Solve the differential equation 2(y + 3)

Question 3:- Solve the differential equation $2(y + 3) - xy\dfrac{dy}{dx} = 0$ ; given y(1) = 2. [CBSE 2025]

Solution:- Given, $2(y + 3) - xy\dfrac{dy}{dx} = 0$

⇒ $- xy\dfrac{dy}{dx} = -2(y + 3)$

⇒ $xy\dfrac{dy}{dx} = 2(y + 3)$

⇒ $y\dfrac{dy}{y + 3} = 2\dfrac{dx}{x}$

Integrate both side

⇒ $\displaystyle \int \dfrac{ydy}{y + 3} = 2\int \dfrac{dx}{x}$

⇒ $\displaystyle \int (y + 3) - 3\dfrac{dy}{y + 3} = 2\int \dfrac{dx}{x}$

⇒ $\displaystyle \int \left[1 - \dfrac{3}{(y + 3)}\right]\dfrac{dy}{y + 3} = 2\int \dfrac{dx}{x}$

⇒ $\left[y - \log (y+3)\right] = 2 \log x + C$ . . . (i)

Given y(1) = 2

$\left[2 - \log(2 + 1)\right] = 2\log 1 + C$

⇒ $2 - \log 3 = 0 + C$

⇒ $2 - \log 3 = C$

Hence, from equation (i)

$y - \log (y + 3) = 2\log x + 2 - \log 3$

⇒ $y - 2 = \log (y + 3) + 2\log x -\log 3$

⇒ $y - 2 = \log (y + 3) + \log x^2 -\log 3$

⇒ $y - 2 = \log \dfrac{x^2(y + 3)}{3}$

⇒ $\dfrac{x^2(y + 3)}{3} = e^{(y - 2)}$

⇒ $x^2 (y + 3) = 3 e^{(y - 2)}$

Question 1:- Using integration, find the area of the region bounded by the line y = 5x + 2, the x- axis and the ordinates x = -2 and x = 2. [CBSE 2025]

Solution:- See full solution

Question 2:- Solve the following differential equation :

$(1+x^2)\dfrac{dy}{dx} + 2xy = 4x^2$ .

Solution:- See full solution

Trigonometry Formulae class 11

Solve the differential equation 2(y + 3) – xydy/dx = 0

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