Solution by Integrating Factor/Examples/y' - 3y = sin x/Proof 1

Theorem

The linear first order ODE:

$\dfrac {\d y} {\d x} - 3 y = \sin x$

has the general solution:

$y = \dfrac 1 {10} \paren {3 \sin x - \cos x} + C e^{3 x}$

Proof

This is a linear first order ODE in the form:

$\dfrac {\d y} {\d x} + \map P x y = \map Q x$

where:

$\map P x y = -3 y$

$\map Q x = \sin x$

Thus:

\(\ds \int \map P x \rd x\)

\(=\)

\(\ds \int -3 \rd x\)

\(\ds \)

\(=\)

\(\ds -3 x\)

\(\ds \leadsto \ \ \)

\(\ds e^{\int P \rd x}\)

\(=\)

\(\ds e^{-3 x}\)

Thus from Solution by Integrating Factor:

\(\ds \dfrac {\d} {\d x} \paren {e^{-3 x} y}\)

\(=\)

\(\ds e^{-3 x} \sin x\)

\(\ds \leadsto \ \ \)

\(\ds e^{-3 x} y\)

\(=\)

\(\ds \int e^{-3 x} \sin x \rd x + C\)

\(\ds \)

\(=\)

\(\ds \frac {e^{3 x} \paren {3 \sin x - \cos x} } {3^2 + 1^2} + C\)

Primitive of $e^{a x} \sin b x$

\(\ds \leadsto \ \ \)

\(\ds y\)

\(=\)

\(\ds \dfrac 1 {10} \paren {3 \sin x - \cos x} + C e^{3 x}\)

$\blacksquare$