91Ó°ÊÓ

: Differentiate the given equation with respect to x to get: \(2cy \frac{dy}{dx} + 4 \frac{dy}{dx} = 4x\) #step2_title#: Find \(y\) and group terms by \(\frac{dy}{dx}\) #step2_content#: To group the terms by \(\frac{dy}{dx}\), factor it out from the left side and isolate the other terms:

: Factor \(\frac{dy}{dx}\) from the left side of the equation: \(\left(2cy + 4\right) \frac{dy}{dx} = 4x\) Now we can divide both sides of the equation by \(2cy + 4\): \(\frac{dy}{dx} = \frac{4x}{(2cy + 4)}\) Since we want to eliminate c, we can substitute \(c = \frac{2x^{2} - 4y}{y^{2}}\) from the original equation: \(\frac{dy}{dx} = \frac{4x}{\left(2(\frac{2x^{2} - 4y}{y^{2}})y + 4\right)}\) The result is the first order differential equation for the family of curves: \(\frac{dy}{dx} = \frac{4x}{4x^{2} - 8y + 4}\) (ii) \(x y = c_{1} e^{x} - c_{2} e^{-x} + x^{2}\) #step1_title#: Differentiate both sides implicitly. #step1_content#: Differentiation can be obtained for both sides of the equation with respect to x. For implicit differentiation, treat y as a function of x (y(x)), and use the chain rule when necessary. In this case, differentiate both sides with respect to x:

: Differentiate the given equation with respect to x to get: \(y + x \frac{dy}{dx} = c_{1} e^{x} + c_{2} e^{-x} + 2x\) #step2_title#: Eliminating the constants \(c_1\) and \(c_2\) #step2_content#: We can eliminate the constants by using substitution. First, write the expression for \(c_1\) and \(c_2\) in terms of x and y: \(c_{1} e^{x} - c_{2} e^{-x} = xy - x^{2}\) Now, differentiate \(c_1 e^x - c_2 e^{-x}\) to find an expression containing the derivatives of \(c_1\) and \(c_2\):

: Differentiate \(c_1 e^x - c_2 e^{-x}\) with respect to x: \(c_1 e^x + c_2 e^{-x}\) Substitute back in to eliminate \(c_1\) and \(c_2\): \(y + x \frac{dy}{dx} = xy - x^2 + 2x\) Now solving for \(\frac{dy}{dx}\):