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# Solve the following differential equation : $(1+x^2)\large\frac{dy}{dx}$$+2xdy=\large\frac{1}{1+x^2}, given y = 0 when x = 1. Can you answer this question? ## 1 Answer 0 votes Toolbox: • To solve the first order linear differential equation of the form \large\frac{dy}{dx}$$ + 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$