y-2x-3y-1-2-find-the-solution-

Question Number 84212 by jagoll last updated on 10/Mar/20

y^{'} = {(2 x + 3 y + 1)}^{2}

find the solution

Commented by niroj last updated on 10/Mar/20

\frac{dy}{dx} = {(2 x + 3 y + 1)}^{2}

put, 2 x + 3 y + 1 = v

2 + 3 \frac{dy}{dx} = \frac{dv}{dx}

3 \frac{dy}{dx} = \frac{dv}{dx} - 2

\frac{dy}{dx} = \frac{1}{3} (\frac{dv}{dx} - 2)

\frac{1}{3} (\frac{dv}{dx} - 2) = v^{2}

\frac{dv}{dx} - 2 = {3 v}^{2}

\frac{dv}{dx} = {3 v}^{2} + 2

\frac{1}{{3 v}^{2} + 2} dv = dx

integrating both side .

\frac{1}{3} \int \frac{1}{{(v)}^{2} + {(\sqrt{\frac{2}{3}})}^{2}} dv = \int dx

\frac{1}{3} . \frac{1}{\sqrt{\frac{2}{3}}} \tan^{- 1} \frac{v}{\sqrt{\frac{2}{3}}} = x + c

\frac{1}{3} . \frac{\sqrt{3}}{\sqrt{2}} . \tan^{- 1} \frac{v \sqrt{3}}{\sqrt{2}} = x + c

\frac{1}{\sqrt{6}} \tan^{- 1} v \frac{\sqrt{6}}{2} = x + c

\tan^{- 1} v . \frac{\sqrt{6}}{2} = \sqrt{6} (x + c)

\frac{v \sqrt{6}}{2} = \tan \sqrt{6} (x + c)

(2 x + 3 y + 1) \sqrt{6} = 2 \tan \sqrt{6} (x + c)

2 x + 3 y + 1 = \frac{\sqrt{6}}{3} \tan \sqrt{6} (x + c) .

Answered by mr W last updated on 10/Mar/20

l e t t = 2 x + 3 y + 1

\frac{d t}{d x} = 2 + 3 \frac{d y}{d x}

y^{'} = \frac{d y}{d x} = \frac{1}{3} (\frac{d t}{d x} - 2)

\frac{1}{3} (\frac{d t}{d x} - 2) = t^{2}

\frac{d t}{d x} = 3 t^{2} + 2

\frac{d t}{3 t^{2} + 2} = d x

\int \frac{d t}{3 t^{2} + 2} = \int d x

\frac{\tan^{- 1} \frac{3 t}{\sqrt{6}}}{\sqrt{6}} = x + C

\Rightarrow t = \frac{\sqrt{6}}{3} \tan (\sqrt{6} x + C)

\Rightarrow 2 x + 3 y + 1 = \frac{\sqrt{6}}{3} \tan (\sqrt{6} x + C)

\Rightarrow y = \frac{1}{3} [\frac{\sqrt{6}}{3} \tan (\sqrt{6} x + C) - 2 x - 1]

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