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  • Solve for particular solution. (1 + x^2) dy/dx + 2xy = 1/(1 + x^2); y = 0 when x = 1

Solve for particular solution.

    Q14.    (1 + x^2)\frac{dy}{dx} + 2xy =\frac{1}{1 + x^2}; \ y = 0 \ when \ x = 1

Answers (1)

best_answer

Given equation is
(1 + x^2)\frac{dy}{dx} + 2xy =\frac{1}{1 + x^2}; \ y = 0 \ when \ x = 1
we can rewrite it as
\frac{dy}{dx}+\frac{2xy}{1+x^2}=\frac{1}{(1+x^2)^2}
This is  \frac{dy}{dx} + py = Q  type where p =\frac{2x}{1+x^2} and Q = \frac{1}{(1+x^2)^2}
Now,
I.F. = e^{\int pdx}= e^{\int \frac{2x}{1+x^2}dx}= e^{\log |1+x^2|}= 1+x^2                     
Now, the solution of given differential equation is given by relation
y(I.F.) =\int (Q\times I.F.)dx +C
y(1+ x^2) =\int (\frac{1}{(1+x^2)^2}\times (1+x^2))dx +C
y(1+x^2) =\int \frac{1}{(1+x^2)}dx +C\\ \\ y(1+x^2) = \tan^{-1}x+ C\\ \\
Now, by using boundary conditions we will find the value of C
It is given that  y = 0 when x = 1
at   x = 1
0.(1+1^2) = \tan^{-1}1+ C\\ \\ C =- \frac{\pi}{4}
Now,
y(1+x^2)= \tan^{-1}x- \frac{\pi}{4}
Therefore, the particular solution is  y(1+x^2)= \tan^{-1}x- \frac{\pi}{4}

Posted by

Gautam harsolia

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