What is the general solution of the differential equation $\frac{d y}{d x} + 3 y = e^{4 x}$ $dy/dx +3y = e^(4x)$ ?

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Answer 1 · 2017-10-06T23:43:45

$y = \frac{1}{7} e^{4 x} + C e^{- 3 x}$ $y = 1/7e^(4x) + Ce^(-3x)$

We can use an integrating factor when we have a First Order Linear non-homogeneous Ordinary Differential Equation of the form;

$\frac{d y}{d x} + P (x) y = Q (x)$ $dy/dx + P(x)y=Q(x)$

Ther equation is already in standard form:

$\frac{d y}{d x} + 3 y = e^{4 x}$ $dy/dx +3y = e^(4x)$ ..... [A}

Then the integrating factor is given by;

$I = e^{\int P (x) d x}$ $I = e^(int P(x) dx)$
$\ \ = exp(int \ 3 \ dx)$
$\ \ = exp( 3x )$
$\ \ = e^(3x)$

And if we multiply the DE [A] by this Integrating Factor, $I$ , we will have a perfect product differential;

$dy/dx +3y = e^(4x)$

$:. e^(3x)dy/dx +3e^(3x)y = e^(3x)e^(4x)$

$:. d/dx( e^(3x)y) = e^(7x)$

Which we can directly integrate to get:

$\ \ \ \ \ e^(3x)y = 1/7e^(7x) + C$

$:. y = 1/7e^(7x)e^(-3x) + Ce^(-3x)$

$:. y = 1/7e^(4x) + Ce^(-3x)$

What is the general solution of the differential equation dy/dx +3y = e^(4x) dydx+3y=e4x dy/dx +3y = e^(4x) ?