Question

A solution of the differential equation $\frac{dy}{dx}$= $\frac{1}{xy\lbrack x^{2}\sin y^{2} + 1\rbrack}$is (C is an arbitrary constant) –

• A. x² (cos y² – sin y² – 2C $e^{–y^{2}}$) = 2
• B. y² (cos x² – (sin y² – 2C $e^{–y^{2}}$) = 2
• C. x² (cos y² – sin y² –$e^{–y^{2}}$) = 4
• D. None of these

Answer 1

Answer: (A)

x² (cos y² – sin y² – 2C $e^{–y^{2}}$) = 2

Answer 2

The given differential equation can be written as

$\frac{dx}{dy}$ = xy [x² sin y² + 1] Ž $\frac{1}{x^{3}}$ $\frac{dx}{dy}$ – $\frac{1}{x^{2}}$ y = y sin y² . This equation is reducible to linear equation, so putting – 1/x² = u, the last equation can be written as $\frac{du}{dy}$ + 2uy = 2y sin y²

The integrating factor of this equation is$e^{y^{2}}$. So required solution is u$e^{y^{2}}$ = $\int_{}^{}{2y}$sin y² .$e^{y^{2}}$dy + C

= $\int_{}^{}{(\sin t)}$e^t dt + C (t = y²) = (1/2)$e^{y^{2}}$ (sin y² – cos y²) + C

Ž 2u = (sin y² – cos y²) + C$e^{- y^{2}}$

Ž 2 = x² [cos y² – sin y² – 2 C$e^{- y^{2}}$]