$y = c x - c^{2}$ is the general solution of the differential equation ( $c$ -arbitrary constant)

{(y^{'})}^{2} - x y^{'} + y = 0

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

y^{''} = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

y^{'} = c

No worries! We‘ve got your back. Try BYJU‘S free classes today!

{(y^{'})}^{2} + y^{'} - x y = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A ${(y^{'})}^{2} - x y^{'} + y = 0$
Given curve is, $y = c x - c^{2}$ ......(1)
Differentiate the above curve,
$y^{'} = c - 0 = c$
Substitute $c = y^{'}$ in (1)
$\Rightarrow y = x y^{'} - (y^{'})^{2}$
$\Rightarrow (y^{'})^{2} - x y^{'} + y = 0$

Suggest Corrections

Similar questions

x^{2} d y - y^{2} d x - x y^{2} (x - y) d y = 0

y = \frac{x}{l n | c x |}

\frac{d y}{d x} = \frac{y}{x} + ϕ (\frac{x}{y})

ϕ (\frac{x}{y})

The general solution of the differential equation $\frac{y d x - x d y}{y} = 0$ is

a) $x y = C$

b) $x = C y^{2}$

c) $y = C x$

d) $y = C x^{2}$

y = c x + c - c^{2}

(y^{2} - x^{3}) d x - x y d y = 0 (x \neq 0)

c

Join BYJU'S Learning Program