Question

(8xy-9y^2)dx+2(x^2-3xy)dy=0​

Answer 1

Answer:

dydx=−y2+2xy2x2+3xy

Let z=yx so y=zx.

dydx=xdzdx+z=−x2z2+2x2z2x2+3x2z=−x2(z2+2z)x2(2+3z)

xdzdx+z=−z2+2z3z+2

xdzdx=−z2+2z+z(3z+2)3z+2=−4z2+4z3z+2

3z+2z(z+1)dz=−4xdx

3z+2z(z+1)=Az+Bz+1

A=3∗0+20+1=2,B=3∗−1+2−1=1

∫2z+1z+1dz=∫−4xdx=ln|z2(z+1)|=−4ln|x|+C

ln|x4z3+x4z2|=ln|xy3+x2y2|=C

xy2(x+y)=C

Answer 2

Answer:

Step-by-step explanation:

From the above question,

       To remedy the differential equation (8xy - 9y^2)dx + 2(x^2 - 3xy)dy = 0, we can use the approach of specific differential equations:

First, we want to test if the equation is precise with the aid of verifying if the partial derivatives of the coefficients with appreciate to y and x are equal:

∂/∂y (8xy - 9y^2) = 8x - 18y

∂/∂x (2(x^2 - 3xy)) = 4x - 6y

Since these partial derivatives are now not equal, the equation is no longer exact. To make it exact, we want to locate a appropriate integrating factor, μ(x,y), such that μ(x,y)(8xy - 9y^2)dx + μ(x,y)(2(x^2 - 3xy)dy) is exact.

We can discover this integrating aspect through the usage of the formula:

μ(x,y) = e^(∫(M_y - N_x)/N dx + C)

where M = 8xy - 9y^2, N = 2(x^2 - 3xy), and C is a consistent of integration.

We first want to locate (M_y - N_x)/N:

(M_y - N_x)/N = (-18y - 4x + 6y)/(2(x^2 - 3xy)) = (-18y - 4x + 6y)/(2y(x - 3/2y))

This can be simplified to:

(M_y - N_x)/N = (-9/y) - (2/(x - (3/4)y))

Now we combine this expression with recognize to x, treating y as a constant:

∫(-9/y) - (2/(x - (3/4)y)) dx = (-9/y)x - 2ln|x - (3/4)y| + C1

where C1 is a consistent of integration.

Next, we exponentiate this expression to acquire μ(x,y):

μ(x,y) = e^(∫(M_y - N_x)/N dx + C) = e^((-9/y)x - 2ln|x - (3/4)y| + C1) = e^(C1)(x - (3/4)y)^(-9/y) * e^(-2ln(x - (3/4)y))

This can be simplified to:

μ(x,y) = k(x - (3/4)y)^(-9/y)(1/(x - (3/4)y)^2)

where okay = e^(C1) is a constant.

Multiplying the given equation by means of this integrating factor, we get:

k(x - (3/4)y)^(-9/y)(1/(x - (3/4)y)^2)(8xy - 9y^2)dx + k(x - (3/4)y)^(-9/y)(1/(x - (3/4)y)^2)(2(x^2 - 3xy))dy = 0

This equation is exact, so we can To remedy the differential equation (8xy - 9y^2)dx + 2(x^2 - 3xy)dy = 0, we can use the approach of specific differential equations:

First, we want to test if the equation is precise with the aid of verifying if the partial derivatives of the coefficients with appreciate to y and x are equal:

∂/∂y (8xy - 9y^2) = 8x - 18y

∂/∂x (2(x^2 - 3xy)) = 4x - 6y

Since these partial derivatives are now not equal, the equation is no longer exact. To make it exact, we want to locate a appropriate integrating factor, μ(x,y), such that μ(x,y)(8xy - 9y^2)dx + μ(x,y)(2(x^2 - 3xy)dy) is exact.

We can discover this integrating aspect through the usage of the formula:

μ(x,y) = $e^{(∫(M_y - N_x)$/N dx + C)

where M = 8xy - 9$y^{2}$, N = 2($x^{2}$ - 3xy), and C is a consistent of integration.

We first want to locate (M_y - N_x)/N:

(M_y - N_x)/N = (-18y - 4x + 6y)/(2($x^{2}$ - 3xy)) = (-18y - 4x + 6y)/(2y(x - 3/2y))

This can be simplified to:

(M_y - N_x)/N = (-9/y) - (2/(x - (3/4)y))

Now we combine this expression with recognize to x, treating y as a constant:

∫(-9/y) - (2/(x - (3/4)y)) dx = (-9/y)x - 2ln|x - (3/4)y| + C1

Where C1 is a consistent of integration.

Next, we exponentiate this expression to acquire μ(x,y):

           μ(x,y) =  $e^{(C1)$(x - (3/4)y)$.^{(-9/y)$ * $e^{(-2ln(x - (3/4)y))$

This can be simplified to:

         μ(x,y) = k(x - (3/4)y) (-9/y)(1/(x - (3/4)y)$.^2$)

         where okay = $e^{(C1)$ is a constant.

This equation is exact, so we can remedy the differential equation (8xy - 9y^2)dx + 2(x^2 - 3xy)dy = 0, we can use the approach of specific differential equations:

First, we want to test if the equation is precise with the aid of verifying if the partial derivatives of the coefficients with appreciate to y and x are equal:

                  ∂/∂y (8xy - 9y2) = 8x - 18y

                  ∂/∂x (2(x2 - 3xy)) = 4x - 6y

Since these partial derivatives are now not equal, the equation is no longer exact.

                 μ(x,y) = e(∫(M_y - N_x)/N dx + C)

where M = 8xy - 9$y^2$, N = 2($x^{2}$ - 3xy), and C is a consistent of integration.

We first want to locate (M_y - N_x)/N:

(M_y - N_x)/N = (-18y - 4x + 6y)/(2(x^2 - 3xy))

                       = (-18y - 4x + 6y)/(2y(x - 3/2y))

This can be simplified to:

(M_y - N_x)/N = (-9/y) - (2/(x - (3/4)y))

Now we combine this expression with recognize to x, treating y as a constant:

∫(-9/y) - (2/(x - (3/4)y)) dx = (-9/y)x - 2ln|x - (3/4)y| + C1

where C1 is a consistent of integration.

Next,

       We exponentiate this expression to acquire μ(x,y)

This can be simplified to:

       μ(x,y) = k(x - (3/4)y) (-9/y)(1/(x - (3/4)y)$.^2$)

where okay = $e^{(C1)$ is a constant.

Multiplying the given equation by means of this integrating factor.

$∂/∂x [k(x - (3/4)y)^(-9/y)(8xy - 9y^2)] = k(x - (3/4)y)^(-9/y)(-9y)$∂/∂y

$[k(x - (3/4)y)^(-9/y)(2(x^2 - 3xy))]$

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