Tutor profile: Vaidehi T.

If $$cos{x} = 4$$, then find the value of $$ cos{2x}.$$

As we know that, $$ cos{2x} = 2cos^2{x} - 1.$$ Plug the value of $$cos{x}$$, $$\therefore cos{2x} = 2(4)^2 - 1$$ $$= 32 - 1$$ $$= 31.$$

Find the equation of the line that passes through two points $$(2, 1),\, (3,4)$$.

When we are given two points and we have to find the equation of the line passing through those points, we can use two point formula. Let's say $$A(x_1, y_1)$$ and $$B(x_2, y_2)$$ are two points then the formula is defined as $$y - y_1 = \dfrac{y_2 - y_1}{x_2 - x_1} (x - x_1) $$. For the given data, we can take $$x_1 = 2,\, x_2 = 3,\, y_1 = 1,\, y_2 = 4.$$ Plugging all these values in the formula to find the equation of a line: $$y - 1 = \dfrac{4 - 1}{2 - 1} (x - 2) $$ $$\therefore y - 1 = 3(x-2)$$ $$\therefore y - 1 = 3x - 6$$ $$ \therefore y = 3x - 5.$$

Solve the following first order differential equation: $$\dfrac{dy}{dx} = x^2 y^2.$$

In order to solve this first order differential equation, we can separate variables like this $$ \dfrac{dy}{y^2} = x^2\, dx.$$ Now, integrating both the sides, we get $$\dfrac{y^{-2+1}}{-2+1} = \dfrac{x^{2+1}}{2+1} + C$$ $$ \therefore \dfrac{-1}{y} = \dfrac{x^3}{3} + C$$ $$ \therefore \dfrac{x^3}{3} + \dfrac{1}{y} + C = 0.$$