Maths

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New answer posted

a year ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

Kindly go through the solution

New answer posted

a year ago

0 Follower 1 View

V
Vishal Baghel

Contributor-Level 10

Kindly go through the solution

New answer posted

a year ago

0 Follower 1 View

V
Vishal Baghel

Contributor-Level 10

Kindly go through the solution

New answer posted

a year ago

0 Follower 7 Views

V
Vishal Baghel

Contributor-Level 10

Kindly go through the solution

New answer posted

a year ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

Kindly go through the solution

New answer posted

a year ago

0 Follower 8 Views

V
Vishal Baghel

Contributor-Level 10

Kindly go through the solution

New question posted

a year ago

0 Follower 3 Views

New answer posted

a year ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

The given differential equation is:

exdy+(yex+2x)dx=0exdydx+yex+2x=0dydx+y=2xex

This is a linear differential equation of the form

dydx+Py=Q,whereP=1&Q=2xexNow,I.F.=ePdx=edx=ex

The general solution of the given differential equation is given by,

y(I.F.)=(Q*I.F.)dx+Cyex=(2xex.ex)dx+Cyex=2xdx+Cyex=x2+Cyex+x2=C

Therefore, option (c) is correct.

New answer posted

a year ago

0 Follower 5 Views

V
Vishal Baghel

Contributor-Level 10

The integrating factor of the given differential equation  dxdy+P1x=Q1

The general solution of the differential equation is given by,

x (I.F.)= (Q*I.F.)dy+Cx.eP1dy= (Q1eP1dy)dy+C

Hence, the correct answer is C.

New answer posted

a year ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

The given differential equation is:

ydxxdyy=0ydxxdyxy=01xdx1ydy=0

Integration both sides, we get:

log|x|log|y|=logklog|xy|=logkxy=ky=1kxy=Cx, where, C=1k

Therefore, option (C) is correct.

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