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$text If end text y equals fraction numerator x over denominator l n invisible function application vertical line c x vertical line end fraction$ (where c is an arbitrary constant) is the general solution of the differential equation $fraction numerator d y over denominator d x end fraction equals fraction numerator y over denominator x end fraction plus$ $phi open parentheses fraction numerator x over denominator y end fraction close parentheses text end text$ then the function $text end text phi open parentheses fraction numerator x over denominator y end fraction close parentheses text is : end text$

$\frac{x^{2}}{y^{2}}$
$\frac{x^{2}}{y^{2}}$
$\frac{y^{2}}{x^{2}}$
$- \frac{y^{2}}{x^{2}}$

The correct answer is: $negative fraction numerator y to the power of 2 end exponent over denominator x to the power of 2 end exponent end fraction$

Related Questions to study

A function y = f (x) satisfies the condition f '(x) sin x + f (x) cos x = 1, f (x) being bounded when x $rightwards arrow$ 0. If $I equals stretchy integral subscript 0 end subscript superscript there exists 2 end superscript f left parenthesis x right parenthesis d x text then end text$

A function y = f (x) satisfies the condition f '(x) sin x + f (x) cos x = 1, f (x) being bounded when x $rightwards arrow$ 0. If $I equals stretchy integral subscript 0 end subscript superscript there exists 2 end superscript f left parenthesis x right parenthesis d x text then end text$

Statement-I : In throwing of two dice, let the events A, B and C be 'the first dice shows an even number', 'the second dice shows an odd numbers' and 'both the dice show an odd numbers or both the dice show an even number’ respectively. Then $text end text P left parenthesis A right parenthesis equals P left parenthesis B right parenthesis equals P left parenthesis C right parenthesis equals fraction numerator 1 over denominator 2 end fraction text end text$ and $P left parenthesis A intersection B right parenthesis equals P left parenthesis B intersection C right parenthesis equals P left parenthesis C intersection A right parenthesis equals fraction numerator 1 over denominator 4 end fraction$ Therefore A, B and C are mutually independent events
Statement-II : Three events A, B and C are mutually independent if and only if P(A Ç B) = P(A) . P(B), P(B Ç C) = P(B) . P(C), $P left parenthesis C intersection A right parenthesis equals P left parenthesis C right parenthesis$ = P(C) . P(A) and $P left parenthesis A intersection B intersection C right parenthesis$ = P(A) . P(B) . P(C)

Statement-I : In throwing of two dice, let the events A, B and C be 'the first dice shows an even number', 'the second dice shows an odd numbers' and 'both the dice show an odd numbers or both the dice show an even number’ respectively. Then $text end text P left parenthesis A right parenthesis equals P left parenthesis B right parenthesis equals P left parenthesis C right parenthesis equals fraction numerator 1 over denominator 2 end fraction text end text$ and $P left parenthesis A intersection B right parenthesis equals P left parenthesis B intersection C right parenthesis equals P left parenthesis C intersection A right parenthesis equals fraction numerator 1 over denominator 4 end fraction$ Therefore A, B and C are mutually independent events
Statement-II : Three events A, B and C are mutually independent if and only if P(A Ç B) = P(A) . P(B), P(B Ç C) = P(B) . P(C), $P left parenthesis C intersection A right parenthesis equals P left parenthesis C right parenthesis$ = P(C) . P(A) and $P left parenthesis A intersection B intersection C right parenthesis$ = P(A) . P(B) . P(C)

The probability that 4th power of a positive integer ends in the digit 6 is:

The probability that 4th power of a positive integer ends in the digit 6 is:

parallel

Let A, B, C be pairwise independent events with P(C)> 0 and $P left parenthesis A intersection B intersection C right parenthesis equals 0$ . Then $P open parentheses A to the power of c end exponent intersection B to the power of c end exponent C close parentheses text is equal to: end text$

Let A, B, C be pairwise independent events with P(C)> 0 and $P left parenthesis A intersection B intersection C right parenthesis equals 0$ . Then $P open parentheses A to the power of c end exponent intersection B to the power of c end exponent C close parentheses text is equal to: end text$

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Let A and B be two events such that $P left parenthesis stack A union B with bar on top right parenthesis equals fraction numerator 1 over denominator 6 end fraction comma P left parenthesis A intersection B right parenthesis equals fraction numerator 1 over denominator 4 end fraction$ and $P left parenthesis stack A with bar on top right parenthesis equals fraction numerator 1 over denominator 4 end fraction comma text where end text stack A with bar on top$ stands for complement of event A. Then events A and B are-

A random variable X has the probability distribution :

For the events E = {X is a prime number} and F = {X < 4}, the probability $P left parenthesis E union F right parenthesis$ is-

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parallel

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Statement - I A,B,C are independent events.
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