# Find the general solution of the differential equation $\large\frac{dy}{dx}$$+2y=\sin x ## 1 Answer Toolbox: • To solve the first order linear differential equation of the form \large\frac{dy}{dx}$$ + Py = Q$
• (i) Write the given equation in the form of $\large\frac{dy}{dx}$$+ Py = Q • (ii) Find the integrating factor (I.F) = e^{\int Pdx}. • (iii) Write the solution as y(I.F) = integration of Q(I.F) dx + C Step 1: Given \large\frac{dy}{dx}$$ + 2y = \sin x$
The given equation is of the form $\large\frac{dy}{dx }$$+ Py = Q, where P = 2 and Q = \sin x I.F = e^{\int 2.dx} = e^{2x} Hence the solution is y.e^{2x }= \int\sin x.e^{2x} dx + C This can be integrated by parts where integration of udv = uv - \int vdu Step 2: let u = e^{2x},du = 2e^{2x}dx and dv = \sin x; v = - \cos x I = - (e^{2x}\cos x) - \int(-\cos x).2e^{2x}dx Again Let u = e^x; du = e^{2x}dx and dv = \cos xdx ;v = \sin x I = - (e^x\cos x) + 2e^{2x}(\sin x) - 2I[\int \sin x 2e^{2x}.dx] \Rightarrow e^{2x}[2\sin x-\cos x]-4I 5I = e^{2x}[2\sin x - \cos x] I = \large\frac{e^{2x}[2\sin x - \cos x]}{5} \therefore y e^{2x} =\large\frac{ e^{2x}[2\sin x - \cos x]}{5}$$ + C$