We #160;have,dydx+y=4x #160; #160; #160; #160; #160; #160; #160; #160; #160; #160; #160;.....1Clearly, #160;it #160;is #160;a #1 ...

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Asymptotes

dydx +y=4 x

Question

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

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Solution

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

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