Verify that y = − x − 1 is a solution of the differential equation (y − x) dy − (y² − x²) dx = 0.

Open in App

Solution

We have,
$y = - x - 1$ ...(1)
Differentiating both sides of (1) with respect to $x$ , we get
$\frac{d y}{d x} = - 1$ ...(2)
Now,
$\frac{d y}{d x} - \frac{y^{2} - x^{2}}{y - x} = \frac{d y}{d x} - (y + x) = - 1 - (- x - 1 + x) [Using (1) and (2)] = - 1 + 1 = 0 \Rightarrow \frac{d y}{d x} = \frac{y^{2} - x^{2}}{y - x} \Rightarrow (y - x) d y = (y^{2} - x^{2}) d x \Rightarrow (y - x) d y - (y^{2} - x^{2}) d x = 0$
Hence, the given function is the solution to the given differential equation.

Suggest Corrections

Similar questions

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

\frac{d^{2} y}{d x^{2}} + 9 y = 0

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

y = c e^{\tan^{- 1} x}

(1 + x^{2}) \frac{d^{2} y}{d x^{2}} + (2 x - 1) \frac{d y}{d x} = 0

y = c x + \frac{1}{x}

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

Join BYJU'S Learning Program