Answer to Question #215948 in Differential Equations for Bashar

Question

( 1 /t + 1/ t ^2 − y /t ^2 + y ^2 ) d t + ( y e ^y + t /t ^2 + y^ 2 ) d y = 0

Answer 1

$( {1 \over t} + {1\over t ^2} − {y \over t ^2 + y ^2 }) d t + ( y e ^y +{ t \over t ^2 + y^ 2 }) d y = 0$

$M(t,y)=$ $( {1 \over t} + {1\over t ^2} − {y \over t ^2 + y ^2 })$

$\implies { \partial M(t,y)\over \partial y}=M_y=-{1(t^2+y^2)-2y(y) \over (t^2+y^2)^2}=-{t^2+y^2-2y^2\over (t^2+y^2)^2}=-{t^2-y^2\over (t^2+y^2)^2}={y^2-t^2\over (t^2+y^2)^2}$

$N(t,y)=( y e ^y +{ t \over t ^2 + y^ 2 })$

$\implies { \partial N(t,y)\over \partial t}=N_t={y^2-t^2\over (t^2+y^2)^2}$

The differential equation is exact since $M_y=N_t$

There exist a function $\mu(t,y)$ such that

${\partial \mu (t,y)\over \partial t}=M(t,y)=( {1 \over t} + {1\over t ^2} − {y \over t ^2 + y ^2 })$

${\partial \mu (t,y)\over \partial y}=N(t,y)=( y e ^y +{ t \over t ^2 + y^ 2 })$

Let $\mu (t,y)=\int^yN(t,y)dy$ $+$ $g(t)$

$=\int^y( y e ^y +{ t \over t ^2 + y^ 2 }) dy$ $+$ $g(t)$

$=ye^y-e^y+tan^{-1}({y\over t})+g(t)$

$\implies {\partial \mu (t,y)\over \partial t}=-{y\over t^2+y^2}+{\partial g(t)\over \partial t}={1 \over t} + {1\over t ^2} − {y \over t ^2 + y ^2 }$

$\implies {\partial g(t)\over \partial t}={1 \over t} + {1\over t ^2}$

$\implies \int dg(t)=\int( {1 \over t} + {1\over t ^2})dt$

$\implies g(t)=ln(t)-{1\over t}$

$\implies \mu(t,y)=ye^y-e^y+tan^{-1}({y\over t})+ln(t)-{1\over t}$

$\therefore$ The general solution is thus

$ye^y-e^y+tan^{-1}({y\over t})+ln(t)-{1\over t}=c$