What is a solution to the differential equation #dy/dx=y^2+1# with y(1)=0?

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Answer 1 · 2016-09-09T13:39:02

#y = tan(x-1)#

#dy/dx=y^2+1#

grouping variables

#dy/(y^2+1) = dx#

Integrating both sides

#arctan(y) = x + C#

then

#y = tan(x+C)#

Now, observing the initial condition

#y(1)=tan(1+C)=0# so

#1+C=0+kpi, k=pm1,pm2,cdots#

and concluding

#y = tan(x-1)#