# Tutor profile: Bindu P.

## Questions

### Subject: Set Theory

Write all possible partitions of S = {x, y , z}.

Let us recall the definition of partition. Partition of S is a collection of nonempty subsets of S which are disjoint and their union is S. 1) {x}, {y}, {z}, We can see, {x}$$\cap$$ {y}$$\cap$$ {z} = $$\emptyset$$ and {x}$$\cup$${y}$$\cup$${z} = S 2) {x, y}, {z} We can see, {x, y}$$\cap$$ {z} = $$\emptyset$$ and {x, y}$$\cup$$ {z} = S 3) {x, z}, {y} We can see, {x, z}$$\cap$$ {y} = $$\emptyset$$ and {x, z}$$\cup$$ {y} = S 4) {y, z}, {x} We can see, {y, z}$$\cap$$ {x} = $$\emptyset$$ and {y, z}$$\cup$$ {x} = S 5) {x, y, z} The set itself is a partition of the set.

### Subject: Discrete Math

For x, y $$\in$$ R . Define the relation x $$\simeq$$ y mean x - y $$\in$$ Z. Prove that $$\simeq$$ is an equivalence relation on R.

First of all, let us recall the definition of an equivalence relation. A relation is an equivalence relation if it is 1) Reflexive i.e x $$\simeq$$ x 2) symmetric i.e x $$\simeq$$ y then y $$\simeq$$ x 3) transitive. i.e x $$\simeq$$ y and y$$\simeq$$ z then x $$\simeq$$ z Since, x - x = 0 $$\in$$ Z, x $$\simeq$$ x Hence, the given relation is reflexive. Now to check whether the given relation is symmetric or not, let us assume that x $$\simeq$$ y. Since, x $$\simeq$$ y, x - y $$\in$$ Z -(x - y) $$\in$$ Z y - x $$\in$$ Z Hence, y $$\simeq$$ x, Hence, the given relation is symmetric. Now to check whether the given relation is reflexive or not, let us assume that x $$\simeq$$ y and y $$\simeq$$ z Since, x $$\simeq$$ y and y $$\simeq$$ z x - y $$\in$$ Z and y - z $$\in$$ Z Since, addition of two whole number gives a whole number, (x - y) +(y - z) $$\in$$ Z x - y + y - z $$\in$$ Z x - z $$\in$$ Z Hence, x $$\simeq$$ z, Hence, the given relation is transitive. Since, given relation is reflexive, symmetric and transitive, it is an equivalence relation.

### Subject: Differential Equations

Find the general solution of the following differential equation. $$\frac{dy}{dx}$$ = 6$$y{^2}{x}$$

$$\frac{dy}{dx}$$ = 6$$y{^2}{x}$$ We can see that the above differential equation is separable. Let us separate above equation. $$y{^-2}$$dy = 6xdx Now integrate both sides. $$ \int \mathrm{y}^{-2}\mathrm{d}y $$ = $$ \int \mathrm{6x}\mathrm{d}x $$ Now use the formula $$ \int \mathrm{x}^{n}\mathrm{d}x $$ = $$\frac{x^{(n+1)}}{n+1}$$. So, we will get $$\frac{y^{(-2+1)}}{-2+1}$$ = 6$$\frac{x^{(1+1)}}{1+1}$$ + C; C is a constant. Simplify above equation. $$\frac{y^{-1}}{-1}$$ = 6$$\frac{x^{2}}{2}$$ + C -$$\frac{1}{y}$$ = 3$${x^{2}}$$ + C Hence, we get the general solution of the given differential equation.