Question

Question #1 under physics.

Answer 1

first off, check whether it is exact or not,
you have the form of \(M_x dx + N_y dy\) so, to check , just take \(M_y ~and~ N_x\)
\(M_y = 0 = N_x\) so, it is exact

Answer 2

now solve it.

Answer 3

\[\psi (x, y ) = \int (2x-1)dx = x^2-x +h(y)\\\text{take derivative respect to y of }\psi (x, y) = h'(y) = N_{x,y}= 3y+7 \]
so, h (y) = \(\dfrac{3}{2}y^2 +7y +C\)
replace that h(y) into \(\psi (x,y)\) you have the result is
\[\psi (x,y) = x^2 -x +\frac{3}{2}y^2 +7y = C\]

Answer 4

wonder why don't you post it in Math or differential equation?

Answer 5

Sorry. Wrong pic was posted.