Ask Questions, Get Answers

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

Solve the following differential equation : $ (1+x^2)\large\frac{dy}{dx}$$+2xdy=\large\frac{1}{1+x^2},$ given y = 0 when x = 1.

Can you answer this question?

1 Answer

0 votes
  • To solve the first order linear differential equation of the form $\large\frac{dy}{dx}$$ + Py = Q$
  • Write the given equation in the form of $\large\frac{dy}{dx}$$ + Py = Q$
  • Find the integrating factor (I.F) = $e^{\int Pdx}$.
Step 1:
Let us rewrite the equation
dividing throughout by $(1+x^2)$ we get,
$\large\frac{dy}{dx } + \frac{2xy}{(1+x^2)}$$ = 1$
Here $P =\large\frac{ 2x}{(1+x^2)}$ and $Q = \large\frac{1}{(1+x^2)^2}$
Let us find the integrating factor(I.F)
$\int Pdx = \int\large\frac{ 2x}{(1+x^2)} $$= \log(1+x^2)$
Hence $I.F = e^{\log(1+x^2)} = (1+x^2)$
Step 2:
Hence the required solution is
$y.(1+x^2) =\int\big[\large\frac{1}{(1+x^2)^2}$$ . (1+x^2)\big] + C$
$\int\large\frac{ 1}{(1+x^2)}$$dx = \tan^{-1}x + C$
$y.(1+x^2) =\tan^{-1}x + C$
Step 3:
To evaluate the value of C
let us substitute the given values of $x = 1$ and $y =0$
$0 = \tan^{-1}(1) + C$
Therefore $C = \large\frac{- \pi}{4}$
substituting this we get,
$y(1+x^2) = \tan^{-1} x -\large\frac{ \pi}{4}$
This is the required solution.
answered Nov 8, 2013 by sreemathi.v

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