Solution by Integrating Factor/Examples/y' - 3y = sin x/Proof 2

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 with constant coefficents in the form:

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

where:

$a = -3$

$\map Q x = \sin x$

Thus from Solution to Linear First Order ODE with Constant Coefficients:

\(\ds y\)

\(=\)

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

\(\ds \)

\(=\)

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

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

\(\ds \)

\(=\)

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

$\blacksquare$

Sources

• 1958: G.E.H. Reuter: Elementary Differential Equations & Operators ... (previous) ... (next): Chapter $1$: Linear Differential Equations with Constant Coefficients: $\S 1$. The first order equation: $\S 1.2$ The integrating factor: Example $1$