Maths · Ordinary differential equations, their order and degree
The order, degree of the differential equation satisfying the relation \) \) is
The order, degree of the differential equation satisfying the relation \( \sqrt{1+x^{2}}+\sqrt{1+y^{2}}=\lambda(x \sqrt{1+y^{2}}) \) \( \left.y \sqrt{1}+x^{2}\right) \) is
- A. 1,1
- B. 2,
- C. 3,2
- D. 0,1
Step-by-step solution
The given relation simplifies to tan⁻¹x + tan⁻¹y = constant after trigonometric substitution. Differentiating gives y' = -(1+x²)/(1+y²), which is a first-order, first-degree differential equation. Thus order = 1 and degree = 1.
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