First slide

Methods of solving first order first degree differential equations

Question

$If \frac{d y}{d x} + x \sin 2 y = x^{3} \cos^{2} y, y (0) = 0 then \tan y (1) =$

Moderate

A

$\frac{1}{e}$
B

$\frac{1}{2 e}$
C

$2 e$
D

$e$

Solution

$\begin{array}{l} \sec^{2} y \frac{d y}{d x} + 2 x \tan y = x^{3} \\ I. F = e^{\int 2 x d x} = e^{x^{2}} \\ Solution is \tan y e^{x^{2}} = \int x^{3} \cdot e^{x^{2}} d x = \frac{1}{2} (x^{2} - 1) e^{x^{2}} + c \\ \Rightarrow \tan y (x) = \frac{1}{2} (x^{2} - 1) + c \cdot e^{- x^{2}} \\ x = 0, y = 0 \Rightarrow c = \frac{1}{2} \\ \Rightarrow \tan y (x) = \frac{1}{2} [x^{2} - 1 + e^{- x^{2}}] \\ \tan y (y (1)) = \frac{1}{2 e} \end{array}$

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