Question

Verify that the given family of functions solves the differential equation.
(i)
dy dt
= (1−2t)y2, y = 1 C −t + t2
.
(ii)
dy dt
= y2 sint, y = 1 C + cost

(c) Evaluate the integralZ1 0
x2 (√4−x2)3dx

Answer 1

(i) Given $\frac{dy}{dt} = (1-2t)y^2$ and  $y = \frac{1}{c-t+t^2}$

Differentiating y with respect to x, 

$\frac{dy}{dx} = -(\frac{1}{c-t+t^2})^2(-1+2t) = (\frac{1}{c-t+t^2})^2(1-2t)$

replacing value of y 

we obtain, $\frac{dy}{dt} = (1-2t)y^2$  

(ii) Given $\frac{dy}{dt} = y^2sin(t)$ $y = \frac{1}{c+cost}$

differentiating both sides with respect to x,

$\frac{dy}{dt} = -(\frac{1}{c+cost})^2(-sint) = (\frac{1}{c+cost})^2(sint)$

replacing value of y with x

we obtain, $\frac{dy}{dt} = y^2sin(t)$

(iii) $\int_0^1 x^2 (\sqrt{4-x^2})^3dx$

let $x = 2sin(u)x=2sin(u)$ then $dx = 2cos(u) du$

Let us solve integration first without limit

$=\int 8cos(u)sin^2(u)(4-4sin^2(u))^{3/2}du$

$=\int 8cos(u)sin^2(u)(4cos^2(u))^{3/2}du$

$= \int 64cos^4(u)sin^2(u)du$

$= \int 64cos^6(u)(1-cos^2(u))du$

integrating it we get,

$=-\frac{x(4-x^2)^{5/2}}{6} + \frac{x(4-x^2)^{3/2}}{6}+x(4-x^2)^{1/2} + 4asin(\frac{x}{2}) + C$

putting the limits, we get

$\int_0^1 x^2 (\sqrt{4-x^2})^3dx = \frac{2\pi}{3}$