Let's solve the differential equation step by step: Given: \[ zydx - z xdy - y^2 dz = 0 \] Let's rewrite it: \[ z y \, dx - z x \, dy - y^2 \, dz = 0 \] Or, \[ z y \, dx - z x \, dy = y^2 \, dz \] Divide both sides by \( y^2 \): \[ \frac{z}{y} \, dx - \frac{z x}{y^2} \, dy = dz \] Bring all terms involving \( dz \) to one side: \[ \frac{z}{y} \, dx - \frac{z x}{y^2} \, dy - dz = 0 \] Let’s use the substitution \( z = v y \Rightarrow dz = v \, dy + y \, dv \): Substitute into the equation: - \( z = v y \) - \( dz = v \, dy + y \, dv \) So, \[ \frac{v y}{y} dx - \frac{v y x}{y^2} dy - (v \, dy + y \, dv) = 0 \] \[ v \, dx - v \frac{x}{y} dy - v \, dy - y \, dv = 0 \] Combine like terms: \[ v \, dx - v \frac{x}{y} dy - v \, dy = y \, dv \] Bring all terms to one side: \[ v \, dx - v \frac{x}{y} dy - v \, dy - y \, dv = 0 \] Let’s collect \( dx \) and \( dy \) terms: \[ v \, dx - v \, dy - v \frac{x}{y} dy = y \, dv \] Now factor out \( v \) from terms with \( dx \) and \( dy \): \[ v [dx - dy - \frac{x}{y} dy] = y \, dv \] \[ v \left[ dx - \left(1 + \frac{x}{y}\right) dy \right] = y \, dv \] \[ v \left[ dx - \left(\frac{y + x}{y}\right) dy \right] = y \, dv \] Now, divide both sides by \( y \): \[ \frac{v}{y} dx - v \frac{y+x}{y^2} dy = dv \] Bring all \( v \) terms to one side: \[ \frac{v}{y} dx - v \frac{y + x}{y^2} dy - dv = 0 \] Alternatively, rearrange the equation: \[ \frac{v}{y} dx - v \frac{y + x}{y^2} dy = dv \] \[ \frac{v}{y} dx - v \frac{y + x}{y^2} dy - dv = 0 \] Let’s factor out \( v \): \[ v \left( \frac{1}{y} dx - \frac{y + x}{y^2} dy \right) - dv = 0 \] Or, \[ v \left( \frac{dx}{y} - \frac{y + x}{y^2} dy \right) = dv \] Now, divide both sides by \( v \): \[ \frac{dx}{y} - \frac{y + x}{y^2} dy = \frac{dv}{v} \] Combine the terms: \[ \frac{dx}{y} - \frac{y}{y^2} dy - \frac{x}{y^2} dy = \frac{dv}{v} \] \[ \frac{dx}{y} - \frac{1}{y} dy - \frac{x}{y^2} dy = \frac{dv}{v} \] Now, rewrite \( -\frac{x}{y^2} dy \): Let’s try to get the variables separated. Let’s bring all \( x \) and \( dx \) terms to one side, \( y \) and \( dy \) to the other: \[ \frac{dx}{y} - \frac{1}{y} dy - \frac{x}{y^2} dy = \frac{dv}{v} \] Now, let's bring all terms involving \( x \) to one side: \[ \frac{dx}{y} - \frac{x}{y^2} dy = \frac{dv}{v} + \frac{1}{y} dy \] Now, the left side can be written as: \[ \frac{dx}{y} - \frac{x}{y^2} dy = d \left( \frac{x}{y} \right ) \] Because, \[ d \left( \frac{x}{y} \right ) = \frac{1}{y} dx - \frac{x}{y^2} dy \] So, \[ d \left( \frac{x}{y} \right ) = \frac{dv}{v} + \frac{1}{y} dy \] Now, integrate both sides: \[ \int d \left( \frac{x}{y} \right ) = \int \frac{dv}{v} + \int \frac{1}{y} dy \] \[ \frac{x}{y} = \ln |v| + \ln|y| + C \] \[ \frac{x}{y} = \ln |v y| + C \] Recall that \( v = \frac{z}{y} \), so \( v y = z \): \[ \frac{x}{y} = \ln |z| + C \] **Final answer:** \[ \boxed{\frac{x}{y} = \ln|z| + C} \] Or, equivalently, \[ \boxed{\frac{x}{y} - \ln|z| = C} \]

Let's carefully solve the given differential equation step by step. --- ### **Given differential equation:** \[ z y \, dx - z x \, dy - y^2 \, dz = 0 \] --- ### **Step 1: Rewrite the equation** Rewrite the differential equation in a more manageable form: \[ z y \, dx - z x \, dy = y^2 \, dz \] --- ### **Step 2: Divide through to simplify** Divide both sides by \( y^2 \): \[ \frac{z y}{y^2} dx - \frac{z x}{y^2} dy = dz \] which simplifies to: \[ \frac{z}{y} dx - \frac{z x}{y^2} dy = dz \] --- ### **Step 3: Substitution \( z = v y \)** Since the equation is homogeneous (each term has degree 