We\'re doing exact-equations in our calculus 2 class which involves us finding t

Question

We're doing exact-equations in our calculus 2 class which involves us finding the partial derivative of P(x,y) with respect to y and Q(x,y) with respect to x. I can do that fine, but now the example questions are asking "Find General Solution" of 4x^3 - 2x + 3y + (3x+2y)y' = 0 So I find the partial derivatives of both equations with respect to the respected variable, but now what? Is that it? Or do I plug the equations back in for P(x,y) and solve for y'? Part 2 asks solve initial value problem for ye^x + sin(y) + [e^x + xcosy + 1]y' = 0 such that y(0) = -3 I'm assuming I solve for the partial derivatives again, but then what?

Explanation / Answer

We're doing exact-equations in our calculus 2 class which involves us finding the partial derivative of P(x,y) with respect to y and Q(x,y) with respect to x.
I can do that fine, but now the example questions are asking
"Find General Solution" of 4x^3 - 2x + 3y + (3x+2y)y' = 0
So I find the partial derivatives of both equations with respect to the respected variable, but now what? Is that it? Or do I plug the equations back in for P(x,y) and solve for y'? Part 2 asks solve initial value problem for ye^x + sin(y) + [e^x + xcosy + 1]y' = 0 such that y(0) = -3 I'm assuming I solve for the partial derivatives again, but then what?

4x^3 - 2x + 3y + (3x+2y)y' = 0

4x^3 - 2x + 3y + (3x+2y) dy/dx= 0

(3x+2)dy + (4x^3-2x+3y) dx = 0

N = (3x+2)

M = (4x^3-2x+3y)

dM/dy = 3

dN/dx = 3

since both are same equation is exact
now

f(x,y) = integral of (4x^3-2x+3y ) dx

f(x,y) = x^4-x^2 + 3xy + f(y)

take derivative with respec to y

N = 3x+ f'(y) = 3x+2

f'(y) = 2
take integrtaion
f(y) = 2y+ C


General solution is
f(x,y) = x^4-x^2 + 3xy + 2y +C

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