Which is a homogeneous differential equation of first order? From (1), we have $\frac{dy}{dx}$ = $\frac{xy + y^{2}}{x^{2} - xy}$.... (2), Now put y = vx, then $\frac{dy}{dx}$ = v + x. Let $$k$$ be a real number. In the above six examples eqn 6.1.6 is non-homogeneous where as the first five equations are homogeneous. In particular, if M and N are both homogeneous functions of the same degree in x and y, then the equation is said to be a homogeneous equation. \dfrac{ky(kx + ky)}{(kx)(ky)} = \dfrac{k^2(y(x + y))}{k^2 xy} = \dfrac{y(x + y)}{xy}. M(x,y) = 3x2+ xy is a homogeneous function since the sum of the powers of x and y in each term is the same (i.e. The degree of this homogeneous function is 2. There are two definitions of the term “homogeneous differential equation.” One definition calls a first‐order equation of the form . From (1), we get v + x  $\frac{dv}{dx}$ = $\frac{x^{2}+y^{2}x^{2}}{2x.vx}$ = $\frac{1 + v^{2}}{2v}$, Or,  x$\frac{dy}{dx}$ = $\frac{1+v^{2}}{2v}$ - v = $\frac{1- v^{2}}{2v}$ or, $\frac{dx}{x}$   = $\frac{2v}{1-v^{2}}$ dv = $\frac{-2vdv}{1-v^{2}}$, log |x| = - $\int$  $\frac{-2vdv}{1-v^{2}}$ + lof C = -log $\left |1 - v^{2} \right |$ + log C, Or,   log |x| = log $\left | \frac{C}{1-v^{2}} \right |$ = log $\left | \frac{Cx^{2}}{x^{2}-y^{2}} \right |$ or, $\frac{cx^{2}}{x^{2}-y^{2}}$ = x, or, $x^{2}$ - $y^{2}$ = Cx……….(2). You da real mvps! He's modelled the situation using the differential equation: First, we need to check that Gus' equation is homogeneous. Solution 2)  We have  ($x^{2}$ + $y^{2}$) dx - 2xy dy = 0 or, $\frac{dy}{dx}$ = $\frac{x^{2} + y^{2}}{2xy}$ … (1), Put y = vx; then $\frac{dy}{dx}$ = v + x$\frac{dv}{dx}$, From, (1), v + x $\frac{dy}{dx}$ = $\frac{x^{2} + y^{2}x^{2}}{2x^{2}v}$ = $\frac{1 + v^{2}}{2v}$, Or,   $\frac{2v}{1-v^{2}}$. \end{align*} \), $$$\frac{dy}{dx}$, From (2), v + x.$\frac{dy}{dx}$ = $\frac{x.vx + v^{2x^{2}}}{x^{2} -x.vx}$ = $\frac{v +v^{2}}{1-v}$, Or, x $\frac{dy}{dx}$ = $\frac{v + v^{2}}{1-v}$ - v = $\frac{v + v^{2} - v + v^{2}}{1-v}$ = $\frac{2v^{2}}{1-v}$, Or, $\frac{1-v}{2v^{2}}$ dv = $\frac{dy}{dx}$ or, $\frac{dx}{x}$ = $\frac{1}{2}$ $\left ( \frac{1}{v^{2}} - \frac{1}{v}\right )$dv, Integrating, log x = $\frac{1}{2}$ $\left ( - \frac{1}{v} - logv \right )$ + $\frac{1}{2}$log C, Or, 2 Log x = - $\frac{1}{v}$ - logv + log C or, log $x^{2}$ + log v - log C = - $\frac{1}{v}$, OR, Log $\left ( \frac{vx^{2}}{C} \right )$ = - $\frac{x}{y}$ [y = vx] or, $\frac{vx^{2}}{C}$ e $\frac{x}{y}$, or, xy = Ce - $\frac{x}{y}$. v + x \; \dfrac{dv}{dx} &= 1 + v\\ It is shown that these transformations allow reducing the order of the quasi-homogeneous ordinary differential equation and that for such an equation the boundary value problems may be simplified. &= 1 + v 2x+5=-1, the solution of which is a number. Step 2: Integrate both sides of the equation. substitution \(y = vx$$. x\; \dfrac{dv}{dx} &= 1 - 2v, \dfrac{d \text{cabbage}}{dt} = \dfrac{ \text{cabbage}}{t}, &= \dfrac{x(vx) + (vx)^2}{x(vx)}\\ \begin{align*} &= \dfrac{x^2 - v x^2 }{x^2}\\ You must be logged in as Student to ask a Question. For example, the differential equation below involves the function $$y$$ and its first derivative $$\dfrac{dy}{dx}$$. 2) Are the vectors in (2) linearly dependent or linearly independent? Calculus is the branch of mathematics that deals with the finding and properties of derivatives and integrals of functions, by methods originally based on the summation of infinitesimal differences. 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