2), use the substitution: \[ z = v y \quad \Rightarrow \quad dz = v \, dy + y \, dv \] --- ### **Step 4: Substitute into the equation** Replace \( z \) and \( dz \): \[ \frac{v y}{y} dx - \frac{v y x}{y^2} dy = v \, dy + y \, dv \] which simplifies to: \[ v \, dx - v \frac{x}{y} dy = v \, dy + y \, dv \] Bring all to one side: \[ v \, dx - v \frac{x}{y} dy - v \, dy - y \, dv = 0 \] --- ### **Step 5: Group terms** Group the \( v \) terms: \[ v \left( dx - \frac{x}{y} dy - dy \right) = y \, dv \] Which simplifies to: \[ v \left( dx - \left( \frac{x}{y} + 1 \right) dy \right) = y \, dv \] --- ### **Step 6: Express in terms of \( x \) and \( y \)** Divide both sides by \( y \): \[ \frac{v}{y} \left( dx - \left( \frac{x}{y} + 1 \right) dy \right) = dv \] Alternatively, rewrite as: \[ \frac{v}{y} dx - v \frac{\frac{x}{y} + 1}{y} dy = dv \] But it's more straightforward to consider the differential: \[ v \left( dx - \left( 1 + \frac{x}{y} \right) dy \right) = y \, dv \] --- ### **Step 7: Recognize the differential form** Note that: \[ dx - \left( 1 + \frac{x}{y} \right) dy \] can be written as: \[ d \left( \frac{x}{y} \right ) \] **Verification:** \[ d \left( \frac{x}{y} \right ) = \frac{1}{y} dx - \frac{x}{y^2} dy \] Multiply both sides by \( y \): \[ dy \, d \left( \frac{x}{y} \right ) = dx - \frac{x}{y} dy \] But our expression involves: \[ dx - \left( 1 + \frac{x}{y} \right) dy = dx - dy - \frac{x}{y} dy \] which can be written as: \[ \left( dx - \frac{x}{y} dy \right) - dy = y \, d \left( \frac{x}{y} \right) - dy \] Therefore, \[ dx - \left(1 + \frac{x}{y}\right) dy = y \, d \left( \frac{x}{y} \right) - dy \] --- ### **Step 8: Rewrite the key relation** Substitute back into the equation: \[ v \left( y \, d \left( \frac{x}{y} \right) - dy \right) = y \, dv \] Divide both sides by \( y \): \[ v \left( d \left( \frac{x}{y} \right) - \frac{dy}{y} \right) = dv \] --- ### **Step 9: Variable substitution** Let: \[ u = \frac{x}{y} \] then \[ d u = d \left( \frac{x}{y} \right) \] Hence, the equation becomes: \[ v \left( du - \frac{dy}{y} \right) = dv \] or \[ v \, du - v \frac{dy}{y} = dv \] --- ### **Step 10: Separate variables** Rearranged: \[ v \, du - dv = v \frac{dy}{y} \] Bring all to one side: \[ v \, du - dv - v \frac{dy}{y} = 0 \] --- ### **Step 11: Recognize as an exact differential** Rewrite as: \[ v \, du - dv = v \frac{dy}{y} \] which can be written as: \[ d (v u) - u \, dv - dv = v \frac{dy}{y} \] But more straightforwardly, note that: \[ v \, du - dv = d (v u) - u \, dv - dv \] Alternatively, let's consider integrating directly. --- ### **Step 12: Final step—integrate** From the previous relation: \[ v \, du - dv = v \frac{dy}{y} \] Divide through by \( v \): \[ du - \frac{1}{v} dv = \frac{dy}{y} \] Now, integrate both sides: \[ \int du - \int \frac{1}{v} dv = \int \frac{dy}{y} \] which yields: \[ u - \ln |v| = \ln |y| + C \] Recall that: \[ u = \frac{x}{y} \quad \text{and} \quad v = \frac{z}{y} \] Thus: \[ \frac{x}{y} - \ln \left| \frac{z}{y} \right| = \ln |y| + C \] Or equivalently: \[ \boxed{ \frac{x}{y} - \ln |z| + \ln |y| = C' } \] where \( C' \) is an arbitrary constant. --- ### **### **Final answer:** \[ \boxed{ \frac{x}{y} - \ln |z| = \text{constant} } \] or, rearranged: \[ \boxed{ \frac{x}{y} = \ln |z| + C } \] --- ## **Summary:** - Used substitution \( z = v y \) - Recognized the differential form involving \( d(x/y) \) - Integrated to find the relation between \( x, y, z \) **This provides the general solution to the differential equation.**