Question

Find the solution of Differential equation.

$\\ \implies \bf y \sin(2x) dx - (1+ {y}^{2} + { \cos}^{2}x)dy = 0$


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Answer 1

Question:-

Find the solution of Differential equation.

$\sf y \sin(2x) dx - (1+ {y}^{2} + { \cos}^{2}x)dy = 0$

Given:-

$\sf \: Differential \: equation. \\ \sf \: y \sin(2x) dx - (1+ {y}^{2} + { \cos}^{2}x)dy = 0$

To Find:-

• Solution of the Given Differential equation.

Solution:-

Given Differential Equation,

$\sf y \sin(2x) dx - (1+ {y}^{2} + { \cos}^{2}x)dy = 0 \: \: \longmapsto➀ \\ \sf is \: of \: the \: form \: Pdx + Qdy = 0.$

Where, P = y sin 2x & Q = – 1 – y² – cos²x.

$\sf \therefore \int Pdx = \int y \: sin \: 2x \: dx \: = \frac{ - 1}{2} y \: cos {}^{2}x + g(y)$

$\sf \: and \int Qdy = \int( - 1 - y {}^{2} - cos {}^{2} x)dy$

$\sf = - y - \frac{y {}^{3} }{3} - (cos {}^{2} x)y + h(x)$

Now, cos²x = 2cos²x – 1.

$\sf \therefore \: g(y) = - y - \frac{y {}^{3} }{3} - \frac{y}{2} \: and \: h(x) = 0.$

∴ Solution of ➀ is

$\sf - \frac{1}{2} y \: cos {}^{2} x - y - \frac{y {}^{3} }{3} - \frac{y}{2} = 0.$

$\sf \therefore - \frac{1}{y} - \frac{1}{2} y \: cos {}^{2} x - \frac{3y}{2} - \frac{y {}^{3} }{3} = 0.$

This is the Required Solution.

Answer:-

$\sf \red{Hence, \: the \: solution \: of \: Given \: Differential \: equation \: is \: }\\ \sf \blue { - \frac{1}{2}y \: cos {}^{2}x - \frac{3y}{y} - \frac{y {}^{3} }{3} = 0.}$

Hope you have satisfied. ⚘

Answer 2

Answer:equation.$\\ \implies \bf y \sin(2x) dx - (1+ {y}^{2} + { \cos}^{2}x)dy = 0$

Step-by-step explanation:

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