The solution of the differential equation dy/dx = (x^2+xy+y^2)/x^2 for x>0, y>0 at x=1, y=2 is Option 1: (y/x) = _e x + 2 Option 2: ^(-1)(x/y) = _e x + ^(-1)2 Option 3: ^(-1)y/x = _e x + ^(-1)2 Option 4: ^(-1)y/x = _e x - ^(-1)2

Answer: ^(-1)y/x = _e x + ^(-1)2. Solution: The equation is homogeneous. Putting y=vx gives v+xdv/dx=1+v+v^2, so dv/(1+v^2)=dx/x and ^(-1)v=_e x+C. Using x=1,y=2 gives C=^(-1)2, hence ^(-1)y/x=_e x+^(-1)2.

Calculus

Differential Equations

Medium

core

The solution of the differential equation $\dfrac{dy}{dx} = \dfrac{x^2+xy+y^2}{x^2}$ for $x>0, y>0$ at $x=1, y=2$ is

$\tan\left(\dfrac{y}{x}\right) = \log_e x + \tan 2$

$\tan^{-1}\left(\dfrac{x}{y}\right) = \log_e x + \tan^{-1}2$

$\tan^{-1}\dfrac{y}{x} = \log_e x + \tan^{-1}2$

$\tan^{-1}\dfrac{y}{x} = \log_e x - \tan^{-1}2$

✅ Correct Option: 3

The equation is homogeneous. Putting $y=vx$ gives $v+x\frac{dv}{dx}=1+v+v^2$ , so $\frac{dv}{1+v^2}=\frac{dx}{x}$ and $\tan^{-1}v=\log_e x+C$ . Using $x=1,y=2$ gives $C=\tan^{-1}2$ , hence $\tan^{-1}\frac{y}{x}=\log_e x+\tan^{-1}2$ .

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Match List-I with List-II

List-I	List-II
(A) $y + \frac{dy}{dx} = \frac{1}{4}\int y\,dx$	(I) Order 1, Degree 1
(B) $y = \frac{dy}{dx} + \frac{C}{dy/dx}$	(II) Order 2, Degree 2
(C) $(xy^2+x)dx + (y-x^2)dy = 0$	(III) Order 2, Degree 1
(D) $\left[\left\{1+\left(\frac{dy}{dx}\right)^2\right\}^{3/2}/\frac{d^2y}{dx^2}\right] = K$	(IV) Order 1, Degree 2

Choose the correct answer from the options given below:

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The solution of the differential equation $\log_e\left(\frac{dy}{dx}\right) = 5x + 2y$ is given by

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Solution of the differential equation $(x + xy)dy - y(1 - x^2)dx = 0$ is

The solution of the differential equation dydx=x2+xy+y2x2\dfrac{dy}{dx} = \dfrac{x^2+xy+y^2}{x^2}dxdy​=x2x2+xy+y2​ for x>0,y>0x>0, y>0x>0,y>0 at x=1,y=2x=1, y=2x=1,y=2 is

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The solution of the differential equation $\dfrac{dy}{dx} = \dfrac{x^2+xy+y^2}{x^2}$ for $x>0, y>0$ at $x=1, y=2$ is