\((y^2+1)=(ysec^2x)dy/dx\)

\((y^2+1)=(ysec^2x)y'\)

\(y/(-y^2-1)y'= (-1)/sec^2x\)

From the above, it is observed that the first order differential is in the form of \(N(x)y'=M(x)\), where \(N(x)= y/(-y^2-1)\) and \(M(x) = (-1)/sec^2x\)

Integrate on both sides of \(y/(-y^2-1)dy= (-1)/sec^2x dx\)

\(=> int y/(-y^2-1) dy= int (-1)/sec^2x dx\)

\(=> -int y/(-y^2-1) dy= -int 1/sec^2x dx\)

\(=> -1/2 ln(-y^2-1)= /1/2(x+1/2sin(2x))+c\)

\(-1/2 ln(0-1)= -1/2(0+1/2sin(0))+c\) \([:' y(0)=0]\)

c=0 Substitute \(c=0 in -1/2 ln(-y^2-1)+ -1/2(x+1/2sin(2x))+c\) and solve for y.

\(-1/2 ln(-y^2-1)= -1/2(x+1/2sin(2x))\)

\(ln(-y^2-1)= x+1/2sin(2x)\)

\(-y^2-1= e^(x+1/2sin(2x))\)

\(-y^2= e^(x+1/2sin(2x))+1\)

\(y^2= e^(x+1/2sin(2x))-1\)

\(y= +- sqrt(e^(x+1/2sin(2x))-1)\)

\((y^2+1)=(ysec^2x)y'\)

\(y/(-y^2-1)y'= (-1)/sec^2x\)

From the above, it is observed that the first order differential is in the form of \(N(x)y'=M(x)\), where \(N(x)= y/(-y^2-1)\) and \(M(x) = (-1)/sec^2x\)

Integrate on both sides of \(y/(-y^2-1)dy= (-1)/sec^2x dx\)

\(=> int y/(-y^2-1) dy= int (-1)/sec^2x dx\)

\(=> -int y/(-y^2-1) dy= -int 1/sec^2x dx\)

\(=> -1/2 ln(-y^2-1)= /1/2(x+1/2sin(2x))+c\)

\(-1/2 ln(0-1)= -1/2(0+1/2sin(0))+c\) \([:' y(0)=0]\)

c=0 Substitute \(c=0 in -1/2 ln(-y^2-1)+ -1/2(x+1/2sin(2x))+c\) and solve for y.

\(-1/2 ln(-y^2-1)= -1/2(x+1/2sin(2x))\)

\(ln(-y^2-1)= x+1/2sin(2x)\)

\(-y^2-1= e^(x+1/2sin(2x))\)

\(-y^2= e^(x+1/2sin(2x))+1\)

\(y^2= e^(x+1/2sin(2x))-1\)

\(y= +- sqrt(e^(x+1/2sin(2x))-1)\)