The degree of differential equation of which $y^{2} = 4 a (x + a)$ is a solution is

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

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

Open in App

Solution

The correct option is A 2
$y^{2} = 4 a x + 4 a^{2}$ .........(1)
Differentiate w.r.t $x \Rightarrow$
$2 y \frac{d y}{d x} = 4 a \Rightarrow y \frac{d y}{d x} = 2 a$
$\frac{d y}{d x} = \frac{2 a}{y} \Rightarrow$ equation (1) is not a solution of this Differentiate equation
$y y^{1} = 2 a$
$\sqrt{4 a (x + a)} y^{1} = 2 a$
$\Rightarrow 4 a (x + a) (y^{1})^{2} = 4 a^{2}$
$\Rightarrow (x + a) (y^{1})^{2} = a$
Equation (1) is a solution of this Differential equation
and for this Degree $= 2$

Suggest Corrections

Similar questions

y^{2} = 4 a (x - a)

a

\frac{d y}{d x} + a \frac{d x}{d y}

y \{1 - {(\frac{d y}{d x})}^{2}\} = 2 x \frac{d y}{d x}

y^{2} = 4 a (x + a)

y \times [1 - {(\frac{d y}{d x})}^{2}] = 2 x \frac{d y}{d x}

y^{2} = 4 a (x + a)

Join BYJU'S Learning Program