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MathematicsDifferential EquationLinear DE / Red. LDEEasy2 minPYQ_2022
MathematicsEasynumerical range

Ifdydx+2ytanx=sinx,0<x<π2andyπ3=0, then the maximum value ofyxis

Options:

Answer:
A
Solution:

We know that,

dydx+y2tanx=sinx is a linear differential equation so,

I.F=e2tanxdx=e2lnsecx=sec2x

The general solution will be 

ysec2x=sinxsec2xdx+C

ysec2x=secx+C

yπ3=0C=-2

Hence the particular solution is

ysec2x=secx-2

y=cosx-2cos2x

y=18-2cosx-142

So, the maximum value of yx is 18

Stream:JEESubject:MathematicsTopic:Differential EquationSubtopic:Linear DE / Red. LDE
2mℹ️ Source: PYQ_2022

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