Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XII  >>  Math  >>  Differential Equations
0 votes

Solve: $x\large \frac{dy}{dx}$$ =y(\log y-\log x+1)$

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

Can you answer this question?

1 Answer

0 votes
  • 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

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App