Suppose that f(x, y) and g(x, y) are homogene | vedclass

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

Question

(D) Given the differential equation $\frac{dy}{dx} = \frac{f(x, y)}{g(x, y)}$.Since $f$ and $g$ are homogeneous functions of the same order,we can wri

Vedclass · Accepted Answer

$\left(\frac{g(V, 1)}{f(V, 1)}-V\right)$

Vedclass · Answer

(D) Given the differential equation $\frac{dy}{dx} = \frac{f(x, y)}{g(x, y)}$.Since $f$ and $g$ are homogeneous functions of the same order,we can write $\frac{dy}{dx} = \frac{f(Vy, y)}{g(Vy, y)} = \frac{y^n f(V, 1)}{y^n g(V, 1)} = \frac{f(V, 1)}{g(V, 1)}$.We are given the substitution $x = Vy$. Differentiating with respect to $y$,we get $\frac{dx}{dy} = V + y\frac{dV}{dy}$.Thus,$\frac{dV}{dy} = \frac{1}{y} \left( \frac{dx}{dy} - V \right)$.Since $\frac{dy}{dx} = \frac{f(x, y)}{g(x, y)}$,we have $\frac{dx}{dy} = \frac{g(x, y)}{f(x, y)} = \frac{g(Vy, y)}{f(Vy, y)} = \frac{g(V, 1)}{f(V, 1)}$.Substituting this into the expression for $\frac{dV}{dy}$,we get $\frac{dV}{dy} = \frac{1}{y} \left( \frac{g(V, 1)}{f(V, 1)} - V \right)$.Comparing this with $\frac{dV}{dy} = \frac{1}{y}(F(V))$,we find $F(V) = \left( \frac{g(V, 1)}{f(V, 1)} - V \right)$.

Suppose that $f(x, y)$ and $g(x, y)$ are homogeneous functions of the same order. If $x=Vy$ reduces the equation $\frac{dy}{dx}=\frac{f(x, y)}{g(x, y)}$ to the form $\frac{dV}{dy}=\frac{1}{y}(F(V))$,then $F(V)=$

Similar Questions

The homogeneous differential equation of the form $\left(1+e^{\frac{x}{y}}\right) dx + e^{\frac{x}{y}}\left(1-\frac{x}{y}\right) dy = 0$ can be solved by making the substitution:

The solution of the differential equation $\frac{dy}{dx} = \frac{x + y}{x}$ satisfying the condition $y(1) = 1$ is:

For the differential equation,the general solution for $x \cos \left( \frac{y}{x} \right) (y dx + x dy) = y \sin \left( \frac{y}{x} \right) (x dy - y dx)$,(where $c$ is the constant of integration) is

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

The solution of the differential equation $x dy - y dx = \sqrt{x^2+y^2} dx$, given that $y=1$ when $x=\sqrt{3}$, is

Vedclass Test Series

Exam Paper Generator

Online Exam Module