\(dx/dy+p(y)x=q(y)\)

Compare the equation \(dx/dy+p(y)x=q(y)\) to \(dx/dy+x/y= 1/(sqrt(1+y^2))\) and obtain \(p(y)=1/y\), \(q(y)= 1/(sqrt(1+y^2))\)

\(I.F.= e^(int p(y)dy)\)

\(= e^(ln y)\)

=y \(xe^(int p(y)dy)= int e^(int p(y)dy)q(y)dy+C\)

where C is arbitrary constant of equation

\(xe^(int p(y)dy)= int e^(int p(y)dy)q(y)dy+C\)

\(xy= int y 1/(sqrt(1+y^2))dy+C\)

\(xy= 1/2 int (2y)/(sqrt(1+y^2)) dy+C\)

\(xy= 1/2 int ((1+y^2)')/(sqrt(1+y^2)) dy+C\) [\(:' (1+y^2)'= 2y\)]

\(xy= 1/2 (2 sqrt(1+y^2))+C\) [\(:' int (f(y))/(sqrt(f(y)))dy= 2 sqrt(f(y))\)]

\(xy= sqrt(1+y^2)+C\)

\(x= (sqrt(1+y^2)+C)/y\)

\(x= (sqrt(1+y^2))/y+C/y\)

Compare the equation \(dx/dy+p(y)x=q(y)\) to \(dx/dy+x/y= 1/(sqrt(1+y^2))\) and obtain \(p(y)=1/y\), \(q(y)= 1/(sqrt(1+y^2))\)

\(I.F.= e^(int p(y)dy)\)

\(= e^(ln y)\)

=y \(xe^(int p(y)dy)= int e^(int p(y)dy)q(y)dy+C\)

where C is arbitrary constant of equation

\(xe^(int p(y)dy)= int e^(int p(y)dy)q(y)dy+C\)

\(xy= int y 1/(sqrt(1+y^2))dy+C\)

\(xy= 1/2 int (2y)/(sqrt(1+y^2)) dy+C\)

\(xy= 1/2 int ((1+y^2)')/(sqrt(1+y^2)) dy+C\) [\(:' (1+y^2)'= 2y\)]

\(xy= 1/2 (2 sqrt(1+y^2))+C\) [\(:' int (f(y))/(sqrt(f(y)))dy= 2 sqrt(f(y))\)]

\(xy= sqrt(1+y^2)+C\)

\(x= (sqrt(1+y^2)+C)/y\)

\(x= (sqrt(1+y^2))/y+C/y\)