Let's solve each of the given problems step by step: --- ### **a) \( y' = x(1+y') + (y')^2 \)** Let \( y' = p \). So, \[ p = x(1+p) + p^2 \] \[ p = x + xp + p^2 \] \[ p - xp - p^2 = x \] \[ p(1 - x - p) = x \] \[ p^2 + (x-1)p + x = 0 \] This is a quadratic in \( p \): \[ p = \frac{-(x-1) \pm \sqrt{(x-1)^2 - 4x}}{2} \] So, \[ p = y' = \frac{-(x-1) \pm \sqrt{(x-1)^2 - 4x}}{2} \] This is the general solution for \( y' \). To get \( y \), integrate \( y' \): \[ \frac{dy}{dx} = \frac{-(x-1) \pm \sqrt{(x-1)^2 - 4x}}{2} \] So, \[ y = \int \frac{-(x-1) \pm \sqrt{(x-1)^2 - 4x}}{2} \, dx + C \] This is the required solution (final answer in terms of an integral). --- ### **b)** \[ \left[\sqrt{x+y} + \sqrt{x-y}\right]dx + \left[\sqrt{x-y} - \sqrt{x+y}\right]dy = 0 \] **Transformation:** Let \( x+y=u \) and \( x-y=v \). Express \( x \) and \( y \) in terms of \( u \) and \( v \): \[ x+y = u\\ x-y = v \] Add: \( 2x = u+v \implies x = \frac{u+v}{2} \) Subtract: \( 2y = u-v \implies y = \frac{u-v}{2} \) **Find \( dx \) and \( dy \):** \[ dx = \frac{1}{2}(du + dv) \] \[ dy = \frac{1}{2}(du - dv) \] Now substitute into the equation: \[ [\sqrt{u} + \sqrt{v}] \cdot \frac{1}{2}(du + dv) + [\sqrt{v} - \sqrt{u}] \cdot \frac{1}{2}(du - dv) = 0 \] Expand: \[ \frac{1}{2}[\sqrt{u} + \sqrt{v}](du + dv) + \frac{1}{2}[\sqrt{v} - \sqrt{u}](du - dv) = 0 \] Multiply out: \[ \frac{1}{2}(\sqrt{u} du + \sqrt{u} dv + \sqrt{v} du + \sqrt{v} dv) + \frac{1}{2}(\sqrt{v} du - \sqrt{v} dv - \sqrt{u} du + \sqrt{u} dv) = 0 \] Group terms: 1. \( \sqrt{u} du \) from first + \( -\sqrt{u} du \) from second: cancels. 2. \( \sqrt{u} dv \) from first + \( \sqrt{u} dv \) from second: \( 2\sqrt{u} dv \). 3. \( \sqrt{v} du \) from first + \( \sqrt{v} du \) from second: \( 2\sqrt{v} du \). 4. \( \sqrt{v} dv \) from first + \( -\sqrt{v} dv \) from second: cancels. So: \[ \frac{1}{2}\left[2\sqrt{u} dv + 2\sqrt{v} du\right] = 0 \] \[ \sqrt{u} dv + \sqrt{v} du = 0 \] \[ \sqrt{u} dv = -\sqrt{v} du \] \[ \frac{dv}{\sqrt{v}} = -\frac{du}{\sqrt{u}} \] Integrate both sides: \[ \int v^{-1/2} dv = -\int u^{-1/2} du \] \[ 2\sqrt{v} = -2\sqrt{u} + C \] \[ \sqrt{v} + \sqrt{u} = C' \] Recall \( u = x+y, v = x-y \): \[ \sqrt{x-y} + \sqrt{x+y} = C \] **Final Answer:** \[ \boxed{\sqrt{x-y} + \sqrt{x+y} = C} \] --- ### **c)** \[ xy' = y(\ln(xy) - 1) \] Use the transformation \( xy = u \). Let \( y = \frac{u}{x} \). Compute \( y' \): \[ y = \frac{u}{x} \] \[ y' = \frac{du}{dx} \cdot \frac{1}{x} - \frac{u}{x^2} \] Now, substitute into the original equation: \[ x \left( \frac{du}{dx} \cdot \frac{1}{x} - \frac{u}{x^2} \right) = \frac{u}{x} [\ln(u) - 1] \] \[ \frac{du}{dx} - \frac{u}{x} = \frac{u}{x} [\ln(u) - 1] \] \[ \frac{du}{dx} - \frac{u}{x} = \frac{u}{x} \ln(u) - \frac{u}{x} \] \[ \frac{du}{dx} = \frac{u}{x} \ln(u) \] Divide both sides by \( u \): \[ \frac{1}{u} \frac{du}{dx} = \frac{\ln(u)}{x} \] Separate variables: \[ \frac{1}{\ln(u)} \cdot \frac{1}{u} du = \frac{1}{x} dx \] Let \( t = \ln(u) \implies du = u dt \): \[ \frac{1}{t} dt = \frac{1}{x} dx \] Integrate both sides: \[ \int \frac{1}{t} dt = \int \frac{1}{x} dx \] \[ \ln|t| = \ln|x| + C \] \[ \ln|\ln(u)| = \ln|x| + C \] Exponentiate: \[ |\ln(u)| = A|x| \quad (A = e^C) \] So, \[ \ln(u) = \pm A x \] \[ u = e^{\pm Ax} \] Recall \( u = xy \): \[ xy = e^{\pm A x} \] \[ y = \frac{e^{\pm A x}}{x} \] **Final Answer:** \[ \boxed{y = \frac{e^{C x}}{x}} \] where \( C \) is an arbitrary constant. --- If you need further clarification on any step, let me know!

Sure! Here are the final answers to each of the given problems: **a)** \[ p = y' = \frac{-(x-1) \pm \sqrt{(x-1)^2 - 4x}}{2} \] and \[ y = \int \frac{-(x-1) \pm \sqrt{(x-1)^2 - 4x}}{2} \, dx + C \] **b)** \[ \boxed{\sqrt{x - y} + \sqrt{x + y} = C} \] **c)** \[ \boxed{y = \frac{e^{C x}}{x}} \] Let me know if you need further assistance!