The solution of \(\rm x {{dy}\over {dx}}=y+x\ tan {y\over x}\) is

Question

Question

Answer 1

Concept:

Some useful formulas are:

\(\rm \int{{dx}\over {x}}=log x+c\)

\(\rm \int{cot\ x\ dx}=log(sin\ x)+c\)

Calculation:

\(\rm x {{dy}\over {dx}}=y+x\ tan {y\over x}\)

Substituting y=vx and \(\rm v+x {{dv}\over{dx}} = {{dy}\over {dx}}\)

\(\rm x [v+x{{dv}\over {dx}}]=vx+x\ tan {v}\)

\(\rm x [v+x{{dv}\over {dx}}]=x(v+ tan {v})\)

\(\rm \int{{dx}\over {x}}=\int{{dv}\over tan {v}}\)

\(\rm \int{{dx}\over {x}}=\int{cotv\ dv}\)

log x = log (sinv) + log c

x = c sinv

Putting the value of v we get,

\(\rm x = c\sin{y\over x}\)

[Where c = constant of integration]