Question

1. (a) Solve the differential equation (x + 1)Dy/dx= xy, = given that y = 2 when x = 0. (b) Find the area between the two curves y = x2 and y = 2x – x2.​

Answer 1

(a) The differential equation is separable, so we separate the variables and integrate:

`(x+1)(dy)/(dx) = xy \implies \frac{dy}y = \frac x{x+1} \, dx = \left(1-\frac1{x+1}\right) \, dx`

`\displaystyle \frac{dy}y = \int \left(1-\frac1{x+1}\right) \, dx`

`\ln|y| = x - \ln|x+1| + C`

When x = 0, we have y = 2, so we solve for the constant C :

`\ln|2| = 0 - \ln|0 + 1| + C \implies C = \ln(2)`

Then the particular solution to the DE is

`\ln|y| = x - \ln|x+1| + \ln(2)`

We can go on to solve explicitly for y in terms of x :

`e^(\ln|y|) = e^(x - \ln|x+1| + \ln(2)) \implies \boxed{y = (2e^x)/(x+1)}`

(b) The curves y = x² and y = 2x - x² intersect for

`x^2 = 2x - x^2 \implies 2x^2 - 2x = 2x (x - 1) = 0 \implies x = 0 \text{ or } x = 1`

and the bounded region is the set

`\left\{(x,y) ~:~ 0 \le x \le 1 \text{ and } x^2 \le y \le 2x - x^2\right\}`

The area of this region is

`\displaystyle \int_0^1 ((2x-x^2)-x^2) \, dx = 2 \int_0^1 (x-x^2) \, dx = 2 \left(\frac{x^2}2 - \frac{x^3}3\right)\bigg|_0^1 = 2\left(\frac12 - \frac13\right) = \boxed{\frac13}`