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)7 In |3x + 3y - 2| + 3x + 6y = kd)7 In |3x + 3y - 2| - 3x + 6y = kCorrect answer is option 'D'. Can you explain this answer?

Question

Answer

Answer

Let V = x + y
Then

implies

or

implies

Answer

This differential equation is a first-order linear equation of the form:

M(x, y) dx + N(x, y) dy = 0

where M(x, y) = x*y - 3, and N(x, y) = -2x - 2y + 1.

To find the general solution of this differential equation, we can use the method of integrating factors. The integrating factor is given by:

IF = e^(∫(N - M')dx)

where M' is the partial derivative of M with respect to y. In this case, M' = x.

So, the integrating factor becomes:

IF = e^(∫(-2x - 2y + 1 - x) dx) = e^(∫(-3x - 2y + 1) dx) = e^(-3x^2/2 - 2xy + x)

Now, multiplying the given differential equation by the integrating factor, we get:

(e^(-3x^2/2 - 2xy + x))(x*y - 3) dx - (e^(-3x^2/2 - 2xy + x))(2x + 2y - 1) dy = 0

Simplifying this expression, we have:

(e^(-3x^2/2 - 2xy + x))(x*y - 3) dx - (e^(-3x^2/2 - 2xy + x))(2x + 2y - 1) dy = 0

Next, we can rewrite this equation as:

(d(e^(-3x^2/2 - 2xy + x)))/(dx) = 0

Integrating both sides with respect to x, we have:

e^(-3x^2/2 - 2xy + x) = C

where C is a constant of integration.

Finally, we can take the natural logarithm of both sides to isolate the exponential term:

-3x^2/2 - 2xy + x = ln(C)

Rearranging this equation, we obtain:

7ln|3x - 3y + 2| - 3x + 6y = k

where k = ln(C).

Therefore, the general solution of the given differential equation is:

7ln|3x - 3y + 2| - 3x + 6y = k, where k is an arbitrary constant.

Hence, the correct answer is option 'D': 7 In |3x - 3y + 2| - 3x + 6y = k.

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