Let's carefully analyze and solve the differential equation: \[ y' = x(y^2 - 1) - \frac{(y - 1)}{x} \] --- ### Step 1: Rewrite the differential equation Express \( y' \) explicitly: \[ \frac{dy}{dx} = x(y^2 - 1) - \frac{(y - 1)}{x} \] --- ### Step 2: Simplify the right side Notice that: \[ x(y^2 - 1) = x y^2 - x \] and the second term is: \[ -\frac{(y - 1)}{x} \] So, the equation becomes: \[ \frac{dy}{dx} = x y^2 - x - \frac{y - 1}{x} \] --- ### Step 3: Find a substitution to simplify the equation The equation involves \( y^2 \) and \( y - 1 \), suggesting a substitution related to \( y - 1 \). Let: \[ z = y - 1 \] then: \[ y = z + 1 \] and: \[ \frac{dy}{dx} = \frac{dz}{dx} \] Now, rewrite the entire equation in terms of \( z \): Calculate \( y^2 \): \[ y^2 = (z + 1)^2 = z^2 + 2z + 1 \] Substitute into the differential equation: \[ \frac{dz}{dx} = x(z^2 + 2z + 1) - x - \frac{z}{x} \] Simplify: \[ \frac{dz}{dx} = x z^2 + 2x z + x - x - \frac{z}{x} \] The \( x \) and \( -x \) cancel: \[ \frac{dz}{dx} = x z^2 + 2x z - \frac{z}{x} \] --- ### Step 4: Factor the right side Factor \( z \) where possible: \[ \frac{dz}{dx} = z (x z + 2x) - \frac{z}{x} \] Alternatively, write as: \[ \frac{dz}{dx} = z (x z + 2x) - \frac{z}{x} \] Note that: \[ x z + 2x = x(z + 2) \] So: \[ \frac{dz}{dx} = z \cdot x(z + 2) - \frac{z}{x} \] --- ### Step 5: Divide through by \( z \) (considering \( z \neq 0 \)) To analyze further, consider the case \( z \neq 0 \): \[ \frac{1}{z} \frac{dz}{dx} = x (z + 2) - \frac{1}{x} \] Define: \[ w = \ln |z| \implies \frac{dw}{dx} = \frac{1}{z} \frac{dz}{dx} \] Thus: \[ \frac{dw}{dx} = x(z + 2) - \frac{1}{x} \] But since \( z = e^{w} \), the problem becomes more complicated here. Alternatively, it's better to rewrite the original in a different way to find an integrating factor. --- ### Step 6: Recognize the form of the differential equation Return to the simplified form: \[ \frac{dz}{dx} = x z^2 + 2x z - \frac{z}{x} \] Divide through by \( z \) (assuming \( z \neq 0 \)): \[ \frac{1}{z} \frac{dz}{dx} = x z + 2x - \frac{1}{x} \] which simplifies to: \[ \frac{d}{dx} (\ln |z|) = x z + 2x - \frac{1}{x} \] Alternatively, since the previous step seems complicated, perhaps it's better to consider the original substitution differently. --- ### Step 7: Alternative substitution Instead of \( z = y - 1 \), consider the structure: \[ y' = x(y^2 - 1) - \frac{(y - 1)}{x} \] Notice that: \[ y^2 - 1 = (y - 1)(y + 1) \] So, \[ y' = x (y - 1)(y + 1) - \frac{(y - 1)}{x} \] Factor out \( (y - 1) \): \[ y' = (y - 1) \left[ x(y + 1) - \frac{1}{x} \right] \] Define: \[ z = y - 1 \] then: \[ y + 1 = z + 2 \] and \[ y' = \frac{dy}{dx} = \frac{dz}{dx} \] The equation becomes: \[ \frac{dz}{dx} = z \left[ x(z + 2) - \frac{1}{x} \right] \] which simplifies to: \[ \frac{dz}{dx} = z \left( x z + 2x - \frac{1}{x} \right) \] --- ### Step 8: Recognize the Bernoulli form The equation: \[ \frac{dz}{dx} = z \left( x z + 2x - \frac{1}{x} \right) \] can be written as: \[ \frac{dz}{dx} = x z^2 + 2x z - \frac{z}{x} \] This is a **Bernoulli differential equation** of the form: \[ \frac{dz}{dx} + P(x) z = Q(x) z^n \] Identify: \[ \frac{dz}{dx} - 2x z + \frac{z}{x} = x z^2 \] but the presence of the \(\frac{z}{x}\) term complicates the standard Bernoulli form. Alternatively, rearranged: \[ \frac{dz}{dx} = x z^2 + 2x z - \frac{z}{x} \] --- ### Step 9: Dividing the entire equation by \( z \) Assuming \( z \neq 0 \): \[ \frac{1}{z} \frac{dz}{dx} = x z + 2x - \frac{1}{x} \] Set: \[ w = \frac{1}{z} \] then: \[ z = \frac{1}{w} \] and: \[ \frac{dz}{dx} = - \frac{1}{w^2} \frac{dw}{dx} \] Substitute into the differential equation: \[ - \frac{1}{w^2} \frac{dw}{dx} = x \left( \frac{1}{w} \right)^2 + 2x \cdot \frac{1}{w} - \frac{1}{x} \] Multiply both sides by \( w^2 \): \[ - \frac{dw}{dx} = x + 2x w - \frac{w^2}{x} \] Rearranged: \[ \frac{dw}{dx} = - x - 2x w + \frac{w^2}{x} \] --- ### Step 10: Recognize a Riccati equation The equation: \[ \frac{dw}{dx} = - x - 2x w + \frac{w^2}{x} \] is a Riccati differential equation in \( w \). Standard Riccati form: \[ \frac{dw}{dx} = a(x) + b(x) w + c(x) w^2 \] Here: \[ a(x) = - x, \quad b(x) = - 2x, \quad c(x) = \frac{1}{x} \] --- ### **Summary so far:** - The original differential equation has been transformed into a Riccati equation in \( w \), which can be solved via substitution if a particular solution is known or via standard methods. --- ### **Step 11: Find a particular solution** Suppose \( w = k x \), where \( k \) is a constant to be determined. Plug into: \[ \frac{dw}{dx} = - x - 2x w + \frac{w^2}{x} \] Calculate \( \frac{dw}{dx} \): \[ \frac{d}{dx} (k x) = k \] Substitute: \[ k = - x - 2x (k x) + \frac{(k x)^2}{x} \] Simplify: \[ k = - x - 2k x^2 + \frac{k^2 x^2}{x} = - x - 2k x^2 + k^2 x \] Now,