Differential Equations
Differential Equations
star_batch_jee_advanced_2025
Grade None
Question:
Identify the statement(s) which is/are true?
The order of differential equation $\sqrt{1 + \frac{d^2y}{dx^2}} = x$ is 1.
solution of the differential equation $xdy - ydx = \sqrt{x^2 + y^2}dx$ is $y + \sqrt{x^2+y^2} = Cy^2$.
$\frac{d^2y}{dx^2} = 2\left(\frac{dy}{dx} - y\right)$ is differential equation of family of curves $y = e^x(A\cos x + B\sin x)$.
The solution of differential equation $(1+y^2)\left(x - 2xe^{\tan^{-1}y}\right)\frac{dy}{dx} = 0$ is $xe^{\tan^{-1}y} = e^{3\tan^{-1}y} + k$.
Step-by-Step Solution
Key Concept: Recognize equation order, apply appropriate integration techniques (exact equations, substitutions), verify solutions by differentiation, and use integrating factors for non-standard forms.
(A) The order of the differential equation is 2 since the highest derivative is $\frac{d^2y}{dx^2}$. (B) Rearranging $\frac{x\,dy - y\,dx}{\sqrt{x^2 + y^2}} = dx$ and manipulating yields $\ln|\frac{y}{x} + \sqrt{1 + \frac{y^2}{x^2}}| = \ln|Cx|$, which simplifies to $y + \sqrt{x^2 + y^2} = Cx^2$. (C) For $y = e^x(A\cos x + B\sin x)$, computing derivatives and verifying shows $\frac{d^2y}{dx^2} - \frac{dy}{dx} - y = e^x(-A\sin x + B\cos x) - y$. (D) Converting the equation $(1+y^2)\frac{dx}{dy} + x = 2e^{\tan^{-1}y}$ to standard form and solving with integrating factor yields $xe^{\tan^{-1}y} = e^{2\tan^{-1}y} + k$.
Correct Answer: 2,3