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Solve the differential equation $ x.\cos\bigg(\large\frac{y}{x} \bigg).\frac{dy}{dx}$$=y\cos \bigg( \large\frac{y}{x} \bigg) $$+x $

1 Answer

  • A function is said to be homogenous function in degree n if $F(kx,ky) = k^nF(x,y)$ for any non zero constant $k.$
  • To solve these type of homogenous functions we make the substitution $y = vx$ hence $\large\frac{dy}{dx}$$ = v+x\large\frac{dv}{dx}$
Step 1:
Clearly this is a homogeneous differential equation.
Let us now substitute $y=vx$
Differentiating with respect to $x$ we get,
Step 2:
Substituting for $y$ and $\large\frac{dy}{dx}$ in equ(1) we get,
$v+x\large\frac{dv}{dx}=\frac{v\cos v+1}{\cos v}$
$x\large\frac{dv}{dx}=\frac{v\cos v+1}{\cos v}$$-v$
$x\large\frac{dv}{dx}=\frac{1}{\cos v}$
Step 3:
Now separating the variables we get,
$\cos v dv=\large\frac{dx}{x}$
Integrating on both sides we get,
$\int\cos vdv=\int\large\frac{dx}{x}$
$\Rightarrow \sin v=\log\mid x\mid+\log c$
$\Rightarrow \sin v=\log cx$
Step 4:
Replace $v$ by $\large\frac{y}{x}$
$\sin\big(\large\frac{y}{x}\big)$$=\log cx$
This is the required solution for the given equation.
answered Oct 3, 2013 by sreemathi.v