1. **Solve the D.E.** \((2x^3 - xy^2 - 2y + 3) dx - (x^2 y + 2x) dy = 0\). - Check if the equation is exact by verifying \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\) for \(M=2x^3 - xy^2 - 2y + 3\) and \(N= - (x^2 y + 2x)\). - \(\frac{\partial M}{\partial y} = -2xy - 2\), \(\frac{\partial N}{\partial x} = -(2xy + 2) = -2xy - 2\), so the equation is exact. - Find potential function \(\psi(x,y)\) such that \(\psi_x = M\). - Integrate \(M\) with respect to \(x\): $$\psi(x,y) = \int (2x^3 - xy^2 - 2y + 3) dx = \frac{x^4}{2} - \frac{x^2 y^2}{2} - 2x y + 3x + h(y)$$ - Differentiate \(\psi\) with respect to \(y\) and equate to \(N\): $$\psi_y = -x^2 y - 2x + h'(y) = N = -x^2 y - 2x$$ - So, \(h'(y) = 0\), thus \(h(y)\) is constant. - Final implicit solution is: $$c = \frac{x^4}{2} - \frac{x^2 y^2}{2} - 2x y + 3x$$ - Multiplying by 2 to match options: $$c = x^4 - x^2 y^2 - 4 x y + 6 x$$ - So correct answer is **C**. 2. **Find the integrating factor for** \((3xy + y^2) dx + (x^2 + xy) dy = 0\). - Let \(M=3xy + y^2\), \(N=x^2 + xy\). - \(\frac{\partial M}{\partial y} = 3x + 2y\), \(\frac{\partial N}{\partial x} = 2x + y\). - Since \(\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x} = (3x + 2y) - (2x + y) = x + y\). - Check for integrating factor dependent on \(x\) only: $$\frac{\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}}{N} = \frac{x + y}{x^2 + x y} = \frac{x + y}{x(x + y)} = \frac{1}{x}$$ - Depends only on \(x\), so integrating factor \(\mu = e^{\int 1/x dx} = e^{\ln x} = x\). - Hence, integrating factor is **A**. 3. **Find solution for** \((3xy + y^2) dx + (x^2 + xy) dy = 0\) with integrating factor \(x\). - Multiply entire equation by \(x\): $$x(3xy + y^2) dx + x(x^2 + xy) dy = (3x^2 y + x y^2) dx + (x^3 + x^2 y) dy = 0$$ - Let new \(M = 3x^2 y + x y^2\), \(N = x^3 + x^2 y\). - Check exactness: $$\frac{\partial M}{\partial y} = 3 x^2 + 2 x y$$ $$\frac{\partial N}{\partial x} = 3 x^2 + 2 x y$$ - Exact, so find potential function \(\psi\) with \(\psi_x = M\). - Integrate \(M\) w.r.t. \(x\): $$\psi = \int (3 x^2 y + x y^2) dx = y x^3 + \frac{x^2 y^2}{2} + h(y)$$ - Differentiate \(\psi\) w.r.t. \(y\): $$\psi_y = x^3 + x^2 y + h'(y) = N = x^3 + x^2 y$$ - So \(h'(y) = 0\), \(h(y)\) constant. - Implicit solution: $$c = x^3 y + \frac{x^2 y^2}{2}$$ - The correct answer is **A**.