Question

Consider the differential equation $y=px+\sqrt{a^2{p}^2+b^2}$,where $p=\frac{dy}{dx}$. The order and degree of the differential equation are

• A. 1, 1
• B. 1, 2
• C. 2, 1
• D. None of these

Answer 1

Answer: (B)

1, 2

$y=px+\sqrt{a^2{p}^2{b}^2}$

${(y-px)}^2=a^2{p}^2+b^2$

$y^2+p^2{x}^2-2pxy=a^2{p}^2+b^2$

$y^2+x^2{\left(\frac{dy}{dx}\right)}^2-2\frac{dy}{dx}xy=a^2{p}^2+b^2$

$y^2+x^2{\left(\frac{dy}{dx}\right)}^2-2\frac{dy}{dx}xy=a^2{\left(\frac{dy}{dx}\right)}^2+b^2$

$x^2{\left(\frac{dy}{dx}\right)}^2-2xy\frac{dy}{dx}-a^2{\left(\frac{dy}{dx}\right)}^2-b^2+y^2=0$

$Order=1$

$Degree=2$