# 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
$$\frac{d y_{2}}{d x}+2 y_{2}=-2, \quad y_{2}(3)=y_{1}(3)$$

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
$$y_{2}=-1+c e^{-2 x}$$
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