First slide

Methods of solving first order first degree differential equations

Question

$The general solution of the differential equation (y^{2} + e^{2 x}) d y - y^{3} d x = 0, (c being the constant of integration), is$

Moderate

A

$y^{2} e^{- x} + 2 \ln |y| = c$
B

$y^{2} e^{- 2 x} - 2 \ln |y| = c$
C

$y^{2} e^{- 2 x} + 2 \ln |y| = c$
D

$y^{2} e^{- x} - 2 \ln |y| = c$

Solution

$\begin{array}{l} W e h a v e \frac{d x}{d y} = \frac{y^{2} + e^{2 x}}{y^{3}} \\ \Rightarrow \frac{d x}{d y} = \frac{1}{y} + \frac{e^{2 x}}{y^{3}} \\ \Rightarrow e^{- 2 x} \frac{d x}{d y} - e^{- 2 x} \frac{1}{y} = \frac{1}{y^{3}} \to (1) \\ L e t \frac{- e^{- 2 x}}{2} = u \to (2) \\ \Rightarrow e^{2 x} \frac{d x}{d y} = \frac{d u}{d y} \to (3) \\ S u b s t i t u t i n g (2), (3) i n (1), w e g e t \\ \frac{d u}{d y} + \frac{2}{y} u = \frac{1}{y^{3}} \\ whose solution is u . y^{2} = \int \frac{1}{y} d y + k \\ \Rightarrow u y^{2} = \ln y + k \\ \Rightarrow - \frac{e^{- 2 x}}{2} y^{2} = \ln y + k (∵ u = - \frac{e^{- 2 x}}{2}) \\ \Rightarrow - e^{- 2 x} y^{2} = 2 \ln y + 2 k \\ ∴ e^{- 2 x} y^{2} + 2 \ln y = c where c is an arbitrary constant \end{array}$

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