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}.

Question

Linear equations of first order.
Solve the initial-value problem on the specified interval

y^{'} - 3 y = e^{2 x} o n (- \infty, + \infty), with y = 0 when x = 0

.

Answer 1

From the given linear differential equation,

P (x) = - 3 and Q (x) = e^{2 x}

.
Therefore, the integrating factor is given by

A (x) = \int_{0}^{x} - 3 t d t = - 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} d t

= e^{3 x} \int_{0}^{x} e^{- t} d t

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

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

= 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.

Expert Answer