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If someone has a chance out there, could you please check my math here and let me know if I am doing this correctly or not.

Solve:

I would like to solve this using the "Exact" method for solving DE's, so:

[tex]\frac{\partial_P}{\partial_y}(2x-1) = 0[/tex]

[tex]\frac{\partial_Q}{\partial_x}(3y+7) = 0[/tex]

[tex]\int{2x-1dx} = x^2-x + g(y)[/tex]

[tex]\frac{\partial_F}{\partial_y}x^2-x+g(y) = g'(y)[/tex]

[tex]g'(y) = 3y+7 -> g(y) = \frac{3}{2}y^2+7y[/tex]

[tex]F(x,y) = x^2-x+\frac{3}{2}y^2+7y[/tex]

How does that look?

**Problem:**Solve:

*(2x-1)dx + (3y+7)dy = 0*I would like to solve this using the "Exact" method for solving DE's, so:

[tex]\frac{\partial_P}{\partial_y}(2x-1) = 0[/tex]

[tex]\frac{\partial_Q}{\partial_x}(3y+7) = 0[/tex]

[tex]\int{2x-1dx} = x^2-x + g(y)[/tex]

[tex]\frac{\partial_F}{\partial_y}x^2-x+g(y) = g'(y)[/tex]

[tex]g'(y) = 3y+7 -> g(y) = \frac{3}{2}y^2+7y[/tex]

[tex]F(x,y) = x^2-x+\frac{3}{2}y^2+7y[/tex]

How does that look?

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