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Solve the following $(x^{2}+5x+7) dy+\sqrt{9+8y-y^{2}}dx=0$

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Toolbox:
  • First order , first degree DE
  • Variable separable : Variables of a DE are rearranged to separate then, ie
  • $f_1(x)g_2(y)dx+f_2(x)g_1(y)dy=0$
  • Can be written as $ \large\frac{g_1 (y)}{g_2(y)}$$dy=-\large\frac{f_1(x)}{f_2(x)}$$dx$
  • The solution is therefore $\int \large\frac{g_1(y)}{g_2(y)}$$dy=-\int \large\frac{f_1(x)}{f_2(x)}$$dx+c$
Step 1:
$(x^2+5x+7)dy+\sqrt {9+8y-y^2}dx=0$ divided by $(x^2+5x+7) \times \sqrt {9+8y-y^2}$
$\large\frac{dy}{\sqrt {9+8y-y^2}}+\large\frac{dx}{x^2+5x+7}$$=0$
Step 2:
The variables are separated
$\large\frac{dy}{\sqrt {9+8y-y^2}}+\large\frac{dx}{x^2+5x+7}$$=C$
$I_1=\int \large\frac{dy}{\sqrt {9-(y^2-8y+16-16)}}=\int \large\frac{dy}{\sqrt {9+16-(y-4)^2}}$
$\quad= \int \large\frac{dy}{\sqrt {5^2-(y-y)^2}}$$=\sin ^{-1} \bigg(\large\frac{y-4}{5}\bigg)$
$I_2= \int \large\frac{dx}{x^2+5x+7}$
$\quad= \int \large\frac{dx}{x^2+5x+\Large\frac{25}{4}-\frac{25}{4}+7}$
$\quad= \int \large\frac{dx}{(x+\Large\frac{5}{2})^2+\bigg(\large\frac{\sqrt 3}{2}\bigg)^2}$
$\quad= \large\frac{2}{\sqrt 3} $$\tan^{-1} \large\frac{x+\Large\frac{5}{2}}{\frac{\sqrt 3}{2}}$
$\quad= \large\frac{2}{\sqrt 3} $$\tan^{-1} \bigg(\large\frac{2x+5}{\sqrt 3}\bigg)$
The GS is
$\sin ^{-1} \bigg(\large\frac{y-4}{5}\bigg)+\large\frac{2}{\sqrt 3} $$\tan ^{-1} \bigg( \large\frac{2x+5}{\sqrt 3}\bigg)=c$
answered Sep 4, 2013 by meena.p
 
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