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Solve the following $\sec 2x dy -\sin 5x \sec^{2} ydx=0$

1 Answer

  • First order , first degree DE
  • Variable separable : Variables of a DE are rearranged to separate them, 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:
$\sec 2x dy - \sin 5x \sec^2 y dx=0$ divided by $\sec 2x \sec^2 y$
$\large\frac{dy}{\sec^2 y}-\frac{\sin 5x}{\sec 2x}$$dx=0$
$ \cos ^2 y dy= \sin 5x \cos 2x dx $
Step 2:
$\int \cos^2 y dy =\int \sin 5x \cos 2x dx +c_1$
$\int \large\frac{1+\cos 2y}{2}dy=\large\frac{1}{2}$$ \int (\sin \large\frac{7x}{2}$$+ \sin 3x)dx+c_1$
$\large\frac{y}{2}+\frac{1}{4} $$ \sin 2y = -\large\frac{1}{2} \times \frac{1}{7} $$\cos 7x-\large\frac{1}{2} \times \large\frac{1}{3} $$\cos 3x +c_1$
Multiply by two and rearrange
$y+ \large\frac{\sin 2y}{2} + \frac{\cos 7x}{7}-\frac{\cos 3x}{3}$$+c$


answered Sep 4, 2013 by meena.p