- \(y = x + C\)
- \(y = \ln(e^x + C)\)
- \(y = e^x + C\)
- \(y = -\ln(e^x + C)\)
Answer: (b) \(y = \ln(e^x + C)\)
- \(\frac{1}{x^2}\)
- \(\frac{1}{y^2}\)
- \(\frac{1}{x^4}\)
- \(\frac{1}{y^4}\)
Answer: (d) \(\frac{1}{y^4}\)
- \(e^{\tan^{-1} y}\)
- \(e^{y}\)
- \(e^{x}\)
- \(e^{\ln y}\)
Answer: (a) \(e^{\tan^{-1} y}\)
- \(\frac{dy}{dx} = \frac{y}{x}\)
- \(\frac{dy}{dx} = -\frac{y}{x}\)
- \(\frac{dy}{dx} = \frac{x}{y}\)
- \(\frac{dy}{dx} = -\frac{x}{y}\)
Answer: (c) \(\frac{dy}{dx} = \frac{x}{y}\)
- \(\frac{dy}{dx} = y \cos x\)
- \(\frac{dy}{dx} = y \sin x\)
- \(\frac{dy}{dx} = -y \cos x\)
- \(\frac{dy}{dx} = -y \sin x\)
Answer: (a) \(\frac{dy}{dx} = y \cos x\)
- \(e^x\)
- \(e^{-x}\)
- \(e^y\)
- \(e^{-y}\)
Answer: (a) \(e^x\)
- \(\alpha = 2, \beta = 1\)
- \(\alpha = 1, \beta = 2\)
- \(\alpha = 2, \beta = -1\)
- \(\alpha = -2, \beta = 1\)
Answer: (a) \(\alpha = 2, \beta = 1\)
- \(y = \frac{C}{x}\)
- \(y = \frac{x}{C}\)
- \(y = Cx^2\)
- \(y = \frac{C}{x^2}\)
Answer: (b) \(y = \frac{x}{C}\)
- \(y = e^{-x}e^{e^x} + C e^{-x}\)
- \(y = e^{-x}e^{e^x} + C e^{x}\)
- \(y = e^{-x}e^{e^x} + C e^{-e^x}\)
- \(y = e^{-x}e^{e^x} + C e^{e^x}\)
Answer: (a) \(y = e^{-x}e^{e^x} + C e^{-x}\)
- \(x\)
- \(\frac{1}{x}\)
- \(x^2\)
- \(\frac{1}{x^2}\)
Answer: (b) \(\frac{1}{x}\)
- \(u = \sin y\)
- \(u = \cos y\)
- \(u = \tan y\)
- \(u = \sec y\)
Answer: (a) \(u = \sin y\)
- \(\frac{d^2y}{dx^2} - y = 0\)
- \(\frac{d^2y}{dx^2} + y = 0\)
- \(\frac{d^2y}{dx^2} - 2y = 0\)
- \(\frac{d^2y}{dx^2} + 2y = 0\)
Answer: (a) \(\frac{d^2y}{dx^2} - y = 0\)
- \(\log x\)
- \(x\)
- \(\frac{1}{\log x}\)
- \(\frac{1}{x}\)
Answer: (a) \(\log x\)
- \(e^{x^2}\)
- \(e^{-x^2}\)
- \(e^{2x^2}\)
- \(e^{-2x^2}\)
Answer: (a) \(e^{x^2}\)
- \(f(x, y) = \sqrt{x} + \sqrt{y}\) is a homogeneous function of degree \(\sqrt{2}\)
- \(y dx - x dy = 0\) is an exact differential equation
- The family of con-focal and coaxial parabolas \(y^2 = 4a(x + a)\) is self-orthogonal
- The number of arbitrary constants present in the general solution of \(y''' - 3y'' + 3y' - 1 = e^x\) is 2
Answer: (c) The family of con-focal and coaxial parabolas \(y^2 = 4a(x + a)\) is self-orthogonal
- \(e^{\tan x}\)
- \(e^{\sec x}\)
- \(e^{\sin x}\)
- \(e^{\cos x}\)
Answer: (a) \(e^{\tan x}\)
- 4 times
- 6 times
- 8 times
- 10 times
Answer: (c) 8 times
- \(\frac{dy}{dx} = \frac{y}{x}\)
- \(\frac{dy}{dx} = -\frac{y}{x}\)
