Differential Equations
Differential Equations
star_batch_jee_advanced_2025
Grade 12
Question:
If $\frac{dy}{dx} = \frac{x^2-y}{x+y}$ such that $y = f(x)$ is a solution of differential equation & $f(0) = 0$, then:
$(f(1)+1)^2 = \frac{5}{3}$
$(f(-1)-1)^2 = \frac{1}{3}$
$(f(3)+3)^{2/3} = 3$
$|f(-3)-3| = 3$
Step-by-Step Solution
Key Concept: Substitution of the form $u = x + y$ can linearize or simplify differential equations involving sums of variables.
Given $\frac{dy}{dx} + 1 = \frac{x^2 - y + x + y}{x+y}$, we simplify to $\frac{dy}{dx} = \frac{x^2 - y + x + y - (x+y)}{x+y} = \frac{x^2 - y}{x+y}$. Let $u = x+y$, so $\frac{du}{dx} = 1 + \frac{dy}{dx}$ and $\frac{dy}{dx} = \frac{u^2 - x^2}{u}$. This leads to $\frac{du}{dx} = \frac{u^2 + x}{u}$ with solution $(x+y)^2 = \frac{2x^3}{3} + \frac{x^2}{2} + c$. Using boundary condition $x=0, y=0$ gives $c=0$.
Correct Answer: 1,2,3