Solve differential equation (y2+1)dx=ysec^2(x)dy, y(0)=0

Solve differential equation (y2+1)dx=ysec^2(x)dy, y(0)=0

Solve differential equation \((y2+1)dx=ysec^2(x)dy\), y(0)=0

Answers (1)

\(y/(-y^2-1)y'= (-1)/sec^2x\)
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.
\(-1/2 ln(-y^2-1)= -1/2(x+1/2sin(2x))\)
\(ln(-y^2-1)= x+1/2sin(2x)\)
\(-y^2-1= e^(x+1/2sin(2x))\)
\(-y^2= e^(x+1/2sin(2x))+1\)
\(y^2= e^(x+1/2sin(2x))-1\)
\(y= +- sqrt(e^(x+1/2sin(2x))-1)\)

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