Question

For the given differential equation find the general solution.

$\frac{dy}{dx}+2y=sinx$

Answer 1

The given differential equation is dy/dx+2y=sinx
Comparing with the form dy/dx +Py=Q
So, P=2 and Q=sinx. $\therefore IF=e^{\int Pdx}\Rightarrow e^{2\int 1dx}\Rightarrow IF=e^{2x}$
Hence, the solutions is given by
$y.IF=\int Q\times IFdx+C\Rightarrow y\times e^{2x}=\int sinx.e^{2x}dx=I\left(say\right)$ ...(i)
$\therefore I=sinx\int e^{2x}dx-\int [\frac{d}{dx}sinx\int e^{2x}dx]dx$ [Integration by parts]
$\Rightarrow I=sinx\frac{e^{2x}}{2}-\frac{1}{2}\int cosx{e}^{2x}dx\Rightarrow I=sinx\frac{e^{2x}}{2}-\frac{1}{2}cosx\frac{e^{2x}}{2}+\frac{1}{2}\int [\frac{d}{dx}cosx\int e^{2x}dx]dx\Rightarrow I=sinx\frac{e^{2x}}{2}-\frac{1}{4}cosx{e}^{2x}+\frac{1}{4}\int (-sinx)e^{2x}dx\Rightarrow I=sinx\frac{e^{2x}}{2}-\frac{1}{4}cosx{e}^{2x}-\frac{1}{4}I+C\Rightarrow \frac{5I}{4}=\frac{e^{2x}}{4}(2sinx-cosx)+C\Rightarrow I=\frac{e^{2x}}{5}(2sinx-cosx)+C$
Therefore, Eq. (i) becomes
$y\times e^{2x}=\frac{e^{2x}}{5}[2sinx-cosx]+C\Rightarrow y=\frac{1}{5}[2sinx-cosx]+e^{-2x}C$