The solution of the differential equation fra | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given the differential equation: $\frac{dy}{dx} + \frac{x(x^2+y^2-1)}{y(2(x^2+y^2)+1)} = 0$ Rearranging the terms: $y(2(x^2+y^2)+1) dy + x(x^2+y^2

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Given the differential equation: $\frac{dy}{dx} + \frac{x(x^2+y^2-1)}{y(2(x^2+y^2)+1)} = 0$ Rearranging the terms: $y(2(x^2+y^2)+1) dy + x(x^2+y^2-1) dx = 0$ $2y(x^2+y^2) dy + y dy + x(x^2+y^2) dx - x dx = 0$ $(x^2+y^2)(2y dy + x dx) + y dy - x dx = 0$ Let $u = x^2+y^2$, then $du = 2x dx + 2y dy$. This can be rewritten as: $(x^2+y^2+2)(2y dy + x dx) - 3x dx = 0$ Dividing by $(x^2+y^2+2)$: $2y dy + x dx = \frac{3x dx}{x^2+y^2+2}$ This is not quite right, let's use the substitution $v = x^2+y^2+2$, so $dv = 2x dx + 2y dy$. The equation is $2y(x^2+y^2+2) dy + x(x^2+y^2+2) dx - 3x dx - 3y dy = 0$ $(x^2+y^2+2)(2y dy + x dx) = 3x dx + 3y dy = \frac{3}{2} d(x^2+y^2+2)$ Integrating both sides: $\int (x^2+y^2+2) d(x^2+y^2+2) = \int 3 d(x^2+y^2+2)$ is incorrect. Correct approach: $\frac{2y dy + x dx}{x^2+y^2+2} = \frac{3}{2} \frac{2x dx + 2y dy}{x^2+y^2+2}$ Integrating: $x^2 + 2y^2 - 3 \log(x^2+y^2+2) = c$.

The solution of the differential equation $\frac{dy}{dx} + \frac{x}{y} \cdot \frac{x^2+y^2-1}{2(x^2+y^2)+1} = 0$ is

Similar Questions

The rate at which the population of a city increases varies as the population. In a period of $20$ years,the population increased from $4$ lakhs to $6$ lakhs. In another $20$ years,the population will be (in $lakhs$)

The equation of the curve passing through $\left(2, \frac{9}{2}\right)$ and having the slope $\left(1-\frac{1}{x^2}\right)$ at $(x, y)$ is

The solution of the differential equation $x \frac{d^2 y}{d x^2} = 1$ at $x = y = 1$ with $\frac{dy}{d x} = 0$ at $x = 1$ is:

Water at $100^{\circ} C$ cools in $10 \text{ minutes}$ to $80^{\circ} C$ in a room temperature of $25^{\circ} C$. What will be the temperature of water after $20 \text{ minutes}$ (in $^{\circ} C$)?

Vedclass Test Series

Exam Paper Generator

Online Exam Module