Final answer to the problem
$y=e^{-3x}\left(\frac{e^{3x}x}{3}+\frac{-e^{3x}}{9}+C_0\right)$
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1
Apply the formula: $a=b$$\to \frac{a}{dx}=extdiff\left(\frac{b}{dx}\right)$, where $a=dy$, $b=\left(x-3y\right)dx$ and $a=b=dy=\left(x-3y\right)dx$
$\frac{dy}{dx}=\left(x-3y\right)\frac{dx}{dx}$
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$\frac{dy}{dx}=\left(x-3y\right)\frac{dx}{dx}$
Learn how to solve problems step by step online. dy=(x-3y)dx. Apply the formula: a=b\to \frac{a}{dx}=extdiff\left(\frac{b}{dx}\right), where a=dy, b=\left(x-3y\right)dx and a=b=dy=\left(x-3y\right)dx. Apply the formula: \frac{a}{a}=1, where a=dx and a/a=\frac{dx}{dx}. Apply the formula: 1x=x, where x=x-3y. Rearrange the differential equation.
Final answer to the problem
$y=e^{-3x}\left(\frac{e^{3x}x}{3}+\frac{-e^{3x}}{9}+C_0\right)$
