dydt tt2yy Initial condition y03 ySolutiondydt tt2yy Initi

dy/dt = t/(t^2y+y). Initial condition: y(0)=3.

y=

dy/dt = t/(t^2y+y). Initial condition: y(0)=3.

dy/dt = t/ y ( t^2+1)

y dy = t/ (t^2+1) dt

take integration

y^2/2 = 1/2 ln(t^2+1) +C

3^2/2 = 1/2 + C

C = 9/2 - 1/2 = 8/2 = 4

y^2/2 = 1/2 ln(t^2+1) +4

dy/dt = t/(t^2y+y). Initial condition: y(0)=3. y=Solutiondy/dt = t/(t^2y+y). Initial condition: y(0)=3. dy/dt = t/ y ( t^2+1) y dy = t/ (t^2+1) dt take integrat

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