JEE MAIN - Mathematics (2020 - 3rd September Evening Slot - No. 11)
If x3dy + xy dx = x2dy + 2y dx; y(2) = e and
x > 1, then y(4) is equal to :
x > 1, then y(4) is equal to :
$${{\sqrt e } \over 2}$$
$${1 \over 2} + \sqrt e $$
$${3 \over 2} + \sqrt e $$
$${3 \over 2}\sqrt e $$
Explanation
$${x^3}dy + xy\,dx = {x^2}dy + 2y\,dx$$
$$ \Rightarrow dy({x^3} - {x^2}) = dx(2y - xy)$$
$$ \Rightarrow $$ $$ - \int {{1 \over y}dx} $$ = $$\int {{{x - 2} \over {{x^2}(x - 1)}}dx} $$
$$ \Rightarrow $$ $$ - \ln y = \int {\left( {{A \over x} + {B \over {{x^2}}} + {C \over {(x - 1)}}} \right)dx} $$
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
$$ \Rightarrow \ln y = - \ln 4 + {1 \over 2} + \ln 3 + \ln 2$$
$$ \Rightarrow {\mathop{\rm lny}\nolimits} = ln\left( {{3 \over 2}} \right) + {1 \over 2}\ln e$$
$$ \Rightarrow y = {3 \over 2}\sqrt e $$
$$ \Rightarrow dy({x^3} - {x^2}) = dx(2y - xy)$$
$$ \Rightarrow $$ $$ - \int {{1 \over y}dx} $$ = $$\int {{{x - 2} \over {{x^2}(x - 1)}}dx} $$
$$ \Rightarrow $$ $$ - \ln y = \int {\left( {{A \over x} + {B \over {{x^2}}} + {C \over {(x - 1)}}} \right)dx} $$
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
$$ \Rightarrow \ln y = - \ln 4 + {1 \over 2} + \ln 3 + \ln 2$$
$$ \Rightarrow {\mathop{\rm lny}\nolimits} = ln\left( {{3 \over 2}} \right) + {1 \over 2}\ln e$$
$$ \Rightarrow y = {3 \over 2}\sqrt e $$
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