Differential Equations
Differential Equations
star_batch_jee_advanced_2025
Grade 12
Question:
The solution of the differential equation $\frac{dy}{dx} = -\frac{1}{xy(x^2 \sin y^2 + 1)}$ is :
x^2(\cos y^2 - \sin y^2 - 2Ce^{-x^2}) = 2
y^2(\cos x^2 - \sin y^2 - 2Ce^{-x^2}) = 2
x^2(\cos y^2 - \sin y^2 - e^{-x^2}) = 4C
None of these
Step-by-Step Solution
Key Concept: A Bernoulli equation is linearized by an appropriate substitution that eliminates nonlinear terms.
Rewrite as $\frac{dx}{dy} = xy[x^2\sin y^2 + 1]$, then $\frac{1}{x^3}\frac{dx}{dy} - \frac{y}{x^2} = \sin y^2$. Using substitution $u = -1/x^2$ gives $\frac{du}{dy} + 2uy = 2y\sin y^2$. The integrating factor is $e^{y^2}$, so $ue^{y^2} = \int 2y\sin y^2 e^{y^2}dy = \frac{1}{2}e^{y^2}(\sin y^2 - \cos y^2) + C$. Thus $2u = (\sin y^2 - \cos y^2) + Ce^{-y^2}$, giving $2x^2[\cos y^2 - \sin y^2 - 2Ce^{-y^2}] = 1$.
Correct Answer: 1