Answer to Question #124157 in Calculus for Amoah Henry

Question #124157
Verify that the given family of functions solves the differential equation. (i)
dy dt
(ii)
dy dt
= (1 − 2t)y2, y = (c) Evaluate the integral
= y2 sin t, y = 1
0 ( 1
C − t + t2 . 1
C + cos t x2
√ . 4 − x2)3 dx
1
Expert's answer
2020-06-29T17:12:44-0400

(i) Given dydt=(12t)y2\frac{dy}{dt} = (1-2t)y^2 and y=1ct+t2y = \frac{1}{c-t+t^2}

Differentiating y with respect to x,

dydx=(1ct+t2)2(1+2t)=(1ct+t2)2(12t)\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, dydt=(12t)y2\frac{dy}{dt} = (1-2t)y^2


(ii) Given dydt=y2sin(t)\frac{dy}{dt} = y^2sin(t) and y=1c+costy = \frac{1}{c+cost}

differentiating both sides with respect to x,

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

replacing value of y with y

we obtain, dydt=y2sin(t)\frac{dy}{dt} = y^2sin(t)


(iii) We need to find integration of 01x2(4x2)3dx\int_0^1 x^2 (\sqrt{4-x^2})^3dx

let x=2sin(u)x = 2sin(u)

then dx=2cos(u)dudx = 2cos(u) du

Let us solve integration first without limit

8cos(u)sin2(u)(44sin2(u))3/2du\int 8cos(u)sin^2(u)(4-4sin^2(u))^{3/2}du


=8cos(u)sin2(u)(4cos2(u))3/2du=\int 8cos(u)sin^2(u)(4cos^2(u))^{3/2}du

=64cos4(u)sin2(u)du= \int 64cos^4(u)sin^2(u)du

=64cos6(u)(1cos2(u))du= \int 64cos^6(u)(1-cos^2(u))du

integrating it we get,

x(4x2)5/26+x(4x2)3/26+x(4x2)1/2+4asin(x2)+C-\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

01x2(4x2)3dx=2π3\int_0^1 x^2 (\sqrt{4-x^2})^3dx = \frac{2\pi}{3}



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