The general solution of the differential equation (x + y -3)dx - (2x+ 2y + 1)dy =0 isa)In | 3x +3y - 2 |+ 3x - 6y = kb)In | 3x +3y - 2 | -3x - 6y = kc)7In | 3x +3y - 2 | +3x +6y = kd)7In | 3x +3y - 2 | -3x +6y = kCorrect answer is option 'D'. Can you explain this answer?

Question

Answer

Given differential equation: (x y -3)dx - (2x 2y 1)dy =0

To solve this differential equation, we can use the method of exact differential equations.

The equation can be rewritten as:
(xdx + ydx - 3dx) - (2xdy + 2ydy - dy) = 0
(x + y - 3)dx - (2x + 2y - 1)dy = 0

Comparing this equation with the standard form Mdx + Ndy = 0, we have:
M = x + y - 3
N = -(2x + 2y - 1)

To check if this equation is exact, we need to verify if the partial derivatives of M and N with respect to y and x, respectively, are equal:
∂M/∂y = 1
∂N/∂x = -2

Since ∂M/∂y ≠ ∂N/∂x, the equation is not exact.

To make the equation exact, we need to find an integrating factor, denoted by μ, such that:
μ(x + y - 3)dx - μ(2x + 2y - 1)dy = 0

To find μ, we use the integrating factor formula:
μ = e^(∫(∂N/∂x - ∂M/∂y)/N dx)

Calculating the integral, we have:
∫(-2 - 1)/(-(2x + 2y - 1)) dx = ∫1/(2x + 2y - 1) dx = ln|2x + 2y - 1|

Therefore, the integrating factor μ is given by:
μ = e^(ln|2x + 2y - 1|) = |2x + 2y - 1|

Multiplying the given equation by the integrating factor μ, we obtain:
|2x + 2y - 1|(x + y - 3)dx - |2x + 2y - 1|(2x + 2y - 1)dy = 0

Simplifying the equation, we have:
(2x + 2y - 1)(x + y - 3)dx - (2x + 2y - 1)(2x + 2y - 1)dy = 0

Expanding and collecting similar terms, we get:
(4x^2 + 6xy - 6x + 4y^2 - 10y + 3)dy - (4x^2 + 6xy - 6x + 4y^2 - 10y + 3)dy = 0

Canceling out the common factors, we obtain:
dy - dy = 0

Integrating both sides, we have:
y - y = C, where C is the constant of integration

Simplifying further, we get:
0 = C

Therefore, the general solution of the given differential equation is:
y = C, where C is a constant.

Answer

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