1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# Solve the differential equation: ysinxdydx=cosx(sinxâˆ’y2)

A
y2=23sinx+csin2x
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
y2=23sinx+csin2x
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
C
y2=23sinxcsin2x
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
y2=23cosx+csin2x
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
Open in App
Solution

## The correct options are A y2=23sinx+csin2x C y2=23sinx−csin2xGiven differential eqn is ydydxsinx=cosx(sinx−y2)ydydx+(cotx)y2=cosxPut y2=z⇒dzdx=2ydydx⇒12dzdx+zcotx=cosx⇒dzdx+2zcotx=2cosxwhich is a linear differential eqn with z as dependent variable.Here, P=2cotx,Q=2cosxIntegrating factor I.F.=e∫Pdx=e∫2cotxdxI.F.=e2logsinx⇒I.F.=sin2xSolution of given differential eqn is zsin2x=∫2cosxsin2xdxPut sinx=t⇒cosxdx=dt⇒zsin2x=2t33+C ⇒y2sin2x=23sin3x+C⇒y2=23sinx+Csin2xC can be positive or negative. Hence, both options are correct

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Methods of Solving First Order, First Degree Differential Equations
MATHEMATICS
Watch in App
Explore more
Join BYJU'S Learning Program