Differential Equations Ncert Solutions Maths class 12th Maths Class 12th Differential Equations

104. Find the particular solution of the differential equation (1 + e2x ) dy + (1 + y2 ) ex dx = 0, given that y = 1 when x = 0.

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Answered by

Vishal Baghel | Contributor-Level 10

4 months ago

$\begin{array}{l} (1 + e^{2 x}) d y + (1 + y^{2}) e^{x} d x = 0 \\ \Rightarrow \frac{d y}{1 + y^{2}} + \frac{e^{x} d x}{1 + e^{2 x}} = 0 \end{array}$

Integrating both sides, we get:

$\begin{array}{l} t a n^{- 1} y + \int \frac{e^{x} d x}{1 + e^{2 x}} = C . . . . . . . . . . (1) \\ L e t, e^{x} = t \Rightarrow e^{2 x} = t^{2} \\ \Rightarrow \frac{d}{d x} (e^{x}) = \frac{d t}{d x} \\ \Rightarrow e^{x} = \frac{d t}{d x} \\ \Rightarrow e^{x} d x = d t \end{array}$

Substituting these values in equation (1), we get:

$\begin{array}{l} t a n^{- 1} y + \int \frac{d t}{1 + t^{2}} = C \\ \Rightarrow t a n^{- 1} y + t a n^{- 1} t = C \\ \Rightarrow t a n^{- 1} y + t a n^{- 1} (e^{x}) = C . . . . . . . . . . (2) \\ N o w, y = 1, a t, x = 0 \end{array}$

Therefore, equation (2) becomes:

$\begin{array}{l} t a n^{- 1} 1 + t a n^{- 1} 1 = C \\ \Rightarrow \frac{π}{4} + \frac{π}{4} = C \\ \Rightarrow C = \frac{π}{2} \end{array}$

Substituting $C = \frac{π}{2}$ in equation (2), we get:

$t a n^{- 1} y + t a n^{- 1} (e^{x}) = \frac{π}{2}$

This is the required solution of the given differential equation.

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104. Find the particular solution of the differential equation (1 + e2x ) dy + (1 + y2 ) ex dx = 0, given that y = 1 when x = 0.

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