- \(\frac{dy}{dx} = \frac{x}{y}\)
- \(\frac{dy}{dx} = -\frac{x}{y}\)
Answer: (a) \(\frac{dy}{dx} = \frac{y}{x}\)
- \(\tan x \tan y = C\)
- \(\sec x \sec y = C\)
- \(\tan x + \tan y = C\)
- \(\sec x + \sec y = C\)
Answer: (a) \(\tan x \tan y = C\)
- \(\tan y - \sec y\)
- \(\cot y+\sec y\)
- \(\sec y - \cot y\)
- \(\sec y + \tan y\)
Answer: (d) \(\sec y + \tan y\)
- \(\frac{1}{2}\)
- \(\frac{-1}{2}\)
- \(\frac{1}{4}\)
- \(\frac{1}{4}\)
Answer: (a) \(\frac{1}{2}\)
- \(\frac{1}{2}\)
- \(\frac{-1}{2}\)
- \(\frac{1}{4}\)
- \(\frac{-1}{4}\)
Answer: (b) \(\frac{-1}{2}\)
- \(y = -\frac{x}{a}\)
- \(y = -\frac{x}{a} + C\)
- \(x^2 + y^2 = C\)
- \(y = Cx^2\)
Answer: (c) \(x^2 + y^2 = C\)
- \(y = -\frac{1}{2}\log x+C\)
- \(y = \frac{1}{4} \log x+C\)
- \(y = -\frac{1}{4} \log x+C\)
- \(y = \frac{1}{2} \log x+C\)
Answer: (a) \(y = -\frac{1}{2}\log x+C\)
- If \(Mdx + Ndy = 0\) is a homogeneous equation, then \((Mx - Ny)^{-1}\) is an integrating factor.
- \(Mdx + Ndy = 0\) is an exact differential equation if \(\frac{\partial N}{\partial y} = \frac{\partial M}{\partial x}\).
- The mathematical formulation for the law of natural growth is \(\frac{dx}{dt} = kx, \, k > 0\).
- \(\frac{dy}{dx} - y \tan x = y^2 \sec^2 x\) is a linear equation in \(y\).
Answer: (c) The mathematical formulation for the law of natural growth is \(\frac{dx}{dt} = kx, \, k > 0\)
- \(\frac{dT}{dt} = -k(T - T_s)\)
- \(\frac{dT}{dt} = k(T + T_s)\)
- \(\frac{dT}{dt} = kT(T - T_s)\)
- \(\frac{dT}{dt} = kT(T + T_s)\)
Answer: (a) \(\frac{dT}{dt} = -k(T - T_s)\)
- \(y = \frac{dy}{dx}x + \left(\frac{dy}{dx}\right)^2\)
- \(y = \frac{dy}{dx}x - \left(\frac{dy}{dx}\right)^2\)
- \(y = \frac{dy}{dx}x + \frac{dy}{dx}\)
- \(y = \frac{dy}{dx}x - \frac{dy}{dx}\)
Answer: (a) \(y = \frac{dy}{dx}x + \left(\frac{dy}{dx}\right)^2\)
- \(\frac{x^2}{a^2 - \lambda} + \frac{y^2}{b^2 - \lambda} = 1\)
- \(\frac{x^2}{a^2 + \lambda} + \frac{y^2}{b^2 + \lambda} = 1\)
- \(\frac{x^2}{a^2 - \lambda} - \frac{y^2}{b^2 - \lambda} = 1\)
- \(\frac{x^2}{a^2 + \lambda} + \frac{y^2}{b^2 - \lambda} = 1\)
Answer: (b) \(\frac{x^2}{a^2 + \lambda} + \frac{y^2}{b^2 + \lambda} = 1\)
- \(\frac{1}{2}\log y = -x + C\)
- \(\frac{1}{2}\log y = x + C\)
- \(\frac{1}{2}\log y = 2x + C\)
- \(\frac{1}{2}\log y = -2x + C\)
Answer: (a) \(\frac{1}{2}\log y = -x + C\)
- \(\frac{1}{x}\)
- \(\frac{1}{x^2}\)
- \(\frac{-1}{x}\)
