wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

The solution of the differential equation dydx+tanyx=tanysinyx2 is


A

2x=siny1+cx2

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B

ysinx+logx=c

No worries! We‘ve got your back. Try BYJU‘S free classes today!
C

logx+x=c

No worries! We‘ve got your back. Try BYJU‘S free classes today!
D

logx+y=c

No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A

2x=siny1+cx2


Explanation for correct option:

Given differential equation is dydx+tanyx=tanysinyx2

Therefore,

dydx+tanyx=tanysinyx2⇒1tanysinydydx+tanyxtanysiny=1x2⇒cotycosecydydx+cosecyx=1x2

Let cosecy=t.

On differentiating both sides we get -cosecycotydy=dt

Now substituting we get

⇒dtdx-tx=-1x2

We know that for the differential equation dydx+Pxy=Qx the integrating factor is fx=e∫Pxdx.

Then the solution of the Differential equation is ∫dyfx=∫Qxfxdx+c

Now comparing with the standard form we have Px=-1x,Qx=-1x2

Therefore Integrating factor

fx=e∫-1xdx=e-logx=1x

Therefore the solution is

∫dt1x=-∫1x2·1xdx+c⇒tx=-∫1x3dx+c⇒tx=12x2+c⇒cosecy=12x+cx2∵t=cosecy⇒2x=siny1+cx2∵1siny=cosecy

Hence, option (A) i.e. 2x=siny1+cx2 is correct


flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
General and Particular Solutions of a DE
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon