The general solution of the differential equation ydx - (x + 2y^2)dy = 0 Option 1: x/y = 2x + C, Where C is constant of integration Option 2: y = 2x^2 + C, Where C is constant of integration Option 3: y/x = 2y + C, Where C is constant of integration Option 4: x = 2y^2 + Cy, Where C is constant of integration

Answer: x = 2y^2 + Cy, Where C is constant of integration. Solution: Given differential equation: ydx - (x + 2y^2)dy = 0 ydx = (x + 2y^2)dy dx/dy = (x + 2y^2)/y dx/dy = x/y + 2y dx/dy - x/y = 2y This is in the standard linear form dx/dy + P(y) * x = Q(y) where P(y) = -1/y and Q(y) = 2y The integrating factor is: IF = e^( P(y)dy) IF = e^( -1/ydy) IF = e^(- y) IF = e^( y^(-1)) IF = 1/y 1/y * dx/dy - 1/y * x/y = 1/y * 2y 1/ydx/dy - x/y^2 = 2 The left side is the derivative of x/y: d/dy(x/y) = 2 d/dy(x/y)dy = 2 dy x/y = 2y + C x = 2y^2 + Cy Therefore, the general solution is x = 2y^2 + Cy where C is the constant of integration.

Calculus

Differential Equations

Medium

core

The general solution of the differential equation $ydx - (x + 2y^2)dy = 0$

$\frac{x}{y} = 2x + C$ , Where $C$ is constant of integration

$y = 2x^2 + C$ , Where $C$ is constant of integration

$\frac{y}{x} = 2y + C$ , Where $C$ is constant of integration

$x = 2y^2 + Cy$ , Where $C$ is constant of integration

✅ Correct Option: 4

Given differential equation: $ydx - (x + 2y^2)dy = 0$

$ydx = (x + 2y^2)dy$

$\frac{dx}{dy} = \frac{x + 2y^2}{y}$

$\frac{dx}{dy} = \frac{x}{y} + 2y$

$\frac{dx}{dy} - \frac{x}{y} = 2y$

This is in the standard linear form $\frac{dx}{dy} + P(y) \cdot x = Q(y)$ where $P(y) = -\frac{1}{y}$ and $Q(y) = 2y$

The integrating factor is:

$IF = e^{\int P(y)dy}$

$IF = e^{\int -\frac{1}{y}dy}$

$IF = e^{-\ln y}$

$IF = e^{\ln y^{-1}}$

$IF = \frac{1}{y}$

$\frac{1}{y} \cdot \frac{dx}{dy} - \frac{1}{y} \cdot \frac{x}{y} = \frac{1}{y} \cdot 2y$

$\frac{1}{y}\frac{dx}{dy} - \frac{x}{y^2} = 2$

The left side is the derivative of $\frac{x}{y}$ :

$\frac{d}{dy}\left(\frac{x}{y}\right) = 2$

$\int \frac{d}{dy}\left(\frac{x}{y}\right)dy = \int 2 \, dy$

$\frac{x}{y} = 2y + C$

$x = 2y^2 + Cy$

Therefore, the general solution is $x = 2y^2 + Cy$ where $C$ is the constant of integration.

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The solution of the differential equation $\dfrac{dy}{dx} = 2^{x+y}$ is

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In the following differential equation $\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 = 2x^2 \log\left(\frac{d^2y}{dx^2}\right)$ order and degree is:

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Match List-I with List-II

List-I	List-II
Differential equation	Order and degree
(A) $y''' + x^2(y'')^3 + y = 0$	(I) $3, 2$
(B) $y''' + y^2 + e^{y'} = 0$	(II) $3, 3$
(C) $(y''')^2 + x^2(y'')^3 + xy = 0$	(III) $3, 1$
(D) $(y''')^3 + x(y'')^2 + y = 0$	(IV) $3, \text{Not defined}$

Choose the correct answer from the options given below:

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The general solution of the differential equation $x^2\dfrac{dy}{dx} + y^2 = xy$ is: (take $\log_e x = \log x$ )

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Choose the correct statements:

A. The order and degree (if defined) of a differential equation are always positive integrals

B. The order of a differential equation is the highest order derivative of the dependent variable with respect to the independent variable involved in a differential equation

C. If $\frac{dy}{dx} + P(x)y = Q(x)$ then Integrating factor $= e^{\int P(x)dx}$

D. The sum of order and degree of differential equation $1 + (y'')^5 = (y''')^3$ is $8$

E. If the solution of a differential equation of order $n$ , contains $n$ arbitrary constant, then it is called a general solution

Choose the correct answer from the options given below:

The general solution of the differential equation ydx−(x+2y2)dy=0ydx - (x + 2y^2)dy = 0ydx−(x+2y2)dy=0

Solution

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The general solution of the differential equation $ydx - (x + 2y^2)dy = 0$