Differential Equations
Differential Equations
star_batch_jee_advanced_2025
Grade 12
Question:
The solution of $\frac{dy}{dx} + x = xe^{(n-1)y}$ is:
$\frac{1}{n-1}\log\left(\frac{e^{(n-1)y}-1}{e^{(n-1)y}}\right) = \frac{x^2}{2} + C$
$e^{(n-1)y} = Ce^{(n-1)y+(n-1)^2/2} + 1$
$\log\left(\frac{e^{(n-1)y}-1}{(n-1)e^{(n-1)y}}\right) = x^2 + C$
$e^{(n-1)y} = ce^{(n-1)x^2/2} + 1$
Step-by-Step Solution
Key Concept: Use substitution $u = e^{(n-1)y}$ to convert the differential equation into a separable form, then apply partial fractions.
Rewriting $\frac{dy}{dx} = x(e^{n-1}y - 1)$ and rearranging gives $\frac{dy}{e^{(n-1)y}-1} = xdx$. Substituting $u = e^{(n-1)y}$ transforms this to $\frac{1}{n-1}\int\frac{du}{u(u-1)} = \frac{x^2}{2} + C$. Using partial fractions and integrating: $\frac{1}{n-1}\log\frac{u-1}{u} = \frac{x^2}{2} + C$, which yields the final solution $e^{(n-1)y} = Ce^{(n-1)y+n-1}x^2/2 + 1$.
Correct Answer: 1,2