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MathematicsDifferential EquationLinear DE / Red. LDEMedium2 minPYQ_2023
MathematicsMediumnumerical range

Let y=yxbe the solution of the differential equationdydx+5xx5+1y=x5+12x7, x>0. Ify1=2, theny2is equal to

Options:

Answer:
C
Solution:

Given:

dydx+5x1+x5y=1+x52x7

This is linear differential equation.

I.F.=e5x1+x5dx

I.F.=e5x5x61+x5dx

I.F.=e5x6x-5+1dx

I.F.=e--5x6x-5+1dx

I.F.=e-dx-5+1x-5+1

I.F.=e-logex-5+1

I.F.=x-5+1-1=x5x5+1

So, solution is given by

yx5x5+1=x5x5+1×x5+12x7dx

yx5x5+1=x5+1x2dx

yx5x5+1=x3+1x2dx

yx5x5+1=x44-1x+C

Since, y1=2, so

22=14-1+CC=74

So,

yx5x5+1=x44-1x+74

Put x=2, then we get

y2×3233=164-12+74

y2=21×33128=693128

Stream:JEESubject:MathematicsTopic:Differential EquationSubtopic:Linear DE / Red. LDE
2mℹ️ Source: PYQ_2023

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