differential equation of (2x cos y + y² cos x) dx +(2y sin x-x² siny) dy = 0 is?
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differential equation of (2x cos y + y² cos x) dx +(2y sin x-x² siny) dy = 0 is?
please give answer with adequate
explanation and working spammers will be reported
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Answer:
To find the differential equation for the given expression:
(2x cos y + y² cos x) dx + (2y sin x - x² sin y) dy = 0
We will rewrite the equation in the standard form M(x, y) dx + N(x, y) dy = 0.
M(x, y) = 2x cos y + y² cos x
N(x, y) = 2y sin x - x² sin y
Now, to find the differential equation, we will apply the following condition:
∂M/∂y = ∂N/∂x
Let's calculate these partial derivatives:
∂M/∂y = -2x sin y + 2y cos x
∂N/∂x = 2y cos x - 2x sin y
Now, equate ∂M/∂y and ∂N/∂x:
-2x sin y + 2y cos x = 2y cos x - 2x sin y
The equation simplifies to:
-2x sin y + 2y cos x - 2y cos x + 2x sin y = 0
-2x sin y + 2x sin y - 2y cos x + 2y cos x = 0
0 = 0
The equation is identically satisfied, which means it's an exact differential equation. Therefore, the differential equation for the given expression is already in standard form, and there's no need for further manipulation.
Step-by-step explanation:
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