Question

The general solution of the homogeneous differential equation of the type $\frac{dy}{dx}=F(x,y)=g\left(\frac{y}{x}\right)$, when $y=v\cdot x$ is

• A. $\int \frac{dv}{g(v)+v}=\int \frac{1}{x}dx+C$
• B. $\int \frac{dv}{g(v)-v}=\int \frac{1}{x}dx+C$
• C. $\int \frac{dv}{g(v)}=\int \frac{1}{x}dx+C$
• D. $\int \frac{dv}{vg(v)}=\int \frac{1}{x}dx+C$

Answer 1

Answer: (B)

To solve a homogeneous differential equation of the type
$\frac{dy}{dx}=F(x,y)=g\left(\frac{y}{x}\right)....$(i)
We make the substitution
$y=v\cdot x....$(ii)
On differentiating Eq. (ii) w.r.t. $x$, we get
$\frac{dy}{dx}=v+x\frac{dv}{dx}....$(iii)
On substituting the value of $\frac{dy}{dx}$ from Eq. (iii) in Eq. (i), we get
$v+x\frac{dv}{dx}=g(v)$
or $x\frac{dv}{dx}=g(v)-v....$(iv)
On separating the variables in Eq. (iv), we get
$\frac{dv}{g(v)-v}=\frac{dx}{x}....$(v)
On integrating both sides of Eq. (v), we get
$\int \frac{dv}{g(v)-v}=\int \frac{1}{x}dx+C.....$(vi)
Eq. (vi) gives general solution (primitive) of the differential Eq. (i) when we replace $v$ by $\frac{v}{x}$.
Note If the homogeneous differential equation is in the form $\frac{dx}{dy}=F(x,y)$ where, $F(x,y)$ is homogeneous function of degree zero, then we make substitution $\frac{x}{y}=v$, i.e., $x=vy$ and we proceed further to find the general solution as discussed above by writing $\frac{dx}{dy}=F(x,y)=h\left(\frac{x}{y}\right)$