Question

a). Find dy/dx for the following integral.

y=Integral from 0 to cosine(x) dt/√1+ t^2 , 0<x<pi

b). Find dy/dx for tthe following integral

y=Integral from 0 to sine^-1 (x) cosine t dt

Answer #1

a)

we have

here,

now,

by the fundamental theorem,

we can say that

b)

we have

here,

now,

by the fundamental theorem,

we can say that

dy/dt=x , dx/dt=6x-8y, x=1 and y=-1 when t=0

Evaluate double integral Z 2 0 Z 1 y/2 cos(x^2 )dx dy
(integral from 0 to 2)(integral from y/2 to 1) for cos(x^2) dx
dy

dx
dt
= y − 1
dy
dt
= −3x + 2y
x(0) = 0, y(0) = 0

dx/dt = 1 - (b+1) x + a x^2 y
dy/dt = bx - a x^2 y
find all the fixed point and classify them with Jacobian.

Solve the system of differential equations using laplace
transformation
dy/dt-x=0,dx/dt+y=1,x(0)=-1,y(0)=1

Solve the 1st order initial value problem:
1+(x/y+cosy)dy/dx=0, y(pi/2)=0

Solve: 1.dy/dx=(e^(y-x)).secy.(1+x^2),y(0)=0.
2.dy/dx=(1-x-y)/(x+y),y(0)=2 .

Let w(x,y,z) = x^2+y^2+z^2 where x=sin(8t), y=cos(8t) , z=
e^t
Calculate dw/dt by first finding dx/dt, dy/dt, and dz/dt and using
the chain rule
dx/dt =
dy/dt=
dz/dt=
now using the chain rule calculate
dw/dt 0=

Find the general solution of the system
dx/dt = 2x + 3y
dy/dt = 5y
Determine the initial conditions x(0) and y(0) such that the
solutions x(t) and y(t) generates a straight line solution. That is
y(t) = Ax(t) for some constant A.

solve the given initial value problem
dx/dt=7x+y x(0)=1
dt/dt=-6x+2y y(0)=0
the solution is x(t)= and y(t)=

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