## Related questions with answers

Find the general solution. Indicate which method in this chapter you are using. Show the details of your work. y'+2.5y=1.6x

Solution

VerifiedThis is $\textbf{linear, non-homogeneous ODE}$.

In this case, we have:

$\begin{align*} p(x) &= 2.5 \implies h = 2.5\int p \; dx = 2.5\int dx =2.5 x \\ r(x) &= 1.6 x \end{align*}$

From this, we can see that:

$e^h = e^{2.5x} \quad \wedge \quad e^{-h} = e^{-2.5x} \quad \wedge \quad re^h = 1.6 x e^{2.5x}$

So, the general solution is:

$\begin{align*} y(x) &= e^{-h} \left(c + \int re^h \; dx \right)\\ & = e^{-2.5x}\Big(c + 1.6\overbrace{\int xe^{2.5x} \; dx}^{\begin{smallmatrix} \text{\textcolor{Fuchsia}{(*)}} \; \text{Solved below} \end{smallmatrix}} \Big) \\ & = e^{-2.5x}\Big(c + 1.6 \cdot 0.4 e^{2.5x}\left( x - 0.4 \right) \Big) \\ & = ce^{-2.5x} + 0. 64\cancel{e^{-2.5x}} \cancel{e^{2.5x}} \left( x - 0.4 \right) \\ & = \boxed{{\color{#4257b2}{ce^{-2.5x} + 0. 64x - 0.256 }}} \end{align*}$

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