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

Question

If x3dy + xy dx = x2dy + 2y dx; y(2) = e and

x > 1, then y(4) is equal to :

Accepted Answer

$${x^3}dy + xy\,dx = {x^2}dy + 2y\,dx$$

$$ \Rightarrow dy({x^3} - {x^2}) = dx(2y - xy)$$

$$ \Rightarrow $$ $$ - \int {{1 \over y}dx} $$ = $$\int {{{x - 2} \over {{x^2}(x - 1)}}dx} $$

$$ \Rightarrow $$ $$ - \ln y = \int {\left( {{A \over x} + {B \over {{x^2}}} + {C \over {(x - 1)}}} \right)dx} $$

Where A = 1, B = +2, C = $-$1

$$ \Rightarrow  - \ln y = \ln x - {2 \over x} - \ln (x - 1) + \lambda $$

As $y(2) = e$

$$ \Rightarrow  - 1 = \ln 2 - 1 - 0 + \lambda $$

$$ \therefore $$ $$\lambda  =  - \ln 2$$

$$ \Rightarrow \ln y =  - \ln x + {2 \over x} + \ln (x - 1) + \ln 2$$

Now put x = 4 in equation

$$ \Rightarrow \ln y =  - \ln 4 + {1 \over 2} + \ln 3 + \ln 2$$

$$ \Rightarrow {\mathop{\rm lny}\nolimits}  = ln\left( {{3 \over 2}} \right) + {1 \over 2}\ln e$$

$$ \Rightarrow y = {3 \over 2}\sqrt e $$

Solutions · 1