Question

Obtain the general solution of
x^{2}d^{2}y/dx^{2} -2x dy/dx
+2y=sin(lnx)

Answer #1

dx/dt=y, dy/dt=2x-2y

Find the particular integral of the differential equation
d2y/dx2 + 3dy/dx + 2y = e −2x
(x + 1). show that the answer is yp(x) = −e −2x ( 1/2
x2 + 2x + 2)

Solve the Initial Value Problem
(y2 cos(x) − 3x2y − 2x) dx + (2y sin(x) −
x3 + ln(y)) dy = 0, y(0) = e

Consider the system [ x' = -2y & y' = 2x] . Use dy/dx to
find the curves y = y(x).
Draw solution curves in the xy phase plane. What type of
equilibrium point is the origin?

Insert K = 3
[2x+ycos(kxy)]dx+[xcos(kxy)-2y]dy=0
Describe whether the first order type: Linear or exact
and solve to find the general Solution.

Consider the initial value problem
dy dx
=
1−2x 2y
, y(0) = − √2
(a) (6 points) Find the explicit solution to the initial value
problem.
(b) (3 points) Determine the interval in which the solution is
deﬁned.

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.

given that y1=xcos(lnx)and y2=xsin(lnx)form a fundamental set of
solutions to x^2y''-xy'+2y=0,find general solution to
x^2y''-xy'+2y=xlnx

Find the general solution to the system:
dx/dt = -4x + 3y
dy/dt = -5x/2 + 2y

Determine the type of below equations and solve it.
a-)(sin(xy)+xycos(xy)+2x)dx+(x2cos(xy)+2y)dy=0
b-)(t-a)(t-b)y’-(y-c)=0 a,b,c are
constant.

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