Kindly Consider the following 25. xa+yb=1

Given: Equation of the family of curves xa+yb=1.......... (i) Differentiating both sides of the given equation with respect to x, we get: 1a+1bdydx=0⇒1a+1by'=0 Again, differentiating both sides with respect to x, we get: 0+1by"=0⇒1by"=0⇒y"=0 Hence, the required differential equation of the given curve is y"=0

Kindly Consider the following

25. $\frac{x}{a} + \frac{y}{b} = 1$

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Vishal Baghel | Contributor-Level 10

8 months ago

Given: Equation of the family of curves $\frac{x}{a} + \frac{y}{b} = 1 . . . . . . . . . . (i)$

Differentiating both sides of the given equation with respect to x, we get:

$\begin{array}{l} \frac{1}{a} + \frac{1}{b} \frac{d y}{d x} = 0 \\ \Rightarrow \frac{1}{a} + \frac{1}{b} y^{'} = 0 \end{array}$

Again, differentiating both sides with respect to x, we get:

$\begin{array}{l} 0 + \frac{1}{b} y^{"} = 0 \\ \Rightarrow \frac{1}{b} y^{"} = 0 \\ \Rightarrow y^{"} = 0 \end{array}$

Hence, the required differential equation of the given curve is $y^{"} = 0$

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dy/dx = 2y/ (xlnx).
dy/y = 2dx/ (xlnx).
ln|y| = 2ln|lnx| + C.
ln|y| = ln (lnx)²) + C.
y = A (lnx)².
(ln2)² = A (ln2)². ⇒ A=1.
y = f (x) = (lnx)².
f (e) = (lne)² = 1² = 1.