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 $y^2 dx + (x^2 - xy + y^2) dy = 0$ is:

$A$ cube having an edge of length $5 \ cm$ is painted on all faces and then it is cut into equal cubes of unit volume. $A$ small cube is selected at random and it is found that at least one of its faces is painted. What is the probability that $2$ more faces of it are also painted?

The general solution of ${y^2}\,dx + ({x^2} - xy + {y^2})\,dy = 0$ is

Which of the following functions are homogeneous?

The solution curve of the differential equation $y \frac{dx}{dy} = x(\log_e x - \log_e y + 1)$,$x > 0, y > 0$ passing through the point $(e, 1)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module