The solution of the differential equation (3x | vedclass

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

Question

(A) The given differential equation is $(3xy + y^2)dx + (x^2 + xy)dy = 0$. Rearranging the terms,we get $\frac{dy}{dx} = -\frac{3xy + y^2}{x^2 + xy}$.

Vedclass · Accepted Answer

$x^2(2xy + y^2) = c^2$

Vedclass · Answer

(A) The given differential equation is $(3xy + y^2)dx + (x^2 + xy)dy = 0$. Rearranging the terms,we get $\frac{dy}{dx} = -\frac{3xy + y^2}{x^2 + xy}$. This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}{dx} = v + x\frac{dv}{dx}$. Substituting these into the equation: $v + x\frac{dv}{dx} = -\frac{3x^2v + x^2v^2}{x^2 + x^2v} = -\frac{3v + v^2}{1 + v}$. $x\frac{dv}{dx} = -\frac{3v + v^2}{1 + v} - v = -\frac{3v + v^2 + v + v^2}{1 + v} = -\frac{2v^2 + 4v}{1 + v} = -\frac{2v(v + 2)}{v + 1}$. Separating the variables: $\frac{v + 1}{v(v + 2)} dv = -\frac{2}{x} dx$. Using partial fractions: $\frac{v + 1}{v(v + 2)} = \frac{A}{v} + \frac{B}{v + 2} \Rightarrow v + 1 = A(v + 2) + Bv$. For $v = 0$,$1 = 2A \Rightarrow A = 1/2$. For $v = -2$,$-1 = -2B \Rightarrow B = 1/2$. So,$\int (\frac{1}{2v} + \frac{1}{2(v + 2)}) dv = -\int \frac{2}{x} dx$. $\frac{1}{2} \ln|v| + \frac{1}{2} \ln|v + 2| = -2 \ln|x| + C'$. $\ln|v(v + 2)| = -4 \ln|x| + 2C' \Rightarrow \ln|v(v + 2)x^4| = C''$. $v(v + 2)x^4 = c^2$. Substituting $v = y/x$: $\frac{y}{x}(\frac{y}{x} + 2)x^4 = c^2 \Rightarrow y(y + 2x)x^2 = c^2 \Rightarrow x^2(y^2 + 2xy) = c^2$.

The solution of the differential equation $(3xy + y^2)dx + (x^2 + xy)dy = 0$ is

Similar Questions

Show that the differential equation $x dy - y dx = \sqrt{x^{2} + y^{2}} dx$ is a homogeneous equation and find its solution.

Show that the differential equation $x \cos \left(\frac{y}{x}\right) \frac{d y}{d x}=y \cos \left(\frac{y}{x}\right)+x$ is homogeneous and solve it.

Let $y=y(x)$ be the solution of the differential equation $(x^2-3y^2)dx+3xydy=0$ with $y(1)=1$. Then $6y^2(e)$ is equal to $......$

The solution of the differential equation $\frac{dy}{dx} = \tan \left(\frac{y}{x}\right) + \frac{y}{x}$ is

$A$ curve passes through the point $\left( 1, \frac{\pi}{4} \right)$ and its slope at any point is given by $\frac{dy}{dx} = \frac{y}{x} - \cos^2 \left( \frac{y}{x} \right)$. Then the equation of the curve is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module