Solve the Differential Equation (12x+5y-9)dx+(5x-2y-4)dy=0
The correct answer and explanation is :
The given differential equation is:
[
(12x + 5y – 9) \, dx + (5x – 2y – 4) \, dy = 0
]
To solve this first-order differential equation, we will determine if it is exact. A differential equation of the form ( M(x, y) \, dx + N(x, y) \, dy = 0 ) is exact if the partial derivatives of ( M(x, y) ) with respect to ( y ) and ( N(x, y) ) with respect to ( x ) are equal:
[
\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}
]
Step 1: Define the functions ( M(x, y) ) and ( N(x, y) )
From the given equation:
- ( M(x, y) = 12x + 5y – 9 )
- ( N(x, y) = 5x – 2y – 4 )
Step 2: Check if the equation is exact
We now calculate the partial derivatives:
- ( \frac{\partial M}{\partial y} = 5 )
- ( \frac{\partial N}{\partial x} = 5 )
Since the partial derivatives are equal, the equation is exact.
Step 3: Solve the equation
For an exact equation, there exists a potential function ( \Psi(x, y) ) such that:
[
\frac{\partial \Psi}{\partial x} = M(x, y) \quad \text{and} \quad \frac{\partial \Psi}{\partial y} = N(x, y)
]
Step 3a: Find ( \Psi(x, y) ) from ( \frac{\partial \Psi}{\partial x} = M(x, y) )
Integrate ( M(x, y) = 12x + 5y – 9 ) with respect to ( x ):
[
\Psi(x, y) = \int (12x + 5y – 9) \, dx = 6x^2 + 5xy – 9x + h(y)
]
where ( h(y) ) is an arbitrary function of ( y ) (since the derivative with respect to ( x ) does not affect ( y )).
Step 3b: Find ( h(y) ) using ( \frac{\partial \Psi}{\partial y} = N(x, y) )
Now, take the partial derivative of ( \Psi(x, y) ) with respect to ( y ):
[
\frac{\partial \Psi}{\partial y} = 5x + h'(y)
]
We know that ( \frac{\partial \Psi}{\partial y} = N(x, y) = 5x – 2y – 4 ). Therefore, we have the equation:
[
5x + h'(y) = 5x – 2y – 4
]
Canceling out ( 5x ) from both sides:
[
h'(y) = -2y – 4
]
Integrating with respect to ( y ):
[
h(y) = -y^2 – 4y + C
]
where ( C ) is a constant of integration.
Step 3c: Combine the results
Thus, the potential function is:
[
\Psi(x, y) = 6x^2 + 5xy – 9x – y^2 – 4y + C
]
Step 4: Set the potential function equal to a constant
The general solution to the differential equation is obtained by setting ( \Psi(x, y) = C_1 ), where ( C_1 ) is a constant:
[
6x^2 + 5xy – 9x – y^2 – 4y = C_1
]
This is the implicit solution to the differential equation.