2.3   ODE No. 3

\[ a y(x)-b \sin (c x)+y'(x)=0 \] Mathematica : cpu = 0.0671842 (sec), leaf count = 40


\[\left \{\left \{y(x)\to \frac {b (a \sin (c x)-c \cos (c x))}{a^2+c^2}+c_1 e^{-a x}\right \}\right \}\] Maple : cpu = 0.041 (sec), leaf count = 38


\[y \relax (x ) = {\mathrm e}^{-a x} c_{1}-\frac {b \left (c \cos \left (c x \right )-\sin \left (c x \right ) a \right )}{a^{2}+c^{2}}\]

Hand solution

\begin {equation} \frac {dy}{dx}+ay\relax (x) =b\sin \left (cx\right ) \tag {1} \end {equation}

Integrating factor \(\mu =e^{\int adx}=e^{ax}\). Hence (1) becomes

\begin {align*} \frac {d}{dx}\left (\mu y\relax (x) \right ) & =\mu b\sin \left ( cx\right ) \\ \mu y\relax (x) & =b\int \mu \sin \left (cx\right ) dx+C \end {align*}

Replacing \(\mu \) by \(e^{ax}\)

\begin {equation} y\relax (x) =be^{-ax}\int e^{ax}\sin \left (cx\right ) dx+Ce^{-ax}\tag {2} \end {equation}

Using \(\sin \left (cx\right ) =\frac {e^{icx}-e^{-icx}}{2i}\) then \begin {align*} \int e^{ax}\sin \left (cx\right ) dx & =\int \frac {e^{\left (ic+a\right ) x}-e^{\left (-ic+a\right ) x}}{2i}dx\\ & =\frac {1}{2i}\left (\frac {e^{\left (ic+a\right ) x}}{ic+a}-\frac {e^{\left (-ic+a\right ) x}}{-ic+a}\right ) \\ & =\frac {1}{2i}e^{ax}\left (\frac {e^{icx}}{ic+a}-\frac {e^{-icx}}{-ic+a}\right ) \\ & =\frac {1}{2i}e^{ax}\left (\frac {e^{icx}\left (-ic+a\right ) -e^{-icx}\left (ic+a\right ) }{\left (ic+a\right ) \left (-ic+a\right ) }\right ) \\ & =\frac {1}{2i}e^{ax}\left (\frac {-ice^{icx}+ae^{icx}-ice^{-icx}-ae^{-icx}}{\left (c^{2}+a^{2}\right ) }\right ) \\ & =\frac {1}{2i}e^{ax}\left (\frac {-ic\left (e^{icx}+e^{-icx}\right ) +a\left (e^{icx}-e^{-icx}\right ) }{\left (c^{2}+a^{2}\right ) }\right ) \\ & =\frac {e^{ax}}{\left (c^{2}+a^{2}\right ) }\left (\frac {-ic\left ( e^{icx}+e^{-icx}\right ) }{2i}+\frac {a\left (e^{icx}-e^{-icx}\right ) }{2i}\right ) \\ & =\frac {e^{ax}}{\left (c^{2}+a^{2}\right ) }\left (-c\cos cx+a\sin cx\right ) \end {align*}

Therefore (2) becomes

\begin {align*} y\relax (x) & =be^{-ax}\left [ \frac {e^{ax}}{\left (c^{2}+a^{2}\right ) }\left (-c\cos cx+a\sin cx\right ) \right ] +Ce^{-ax}\\ & =\frac {b}{\left (c^{2}+a^{2}\right ) }\left (-c\cos cx+a\sin cx\right ) +Ce^{-ax} \end {align*}