Question

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}.

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asked 2021-03-08
Linear equations of first order.
Solve the initial-value problem on the specified interval \(\displaystyle{y}'-{3}{y}={e}^{{{2}{x}}}{o}{n}{\left(-\infty,+\infty\right)},\ \text{with}\ {y}={0}\ \text{when}\ {x}={0}\).

Answers (1)

2021-03-09
From the given linear differential equation, \({P}{\left({x}\right)}=-{3}{\quad\text{and}\quad}{Q}{\left({x}\right)}={e}^{{{2}{x}}}\).
Therefore, the integrating factor is given by
\(A{\left({x}\right)}={\int_{{0}}^{{x}}}-{3}{t}{\left.{d}{t}\right.}=-{3}\)
So that, the solution is given by
\(y={0}{e}^{{{3}{x}}}+{e}^{{{3}{x}}}{\int_{{0}}^{{x}}}{e}^{{-{3}{t}}}{e}^{{{2}{t}}}{\left.{d}{t}\right.}\)
\(={e}^{{{3}{x}}}{\int_{{0}}^{{x}}}{e}^{ -{{t}}}{\left.{d}{t}\right.}\)
\(={e}^{{{3}{x}}}{{\left[\frac{{e}^{ -{{t}}}}{ -{{1}}}\right]}_{{0}}^{{x}}}\)
\(={e}^{{{3}{x}}}{\left({1}-{e}^{ -{{x}}}\right)}\)
\(={e}^{{{3}{x}}}-{e}^{{{2}{x}}}\)
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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