The correct solution $y_p$ for $y'' + 4y' + 3y = 6e^{-x}$ is $y_p = Ce^{-x}$. False True
Added by Bruce W.
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First, let's simplify the equation: y + 4y + 3y = 6e 8y = 6e Now, let's solve for y by dividing both sides of the equation by 8: y = 6e/8 Simplifying further: y = 3e/4 So, the correct solution for y is y = 3e/4. Show more…
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$$\begin{aligned}8 y-6 =-2+10 y \\8 y-6- \underline{\quad}=-2+10 y- \underline{\quad}\\-6 &=-2+ \underline{\quad}\\ -6+ \underline{\quad}=-2+2 y+ \underline{\quad} \\-4 = \underline{\quad}\\ \frac{-4} {\quad} = \frac{2 y}{\quad} \\ \underline{\quad}=y \end{aligned}$$ $$\begin{aligned}\text{Check:} \\8 y-6=-2+10 y\\8 \underline{\quad}-6-2+10 \underline{\quad}\\ -16-6 \underline{?}-2+ \underline{\quad} \\ \underline{\quad}=-22 \quad \text { True } \\ \text{The solution is} \underline{\quad\quad}\end{aligned}$$
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