The solution of (x^2+y^2) dx = 2xy dy is: | vedclass

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

Question

(A) Given the differential equation: $(x^2+y^2) dx = 2xy dy$Rearranging,we get: $\frac{dy}{dx} = \frac{x^2+y^2}{2xy}$Since this is a homogeneous diffe

Vedclass · Accepted Answer

$c(x^2-y^2)=x$

Vedclass · Answer

(A) Given the differential equation: $(x^2+y^2) dx = 2xy dy$Rearranging,we get: $\frac{dy}{dx} = \frac{x^2+y^2}{2xy}$Since this is a homogeneous differential equation,substitute $y = vx$,which implies $\frac{dy}{dx} = v + x \frac{dv}{dx}$.Substituting these into the equation: $v + x \frac{dv}{dx} = \frac{x^2 + (vx)^2}{2x(vx)} = \frac{x^2(1+v^2)}{2x^2v} = \frac{1+v^2}{2v}$Subtracting $v$ from both sides: $x \frac{dv}{dx} = \frac{1+v^2}{2v} - v = \frac{1+v^2-2v^2}{2v} = \frac{1-v^2}{2v}$Separating the variables: $\frac{2v}{1-v^2} dv = \frac{1}{x} dx$Integrating both sides: $\int \frac{2v}{1-v^2} dv = \int \frac{1}{x} dx$Let $u = 1-v^2$,then $du = -2v dv$. The integral becomes: $-\int \frac{1}{u} du = \ln|x| + C$$-\ln|1-v^2| = \ln|x| + \ln|c|$$-\ln|1 - (y/x)^2| = \ln|cx|$$-\ln|\frac{x^2-y^2}{x^2}| = \ln|cx|$$\ln|\frac{x^2}{x^2-y^2}| = \ln|cx|$$\frac{x^2}{x^2-y^2} = cx$$x = c(x^2-y^2)$

The solution of $(x^2+y^2) dx = 2xy dy$ is:

Similar Questions

Let $y=y(x)$ be the solution of the differential equation $\frac{dy}{dx}=\frac{4y^3+2yx^2}{3xy^2+x^3}$ with $y(1)=1$. If for some $n \in N$,$y(2) \in [n-1, n)$,then $n$ is equal to...

Solve the differential equation $y e^{\frac{x}{y}} dx = \left( x e^{\frac{x}{y}} + y^2 \right) dy$ where $y \neq 0$.

The general solution of the differential equation $(x^3-3xy^2)dx = (y^3-3x^2y)dy$ is,where $c$ is an arbitrary constant:

The solution of the differential equation $x y^{\prime} = 2 x e^{-y / x} + y$ is

The general solution of the differential equation $\frac{dy}{dx} = \frac{2x^2 - xy - y^2}{x^2 - y^2}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module