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2021-03-08

Linear equations of first order.

Solve the initial-value problem on the specified interval${y}^{\prime}-3y={e}^{2x}on(-\mathrm{\infty},+\mathrm{\infty}),\text{}\text{with}\text{}y=0\text{}\text{when}\text{}x=0$ .

Solve the initial-value problem on the specified interval

Nola Robson

Skilled2021-03-09Added 94 answers

From the given linear differential equation, $P\left(x\right)=-3{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}Q\left(x\right)={e}^{2x}$ .

Therefore, the integrating factor is given by

$A\left(x\right)={\int}_{0}^{x}-3tdt=-3$

So that, the solution is given by

$y=0{e}^{3x}+{e}^{3x}{\int}_{0}^{x}{e}^{-3t}{e}^{2t}dt$

$={e}^{3x}{\int}_{0}^{x}{e}^{-t}dt$

$={e}^{3x}{\left[\frac{{e}^{-t}}{-1}\right]}_{0}^{x}$

$={e}^{3x}(1-{e}^{-x})$

$={e}^{3x}-{e}^{2x}$

We can verify the function is the solution of the initial-value periblem since it satisfies the differential equation and the initial condition.

Therefore, the integrating factor is given by

So that, the solution is given by

We can verify the function is the solution of the initial-value periblem since it satisfies the differential equation and the initial condition.

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