Find the particular solution of e^dy^/dx = x + 1, given that y = 3, when x = 0.

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Solution

$We have, e^{\frac{d y}{d x}} = x + 1 \Rightarrow \frac{d y}{d x} = \log (x + 1) \Rightarrow d y = \log (x + 1) d x Integrating both sides, we get \int d y = \int \log (x + 1) d x \Rightarrow y = \log (x + 1) \int 1 d x - \int [\frac{d}{d x} \{\log (x + 1)\} \int 1 d x] d x \Rightarrow y = x \log (x + 1) - \int \frac{1}{x + 1} \times x d x \Rightarrow y = x \log (x + 1) - \int (1 - \frac{1}{x + 1}) d x \Rightarrow y = x \log (x + 1) - \int d x + \int \frac{1}{x + 1} d x \Rightarrow y = x \log (x + 1) - x + \log |x + 1| + C \Rightarrow y = (x + 1) \log |x + 1| - x + C . . . . . (1) It is given that at x = 0 and y = 3 . Substituing the values of x and y in (1), we get C = 3 Therefore, substituting the value of C in (1), we get y = (x + 1) \log |x + 1| - x + 3 Hence, y = (x + 1) \log |x + 1| - x + 3 is the required solution .$

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Similar questions

e^{\frac{d y}{d x}} = x

x = 1

y = 0

e^{d y / d x} = x + 1

x = 0, y = 3

e^{\frac{d y}{d x}} = x; y (1) = 3; x > 0

e^{(d y / d x)} = x + 1

x = 0, y = 3

y = f (x)

e^{d y / d x} = x + 1

x = 0, y = 3

y = f (x)

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