3. Solve the following first order differential equation. \[ \begin{array}{l}\text { (a) }\left(e^{x} \sin y+2 x y\right)+\left(e^{x} \cos y\right) \frac{d y}{d x}=0 \\ \text { (b) }\left(x^{3}+y^{3}\right) d x-3 x y^{2} d y=0 \\ \text { (c) } x\left(6 x^{2}-y-1\right) d y+2 y d x=0 \\ \text { (d) } x \frac{d y}{d y}-2 y=x^{4} e^{x}\end{array} \]

To solve the given first-order differential equations, we will tackle each part one by one. ### (a) The equation is given as: \[ \left(e^{x} \sin y + 2xy\right) + \left(e^{x} \cos y\right) \frac{dy}{dx} = 0 \] Rearranging gives: \[ \left(e^{x} \cos y\right) \frac{dy}{dx} = -\left(e^{x} \sin y + 2xy\right) \] This can be rewritten as: \[ \frac{dy}{dx} = -\frac{e^{x} \sin y + 2xy}{e^{x} \cos y} \] This is a separable equation. We can separate variables: \[ \frac{dy}{e^{x} \sin y + 2xy} = -\frac{dx}{e^{x} \cos y} \] Integrating both sides will yield the solution. However, the integration may not yield a simple closed form, and numerical methods or specific initial conditions may be needed for a complete solution. ### (b) The equation is: \[ \left(x^{3} + y^{3}\right) dx - 3xy^{2} dy = 0 \] Rearranging gives: \[ \frac{dy}{dx} = \frac{x^{3} + y^{3}}{3xy^{2}} \] This is also separable: \[ 3xy^{2} dy = (x^{3} + y^{3}) dx \] Separating variables: \[ \frac{3y^{2}}{x^{3} + y^{3}} dy = \frac{1}{x} dx \] Integrating both sides will yield the solution. The left side may require substitution or partial fractions, while the right side integrates to \(\ln |x|\). ### (c) The equation is: \[ x(6x^{2} - y - 1) dy + 2y dx = 0 \] Rearranging gives: \[ x(6x^{2} - y - 1) \frac{dy}{dx} = -2y \] This can be rewritten as: \[ \frac{dy}{dx} = -\frac{2y}{x(6x^{2} - y - 1)} \] This is separable: \[ \frac{dy}{2y} = -\frac{dx}{x(6x^{2} - y - 1)} \] Integrating both sides will yield the solution. The left side integrates to \(\ln |y|\), while the right side may require substitution. ### (d) The equation is: \[ x \frac{dy}{dy} - 2y = x^{4} e^{x} \] This seems to be a typo; it should likely be: \[ x \frac{dy}{dx} - 2y = x^{4} e^{x} \] Rearranging gives: \[ \frac{dy}{dx} = \frac{2y}{x} + x^{3} e^{x} \] This is a linear first-order differential equation. We can use an integrating factor: \[ \mu(x) = e^{\int \frac{2}{x} dx} = e^{2 \ln |x|} = x^{2} \] Multiplying through by the integrating factor: \[ x^{2} \frac{dy}{dx} - 2y = x^{5} e^{x} \] This can be solved using the integrating factor method, leading to: \[ \frac{d}{dx}(x^{2}y) = x^{5} e^{x} \] Integrating both sides will yield the solution. ### Summary Each part of the problem involves different techniques for solving first-order differential equations, including separation of variables, integrating factors, and recognizing linear forms. The specific integrals may require additional techniques or numerical methods for complete solutions.