- \(\frac{-1}{x^2}\)
Answer: (b) \(\frac{1}{x^2}\)
- \(p = 3\)
- \(p = \frac{1}{\sqrt{3}}\)
- \(p = \sqrt{3}\)
- \(p = -\sqrt{3}\)
Answer: (c) \(p = \sqrt{3}\)
- \(e^x\)
- \(e^{-x}\)
- \(e^{x^2}\)
- \(e^{-x^2}\)
Answer: (c) \(e^{x^2}\)
- 1
- 2
- 3
- 4
Answer: (c) 3
- \(\frac{1}{x^3}\)
- \(\frac{1}{y^2}\)
- \(\frac{1}{x^2y}\)
- \(\frac{-1}{x^2 y^2}\)
Answer: (b) \(\frac{1}{y^2}\)
- \(y = e^{-x^2}(x + C)\)
- \(y = e^{x^2}(x + C)\)
- \(y = e^{-x^2}(x^2 + C)\)
- \(y = e^{x^2}(x^2 + C)\)
Answer: (a) \(y = e^{-x^2}(x + C)\)
- \(\frac{-1}{xy\cos xy}\)
- \(\frac{-1}{2xy\cos xy}\)
- \(\frac{1}{2xy\cos xy}\)
- \(\frac{1}{xy\cos xy}\)
Answer: (c) \(\frac{1}{2xy\cos xy}\)
- 79.24 years
- 80.24 years
- 85.24 years
- 78.24 years
Answer: (a) 79.24 years
- 1
- 2
- 3
- 4
Answer: (d) 4
- \(3e^{2x}+2e^{y} = C\)
- \(3e^{2x}+2e^{3y} = C\)
- \(3e^{x}+2e^{3y} = C\)
- \(3e^{2x}+2e^{y} = C\)
Answer: (b) \(3e^{2x}+2e^{3y} = C\)
- \(Ce^{2x}\)
- \(Ce^{-2x}\)
- \(Cx^2\)
- \(Cx^{-2}\)
Answer: (a) \(Ce^{2x}\)
- \(\frac{\partial \phi}{\partial y} = \frac{\partial \psi}{\partial x}\)
- \(\frac{\partial \phi}{\partial x} = \frac{\partial \psi}{\partial y}\)
- \(\frac{\partial \phi}{\partial y} = -\frac{\partial \psi}{\partial x}\)
- \(\frac{\partial \phi}{\partial x} = -\frac{\partial \psi}{\partial y}\)
Answer: (a) \(\frac{\partial \phi}{\partial y} = \frac{\partial \psi}{\partial x}\)
- \(\lambda = 2\)
- \(\lambda = 6\)
- \(\lambda = 3\)
- \(\lambda = 4\)
Answer: (c) \(\lambda = 3\)
- \(y^2 = -4a(x + a)\)
- \(y^2 = 4a(x - a)\)
- \(y^2 = -4a(x - a)\)
- \(y^2 = 4a(x + a)\)
Answer: (d) \(y^2 = 4a(x + a)\)
- 500 grams
- 625 grams
- 750 grams
- 875 grams
Answer: (b) 625 grams
- 300K
- 310K
- 320K
- 330K
Answer: (b) 310K
- \(\frac{1}{y}\)
- \(\frac{1}{y^2}\)
- \(\frac{1}{y^3}\)
- \(\frac{1}{y^4}\)
Answer: (a) \(\frac{1}{y}\)
- \(\frac{x}{y} + e^{x^3}+C\)
- \(\frac{y}{x} + e^{-x^3}+C\)
- \(\frac{x}{y} + e^{-x^3}+C\)
- \(\frac{y}{x} + e^{x^3}+C\)
Answer: (a) \(\frac{x}{y} + e^{x^3}+C\)
- \(\left(2x + e^x \log y\right)dx + \left(\frac{e^x}{y} + 1\right)dy = 0\)
- \(\left(y + \log x\right)dx - x dy = 0\)
- \(\left(ax + hy + g\right)dx + \left(hx + by + f\right)dy = 0\)
- \(\left(e^y + 1\right)\cos x dx + e^y \sin x dy = 0\)
Answer: (b) \(\left(y + \log x\right)dx - x dy = 0\)
- \(e^{kx}\)
- \(e^{ky}\)
- \(e^{kxy}\)
- \(e^{k(x + y)}\)
Answer: (a) \(e^{kx}\)