61. Kindly Consider the following y'=x+yx

The given D.E. is y1=x+yx=dydx=1+yx=f (yx) Hence, the given D.E is homogenous. Let, y=vx→yx=v, sothat, dydx=v+xdvdx So, the D.E. becomes v+xdvdx=1+v⇒xdvdx=1⇒dv=dxx Integrating both sides, ∫dv=∫dxxv=log|x|+c Putting v=yx back we get, yx=log|x|+c

61. Kindly Consider the following

$y' = \frac{x + y}{x}$

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Vishal Baghel | Contributor-Level 10

8 months ago

The given D.E. is

$\begin{array}{l} y^{1} = \frac{x + y}{x} \\ = \frac{d y}{d x} = 1 + \frac{y}{x} = f (\frac{y}{x}) \end{array}$

Hence, the given D.E is homogenous.

Let, $y = v x \to \frac{y}{x} = v, s o t h a t, \frac{d y}{d x} = v + \frac{x d v}{d x}$

So, the D.E. becomes

$\begin{array}{l} v + \frac{x d v}{d x} = 1 + v \\ \Rightarrow \frac{x d v}{d x} = 1 \\ \Rightarrow d v = \frac{d x}{x} \end{array}$

Integrating both sides,

$\begin{array}{l} \int d v = \int \frac{d x}{x} \\ v = l o g | x | + c \end{array}$

Putting $v = \frac{y}{x}$ back we get,

$\frac{y}{x} = l o g | x | + c$

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