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Home  >>  CBSE XII  >>  Math  >>  Differential Equations
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Solve: $x\large \frac{dy}{dx}$$ =y(\log y-\log x+1)$

$\begin{array}{1 1} (A)\;\log \large|\frac{y}{x}|=cx \\ (B)\;\log \large|\frac{y}{x}|=c \\ (C)\;\log \large|\frac{x}{y}|=cy \\ (D)\;\log \large|\frac{x}{y}|=c\end{array}$

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  • If a linear differential equation is of the form $\large\frac{dy}{dx}$$=f(x,y)$ then it is a homologous linear differential equation.
  • It can be solved by substituting $y=vx$ and $\large\frac{dy}{dx}$$=v+x \large\frac{dv}{dx}$
  • $ \int \large\frac{dx}{x}$$=\log x+c$
Step 1:
$x \large\frac{dy}{dx}$$=y(\log y-\log x+1)$
Dividing by x on both sides,
$\large \frac{dy}{dx}=\frac{y}{x}$$(\log y-\log x+1)$
Now this is a homologous differential equation, and it can be solve by putting $y=vx$ and $v+x \large\frac{dv}{dx}=\frac{dy}{dx}$
$v+x \large\frac{dv}{dx}=\frac{vx}{x}$$(\log vx-\log x+1)$
but $ \log vx-\log x=\log \large\frac{vx}{x}$$=\log v $
Therefore $v+x \large\frac{dv}{dx}$$=v(\log v+1)$
$x \large\frac{dv}{dx}$$=v \log v+v-v$
$x \large\frac{dv}{dx}$$=v \log v$
Now seperating the variables we get,
$\large\frac{dv}{v \log v}=\frac{dx}{x}$
Step 2:
On integrating on both sides we get,
$\large \int \frac{dv}{v \log v}=\int \frac{dx}{x}$
Let $\log v =t$. Differentiating w.r.t x on both sides we get
Now substituting this we get,
$\large\int \frac{dt}{t}=\int \frac{dx}{x}$
$=\log (t)=\log x+\log c$
Substituting for t we get,
$\log (\log v)=\log cx$
$\log v=cx$
Substituting for $v=\large\frac{y}{x}$
$\log \large|\frac{y}{x}|$$=cx$
Hence the required solution is $\log \large|\frac{y}{x}|$$=cx$
answered May 23, 2013 by meena.p

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