If x³dy + xy dx = x²dy + 2y dx; y(2) = e and
x > 1, then y(4) is equal to :

${x^3}dy + xy\,dx = {x^2}dy + 2y\,dx$<br><br>$\Rightarrow dy({x^3} - {x^2}) = dx(2y - xy)$<br><br>$\Rightarrow$ $- \int {{1 \over y}dx}$ = $\int {{{x - 2} \over {{x^2}(x - 1)}}dx}$<br><br>$\Rightarrow$ $$ - \ln y = \int {\left( {{A \over x} + {B \over {{x^2}}} + {C \over {(x - 1)}}} \right)dx} $$<br><br>Where A = 1, B = +2, C = $-$1<br><br>$\Rightarrow - \ln y = \ln x - {2 \over x} - \ln (x - 1) + \lambda$<br><br>As $y(2) = e$<br><br>$\Rightarrow - 1 = \ln 2 - 1 - 0 + \lambda$<br><br>$\therefore$ $\lambda = - \ln 2$<br><br>$\Rightarrow \ln y = - \ln x + {2 \over x} + \ln (x - 1) + \ln 2$<br><br>Now put x = 4 in equation<br><br>$\Rightarrow \ln y = - \ln 4 + {1 \over 2} + \ln 3 + \ln 2$<br><br>$$ \Rightarrow {\mathop{\rm lny}\nolimits} = ln\left( {{3 \over 2}} \right) + {1 \over 2}\ln e$$<br><br>$\Rightarrow y = {3 \over 2}\sqrt e$