_{1}(3)
$$

1 Answer

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Best answer
For $x>3$, we have
$$
\frac{d y_{2}}{d x}+2 y_{2}=-2, \quad y_{2}(3)=y_{1}(3)
$$
Solving this is similar to solving the IVP for $y_{1}$. Here we have
$$
y_{2}=-1+c e^{-2 x}
$$
by Bronze Status
(7,476 points)

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