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Equation of Normal at a Point (x,y) in Terms of f'(x)

dydx+2 y=sin ...

Question

$\frac{d y}{d x} + 2 y = \sin 3 x$

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Solution

$We have, \frac{d y}{d x} + 2 y = \sin 3 x . . . . . (1) Clearly, it is a linear differential equation of the form \frac{d y}{d x} + P y = Q where P = 2 and Q = \sin 3 x . ∴ I . F . = e^{\int P d x} = e^{\int 2 d x} = e^{2 x} Multiplying both sides of (1) by I . F . = e^{2 x}, we get e^{2 x} (\frac{d y}{d x} + 2 y) = e^{2 x} \sin 3 x \Rightarrow e^{2 x} \frac{d y}{d x} + 2 e^{2 x} y = e^{2 x} \sin 3 x Integrating both sides with respect to x, we get y e^{2 x} = \int e^{2 x} \sin 3 x d x + C \Rightarrow y e^{2 x} = I + C . . . . . (1) Where, I = \int e^{2 x} \sin 3 x d x . . . . . (2) \Rightarrow I = e^{2 x} \int \sin 3 x d x - \int [\frac{d e^{2 x}}{d x} \int \sin 3 x d x] d x \Rightarrow I = - \frac{e^{2 x} \cos 3 x}{3} + \frac{2}{3} \int e^{2 x} \cos 3 x d x \Rightarrow I = - \frac{e^{2 x} \cos 3 x}{3} + \frac{2}{3} [e^{2 x} \int \cos 3 x d x - \int (\frac{d e^{2 x}}{d x} \int \cos 3 x d x) d x] \Rightarrow I = - \frac{e^{2 x} \cos 3 x}{3} + \frac{2}{3} [\frac{e^{2 x} \sin 3 x}{3} - \frac{2}{3} \int e^{2 x} \sin 3 x d x] \Rightarrow I = - \frac{e^{2 x} \cos 3 x}{3} + \frac{2 e^{2 x} \sin 3 x}{9} - \frac{4}{9} \int e^{2 x} \sin 3 x d x \Rightarrow I = - \frac{e^{2 x} \cos 3 x}{3} + \frac{2 e^{2 x} \sin 3 x}{9} - \frac{4}{9} I [Using (2)] \Rightarrow \frac{13 I}{9} = - \frac{e^{2 x} \cos 3 x}{3} + \frac{2 e^{2 x} \sin 3 x}{9} \Rightarrow I = \frac{9}{13} (\frac{2 e^{2 x} \sin 3 x}{9} - \frac{e^{2 x} \cos 3 x}{3}) \Rightarrow I = \frac{e^{2 x}}{13} (2 \sin 3 x - 3 \cos 3 x) . . . . . (3) From (1) and (3), we get y e^{2 x} = \frac{e^{2 x}}{13} (2 \sin 3 x - 3 \cos 3 x) + C \Rightarrow y = \frac{3}{13} (\frac{2}{3} \sin 3 x - \cos 3 x) + C e^{- 2 x} Hence, y = \frac{3}{13} (\frac{2}{3} \sin 3 x - \cos 3 x) + C e^{- 2 x} is the required solution .$

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