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

If a body cools from $80^{\circ} C$ to $60^{\circ} C$ in a room temperature of $30^{\circ} C$ in $30 \text{ min}$,then the temperature of the body after one hour is (in $^{\circ} C$)

The solution of the differential equation $\frac{dy}{dx} = \frac{a + bx}{c + dy}$ represents a family of circles centered at the origin if...

In a bank,the principal increases continuously at a rate of $x \%$ per year. Then the rate,$x$,if ₹$100$ doubles itself in $10$ years,is (Given $\log 2 = 0.6931$) (in $\%$)

$A$ curve passes through the point $(3,2)$ for which the segment of the tangent line contained between the coordinate axes is bisected at the point of contact. The equation of the curve is

$A$ radioactive substance has a half-life of $h$ days. Its initial decay rate is given by (where the radioactive substance has an initial mass $m_0$).

Vedclass Test Series

Exam Paper Generator

Online Exam Module