a) When we say the differential equation is in variable separable form? b) Find the general solution of xy2 + (1-x)(dy)/(dx) = 0 when x!= 1 and y!= 0. Also, compute a particular solution when y(2)=1.

*a) When we say the differential equation is in variable separable form?

A first-order differential equation is in variable separable form if it can be rearranged such that all terms involving the dependent variable ($y$) and its differential ($dy$) are on one side of the equation, and all terms involving the independent variable ($x$) and its differential ($dx$) are on the other side. This means it can be written in the form $f(y) dy = g(x) dx$, where $f(y)$ is a function of $y$ only and $g(x)$ is a function of $x$ only.

*b) Find the general solution of $xy^2 + (1-x)\frac{dy}{dx} = 0$ when $x \neq 1$ and $y \neq 0$. Also, compute a particular solution when $y(2)=1$.

Step 1: Separate the variables. Given the differential equation: $xy^2 + (1-x)\frac{dy}{dx} = 0$ Subtract $xy^2$ from both sides: $(1-x)\frac{dy}{dx} = -xy^2$ Divide both sides by $(1-x)$ and $y^2$ to separate the variables. Since $x \neq 1$ and $y \neq 0$, these divisions are valid. $\frac{1}{y^2} dy = \frac{-x}{1-x} dx$

Step 2: Integrate both sides of the equation. $\int \frac{1}{y^2} dy = \int \frac{-x}{1-x} dx$ For the left-hand side (LHS): $\int y^{-2} dy = \frac{y^{-1}}{-1} = -\frac{1}{y}$ For the right-hand side (RHS), rewrite the integrand $\frac{-x}{1-x}$ as $\frac{x}{x-1}$. $\int \frac{x}{x-1} dx$ We can perform polynomial division or add and subtract 1 in the numerator: $\int \frac{x-1+1}{x-1} dx = \int \left( \frac{x-1}{x-1} + \frac{1}{x-1} \right) dx$ $= \int \left( 1 + \frac{1}{x-1} \right) dx$ $= x + \ln|x-1| + C$ Combining the results from both sides, the general solution is: $-\frac{1}{y} = x + \ln|x-1| + C$

Step 3: Find the particular solution using the initial condition $y(2)=1$. Substitute $x=2$ and $y=1$ into the general solution to find the value of $C$: $-\frac{1}{1} = 2 + \ln|2-1| + C$ $-1 = 2 + \ln(1) + C$ Since $\ln(1) = 0$: $-1 = 2 + 0 + C$ $-1 = 2 + C$ $C = -1 - 2$ $C = -3$

Step 4: Write the particular solution. Substitute $C=-3$ back into the general solution: $-\frac{1}{y} = x + \ln|x-1| - 3$ We can express $y$ explicitly: $\frac{1}{y} = 3 - x - \ln|x-1|$ $y = \frac{1}{3 - x - \ln|x-1|}$

The general solution is $\boxed{-\frac{1}{y} = x + \ln|x-1| + C}$. The particular solution is $\boxed{y = \frac{1}{3 - x - \ln|x-1|}}$.