JEE MAIN - Mathematics (2019 - 12th April Evening Slot - No. 3)
The general solution of the differential equation (y2
– x3)dx – xydy = 0 (x $$ \ne $$ 0) is :
(where c is a constant of integration)
y2
+ 2x3
+ cx2
= 0
y2
+ 2x2
+ cx3
= 0
y2
– 2x + cx3
= 0
y2
– 2x3
+ cx2
= 0
Explanation
$$({y^2} - {x^3})dx - xydy = 0$$
$$ \Rightarrow y(ydx - xdy) = {x^3}dx$$
$$ \Rightarrow {y \over x}\left( {{{ydx - xdy} \over {{x^2}}}} \right) = dx$$
$$ \Rightarrow - {y \over x}d\left( {{y \over x}} \right) = dx$$
$$ \Rightarrow - {1 \over 2}{\left( {{y \over x}} \right)^2} = x + k$$
$$ \Rightarrow - {y^2} = 2{x^3} + 2{x^2}k$$
$$ \Rightarrow {y^2} + 2{x^3} + c{x^2} = 0$$
$$ \Rightarrow y(ydx - xdy) = {x^3}dx$$
$$ \Rightarrow {y \over x}\left( {{{ydx - xdy} \over {{x^2}}}} \right) = dx$$
$$ \Rightarrow - {y \over x}d\left( {{y \over x}} \right) = dx$$
$$ \Rightarrow - {1 \over 2}{\left( {{y \over x}} \right)^2} = x + k$$
$$ \Rightarrow - {y^2} = 2{x^3} + 2{x^2}k$$
$$ \Rightarrow {y^2} + 2{x^3} + c{x^2} = 0$$
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