# Linear equations of first order. Solve the initial-value problem on the specified interval displaystyle{y}'-{3}{y}={e}^{{{2}{x}}}{o}{n}{left(-infty,+inftyright)}, text{with} {y}={0} text{when} {x}={0}.

Linear equations of first order.
Solve the initial-value problem on the specified interval .
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Nola Robson
From the given linear differential equation, $P\left(x\right)=-3\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}Q\left(x\right)={e}^{2x}$.
Therefore, the integrating factor is given by
$A\left(x\right)={\int }_{0}^{x}-3tdt=-3$
So that, the solution is given by
$y=0{e}^{3x}+{e}^{3x}{\int }_{0}^{x}{e}^{-3t}{e}^{2t}dt$
$={e}^{3x}{\int }_{0}^{x}{e}^{-t}dt$
$={e}^{3x}{\left[\frac{{e}^{-t}}{-1}\right]}_{0}^{x}$
$={e}^{3x}\left(1-{e}^{-x}\right)$
$={e}^{3x}-{e}^{2x}$
We can verify the function is the solution of the initial-value periblem since it satisfies the differential equation and the initial condition.