Question

If a curve y = f(x) passes through the point (1, –1) and satisfies the differential equation, y(1 + xy) dx = x dy, then find the value of $\small\tt{\frac{-1}{2}$​

Answer 1

Step-by-step explanation:

We have,

$y(1 + xy) \: dx = x \: dy$

$\implies \: \frac{dy}{dx} = \frac{y(1 + xy) }{x}$

Let $\:y=vx$

$\implies\:\frac{dy}{dx}=v+x\frac{dv}{dx}$

So,

$\implies \: v + x\frac{dv}{dx} = \frac{vx(1 + x(vx)) }{x}$

$\implies \: v + x\frac{dv}{dx} = v(1 + vx^{2} )$

$\implies \: x\frac{dv}{dx} = v + v ^{2} x^{2} - v$

$\implies \: x\frac{dv}{dx} = v ^{2} x^{2}$

$\implies \: \frac{dv}{dx} = v ^{2} x$

$\implies \: \frac{dv}{ {v}^{2} } = x \: dx$

$\implies \: \int \frac{dv}{ {v}^{2} } = \int x \: dx$

$\implies \: - \frac{1}{ {v} } = \frac{ {x}^{2} }{2} +C$

$\implies \: - \frac{x}{y } = \frac{ {x}^{2} }{2} +C$

It passes through $(1,-1)$

$\implies \: - \frac{(1)}{( - 1) } = \frac{ {(1)}^{2} }{2} +C$

$\implies \: 1= \frac{1 }{2} +C$

$\implies \: C = \frac{1}{2}$

So, required curve,

$\implies \: - \frac{x}{y } = \frac{ {x}^{2} }{2} + \frac{1}{2}$

$\implies \: - \frac{x}{y } = \frac{ {x}^{2} + 1 }{2}$

$\implies \: - \frac{2x}{ {x}^{2} + 1 } = y$

$\implies \: y =f(x) = - \frac{2x}{ {x}^{2} + 1 }$

So,

$f \bigg(-\frac{1}{2} \bigg) = - \frac{2 \times \frac{1}{2} }{( \frac{1}{2} )^{2} + 1}$

$\implies \: f \bigg(-\frac{1}{2} \bigg) = - \frac{1}{\frac{1}{4} + 1}$

$\implies \: f \bigg(-\frac{1}{2} \bigg) = - \frac{1}{\frac{1 + 4}{4} }$

$\implies \: f \bigg(-\frac{1}{2} \bigg) = - \frac{4}{1 + 4 }$

$\implies \: f \bigg(-\frac{1}{2} \bigg) = - \frac{4}{5}$

Answer 2

Answer:

if a body starts from rest what can be said about the acceleration of body

Step-by-step explanation:

Sorry but can I know how you are able to give multiple thanks? (-^〇^-)