First slide

Methods of solving first order first degree differential equations

Question

$The solution of 2 y \cos y^{2} \frac{d y}{d x} - \frac{2}{x + 1} \sin y^{2} = (x + 1)^{3} is$

Moderate

A

$\sin y^{2} = {(x + 1)}^{2} [{(x + 1)}^{2} + c]$
B

$\sin y^{2} = {(x + 1)}^{2} [\frac{{(x + 1)}^{2}}{2} + c]$
C

$\sin y^{2} = {(x + 1)}^{2} [\frac{{(x + 1)}^{3}}{3} + c]$
D

$\sin y^{2} = {(x + 1)}^{2} [\frac{{(x + 1)}^{4}}{4} + c]$

Solution

$\begin{array}{l} Let \sin y^{2} = t \Rightarrow 2 y \cos y^{2} \frac{d y}{d x} = \frac{d t}{d x} \Rightarrow \frac{d t}{d x} - \frac{2 t}{(x + 1)} = (x + 1)^{3} \\ Solve the linear equation I. F = \frac{1}{(x + 1)^{2}} \\ G . S i s t I . F . = \int Q (x) I F d x \\ \frac{t}{{(x + 1)}^{2}} = \int (x + 1) d x \\ \frac{\sin y^{2}}{{(x + 1)}^{2}} = \frac{{(x + 1)}^{2}}{2} + c \end{array}$

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