$$\frac{dy}{dx} – 2y = 3x + 1$$
$$e^{ \int \! -2 \ dx } = e^{-2x}$$
$$e^{-2x} \frac{dy}{dx} – 2 e^{-2x} y = e^{-2x} \left( 3x + 1 \right)$$
$$\frac{d}{dx} \left( e^{-2x} y \right) = 3 x e^{-2x} + e^{-2x}$$
$$e^{-2x} y + C = \int \! 3 x e^{-2x} \ dx + \int \! e^{-2x} \ dx$$
$$e^{-2x} y = \int \! 3 x e^{-2x} \ dx – \frac{1}{2}e^{-2x} + K \ \ where \ \ K = \left( C_0 – C \right)$$
\( \int \! 3 x e^{-2x} \ dx \)
\( u = 3x \)
\( du = 3 \ dx \)
\( dv = e^{-2x} \ dx \)
\( v = – \frac{1}{2}e^{-2x}\)
\( \int \! 3 x e^{-2x} \ dx = – \frac{3}{2}xe^{-2x} + \int \! \frac{3}{2}e^{-2x} \ dx \)
\( \int \! 3 x e^{-2x} \ dx = – \frac{3}{2}xe^{-2x} – \frac{3}{4}e^{-2x} + C_1 \)
$$e^{-2x} y = – \frac{3}{2}xe^{-2x} – \frac{5}{4}e^{-2x} + C_2 \ \ where \ \ C_2 = \left( K + C_1 \right)$$
$$y \left( x \right) = – \frac{3}{2}x – \frac{5}{4} + C_2e^{2x}$$
$$\frac{dy}{dx} + \cos \left(x \right) y = \cos x$$
$$\mu = e^{ \int \! \cos x \ dx } = e^{ \sin x }$$
$$e^{ \sin x } \frac{dy}{dx} + e^{ \sin x } \cos \left(x \right) y = e^{ \sin x } \cos x$$
$$\frac{d}{dx} \left( e^{ \sin x } y \right) = e^{ \sin x } \cos x$$
$$\left( e^{ \sin x } y \right) = \int \! e^{ \sin x } \cos x \ dx + C$$
\( v = \sin x \)
\( dv = \cos x \ dx \)
\( \int \! e^v \ dv = e^v + K = e^{ \sin x } + C_1\)
$$\left( e^{ \sin x } y \right) = e^{ \sin x } + C_1 + C = e^{ \sin x } + C_2$$
$$y \left( x \right) = 1 + C_2 e^{ – \sin x }$$