First slide

Methods of solving first order first degree differential equations

Question

$The solution of differential equation \{x y^{3} (1 + \cos x) - y\} d x + x d y = 0 is$

Moderate

A

$y = x^{3} + 2 x \sin x + x \cos x - 2 \sin x + c$
B

$\frac{x^{3}}{3} + 2 \cos x - x \sin x + c$
C

$\frac{x^{3}}{3} + x^{2} \sin x - 2 x \sin x + \cos x + c$
D

$\frac{x^{2}}{2 y^{2}} = \frac{x^{3}}{3} + x^{2} \sin x + 2 x \cos x - 2 \sin x + c$

Solution

$\begin{array}{l} \frac{d y}{d x} + y^{3} (1 + \cos x) - \frac{y}{x} = 0 \\ \frac{- 1}{y^{3}} \frac{d y}{d x} + \frac{1}{y^{2} x} = (1 + \cos x) \\ Put \frac{+ 1}{y^{2}} = u \Rightarrow \frac{- 2}{y^{3}} \frac{d y}{d x} = \frac{d u}{d x} \\ \frac{1}{2} \frac{d u}{d x} + \frac{1}{x} u = 1 + \cos x \\ \frac{d u}{d x} + \frac{2}{x} u = 2 (1 + \cos x) \\ I. F = e^{2 \int \frac{1}{x} d x} = x^{2} \\ G.S is \frac{1}{y^{2}} (x^{2}) = \int 2 (1 + \cos x) x^{2} d x \\ \frac{x^{2}}{2 y^{2}} = \int x^{2} d x + \int x^{2} \cos x d x \\ \frac{x^{2}}{2 y^{2}} = \frac{x^{3}}{3} + x^{2} \sin x + 2 x \cos x - 2 \sin x + c \end{array}$

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