If (x^2+y^2) dy = xy dx,with y(x_0) = e and y | vedclass

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

Question

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

Vedclass · Accepted Answer

$\sqrt{3} e$

Vedclass · Answer

(A) Given the differential equation: $(x^2+y^2) dy = xy dx$. Rearranging,we get: $\frac{dy}{dx} = \frac{xy}{x^2+y^2}$. This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}{dx} = v + x \frac{dv}{dx}$. Substituting into the equation: $v + x \frac{dv}{dx} = \frac{x(vx)}{x^2+(vx)^2} = \frac{vx^2}{x^2(1+v^2)} = \frac{v}{1+v^2}$. $x \frac{dv}{dx} = \frac{v}{1+v^2} - v = \frac{v - v - v^3}{1+v^2} = -\frac{v^3}{1+v^2}$. Separating variables: $\int \frac{1+v^2}{v^3} dv = -\int \frac{dx}{x}$. $\int (v^{-3} + v^{-1}) dv = -\ln|x| + C$. $-\frac{1}{2v^2} + \ln|v| = -\ln|x| + C$. $-\frac{x^2}{2y^2} + \ln|y/x| = -\ln|x| + C$. $-\frac{x^2}{2y^2} + \ln|y| - \ln|x| = -\ln|x| + C$. $-\frac{x^2}{2y^2} + \ln|y| = C$. Using $y(1) = 1$: $-\frac{1^2}{2(1)^2} + \ln(1) = C \Rightarrow C = -\frac{1}{2}$. So,$-\frac{x^2}{2y^2} + \ln|y| = -\frac{1}{2}$. For $y(x_0) = e$: $-\frac{x_0^2}{2e^2} + \ln(e) = -\frac{1}{2}$. $-\frac{x_0^2}{2e^2} + 1 = -\frac{1}{2} \Rightarrow \frac{x_0^2}{2e^2} = \frac{3}{2}$. $x_0^2 = 3e^2 \Rightarrow x_0 = \sqrt{3}e$.

If $(x^2+y^2) dy = xy dx$,with $y(x_0) = e$ and $y(1) = 1$,then $x_0$ has the value:

Similar Questions

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

The general solution of the differential equation $\frac{dy}{dx} = \frac{x+2y-3}{2x+y-3}$ is

Solve the differential equation $(x \,dy-y \,dx) y\, \sin \left(\frac{y}{x}\right)=(y \,dx+x\, dy) x\, \cos \left(\frac{y}{x}\right)$.

For the differential equation $(x^2 + y^2) dy = xy dx$, it is given that $y(1) = 1$ and $y(x_0) = e$. Find the value of $x_0$.

If $\frac{dy}{dx} = \frac{y + x \tan(\frac{y}{x})}{x}$, then $\sin(\frac{y}{x})$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module