Find the particular solution of the following differential equation:
$x + 1 \frac{dy}{dx} = 2 e^{- y} - 1; y = 0 when x = 0 .$

Solution

$(x + 1) \frac{dy}{dx} = 2 e^{- y} - 1 \Rightarrow \frac{dy}{2 e^{- y} - 1} = \frac{dx}{x + 1} \Rightarrow \frac{e^{y}}{2 - e^{y}} = \frac{dx}{x + 1}$

Integrating both sides, we get:

$\int \frac{e^{y} dy}{2 - e^{y}} = \log |x + 1| + \log C . . . . . . . . . . . . (i) Let 2 - e^{y} = t . ∴ \frac{d}{dy} = (2 - e^{y}) = \frac{dt}{dy} \Rightarrow - e^{y} = \frac{dt}{dy} \Rightarrow e^{y} dy = - dt$

substituting this value in equation ( i ), we get:

$\int \frac{- dt}{t} = \log |x + 1| + \log C \Rightarrow - \log |t| = \log |C (x + 1)| \Rightarrow - \log |2 - e^{y}| = \log |C (x + 1)| \Rightarrow \frac{1}{2 - e^{y}} = C (x + 1) \Rightarrow 2 - e^{y} = \frac{1}{C (x + 1)} . . . . . . . . . . . . . . (ii)$

Now, at x = 0 and y = 0, equation ( ii ) becomes:

$\Rightarrow 2 - e^{y} = \frac{1}{x + 1} \Rightarrow e^{y} = 2 - \frac{1}{x + 1} \Rightarrow e^{y} = \frac{2 x + 2 - 1}{x + 1} \Rightarrow e^{y} = \frac{2 x + 1}{x + 1} \Rightarrow y = \log |\frac{2 x + 1}{x + 1}|; (x \neq - 1)$

This is the required particular solution of the given differential equation.

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Vector Algebra

Find the particular solution of the following differential equation:
$x + 1 \frac{dy}{dx} = 2 e^{- y} - 1; y = 0 when x = 0 .$

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For Daily Free Study Material Join wiredfaculty

Vector Algebra

Find the particular solution of the following differential equation:x + 1 dydx = 2 e- y - 1; y = 0 when x = 0.

Some More Questions From Vector Algebra Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Find the particular solution of the following differential equation:
$x + 1 \frac{dy}{dx} = 2 e^{- y} - 1; y = 0 when x = 0 .$