If x^(p) y^(q) = (x + y)^((p + q)) " the...

If `x^(p) y^(q) = (x + y)^((p + q)) " then " (dy)/(dx)=` ?

Text Solution

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`x^(p).y^(q)=(x+y)^(p+q)`,
`therefore log(x^(p).y^(q))=log(x+y)^(p+q)`
`thereforelogx^(p)+logy^(q)=log(x+y)^(p+q)`
`therefore plogx+qlogy=(p+q)log(x+y)`
Differentiating both sides w.r.t. x, we get
`pxx(1)/(x)+qxx(1)/(y).(dy)/(dx)=(p+q)xx(1)/(x+y).(d)/(dx)(x+y)`
`therefore (p)/(x)+(q)/(y).(dy)/(dx)=(p+q)/(x+y)(1+(dy)/(dx))`
`therefore(p)/(x)+(q)/(y).(dy)/(dx)=(p+q)/(x+y)+(p+q)/(x+y).(dy)/(dx)`
`therefore((q)/(y)-(p+q)/(x+y))(dy)/(dx)=(p+q)/(x+y)-(p)/(x)`
`therefore[(qx+qy-py-qy)/(y(x+y))](dy)/(dx)=(px+qx-px-py)/(x(x+y))`
`therefore[(qx-py)/(y(x+y))](dy)/(dx)=(qx-py)/(x(x+y))`
`therefore(1)/(y).(dy)/(dx)=(1)/(x)`
`therefore(dy)/(dx)=(y)/(x)`

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If y=(1)/(1+^(q-p) +x^(r-p) ) +(1)/( 1+x^(r-p) + x^(p-q)) + (1)/( 1+x^(p-r) +x^(q-r)) then (dy)/(dx)=

`(p+q+r) x^(p+q+r)`

None

If y = (1)/(1+x^( p-r) +x^(q-r)) +(1)/( 1+ x^(p-q) +x^(r-q) )+ (1) /( 1+x^(q-p) +x^(r-p)),then (dy)/(dx) =

` x^(p+q+r)`

` x^(p+q+r-1)`

If the integrating factor of the differential equation (dy)/(dx) +P (x) y = Q (x) is x , then P (x) is

`x^(2)//2`

`1//x`

`1//x^(2)`