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Asymptotes

dydx +2 y=4 x

Question

$\frac{d y}{d x} + 2 y = 4 x$

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Solution

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

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