Solve the following differential equation:
(x² − y²⁾ dx + 2xy dy = 0 given that y = 1 when x = 1

Solution

( x² - y² ) dx + 2xy dy = 0

$\Rightarrow \frac{dy}{dx} = \frac{y^{2} - x^{2}}{2 xy} . . . . . . . . . (1)$

It is a homogeneous differential equation.

Let y = vx ..........(2)

$∴ \frac{dy}{dx} = v + x \frac{dv}{dx} . . . . . . . . . . . (3)$

Substituting (2) and (3) in (1), we get:

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

Integrating both sides, we get:

$\int \frac{2 v}{v^{2} + 1} dv = - \int (\frac{1}{x}) dx \log |v^{2} + 1| = - \log |x| + \log C \log |v^{2} + 1| = \log |\frac{C}{x}| v^{2} + 1 = \frac{C}{x} x (v^{2} + 1) = C$

$x [{(\frac{y}{x})}^{2} + 1] = C y^{2} + x^{2} = Cx . . . . . . . . . (4)$

It is given that when x = 1, y = 1

(1)² + (2)² = C(1)

$\Rightarrow$ C = 2

Thus, the required solution is y² + x² = 2x.

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Differential Equations

Solve the following differential equation:
(x² − y²⁾ dx + 2xy dy = 0 given that y = 1 when x = 1

Some More Questions From Differential Equations Chapter

Mock Test Series

NCERT Sample Papers

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For Daily Free Study Material Join wiredfaculty

Differential Equations

Solve the following differential equation:(x2 − y2) dx + 2xy dy = 0 given that y = 1 when x = 1

Some More Questions From Differential Equations Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Solve the following differential equation:
(x² − y²⁾ dx + 2xy dy = 0 given that y = 1 when x = 1