JEE MAIN - Mathematics (2006 - No. 11)
The differential equation whose solution is $$A{x^2} + B{y^2} = 1$$
where $$A$$ and $$B$$ are arbitrary constants is of
where $$A$$ and $$B$$ are arbitrary constants is of
second order and second degree
first order and second degree
first order and first degree
second order and first degree
Explanation
$$A{x^2} + B{y^2} = 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$
$$Ax + by{{dy} \over {dx}} = 0\,\,\,\,\,\,\,\,\,\,\,\,...\left( 2 \right)$$
$$A + By{{{d^2}y} \over {d{x^2}}} + B{\left( {{{dy} \over {dx}}} \right)^2} = 0\,\,\,\,\,\,\,\,\,\,\,\,...\left( 3 \right)$$
From $$(2)$$ and $$(3)$$
$$x\left\{ { - By{{{d^2}y} \over {d{x^2}}} - B{{\left( {{{dy} \over {dx}}} \right)}^2}} \right\} + By{{dy} \over {dx}} = 0$$
Dividing both sides by $$-B,$$ we get
$$ \Rightarrow xy{{{d^2}y} \over {d{x^2}}} + x{\left( {{{dy} \over {dx}}} \right)^2} - y{{dy} \over {dx}} = 0$$
Which is $$DE$$ of order $$2$$ and degree $$1.$$
$$Ax + by{{dy} \over {dx}} = 0\,\,\,\,\,\,\,\,\,\,\,\,...\left( 2 \right)$$
$$A + By{{{d^2}y} \over {d{x^2}}} + B{\left( {{{dy} \over {dx}}} \right)^2} = 0\,\,\,\,\,\,\,\,\,\,\,\,...\left( 3 \right)$$
From $$(2)$$ and $$(3)$$
$$x\left\{ { - By{{{d^2}y} \over {d{x^2}}} - B{{\left( {{{dy} \over {dx}}} \right)}^2}} \right\} + By{{dy} \over {dx}} = 0$$
Dividing both sides by $$-B,$$ we get
$$ \Rightarrow xy{{{d^2}y} \over {d{x^2}}} + x{\left( {{{dy} \over {dx}}} \right)^2} - y{{dy} \over {dx}} = 0$$
Which is $$DE$$ of order $$2$$ and degree $$1.$$
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