Find the particular solution of the differential equation \(\frac{dy}{dx}+8x=16x^2+4\) given that y=\(\frac{1}{3}\) when x=1.(a) y=\(\frac{(2x+1)^2}{3}\)(b) y=\(\frac{(4x+1)^2}{12}\)(c) y=\(\frac{(4x-2)^2}{3}\)(d) y=\(\frac{(2x-1)^2}{3}\)The question was asked during an interview.I would like to ask this question from Methods of Solving First Order & First Degree Differential Equations topic in division Differential Equations of Mathematics – Class 12

Find the particular solution of the differential equation \(\frac{dy}{

1.	Find the particular solution of the differential equation \(\frac{dy}{dx}+8x=16x^2+4\) given that y=\(\frac{1}{3}\) when x=1.(a) y=\(\frac{(2x+1)^2}{3}\)(b) y=\(\frac{(4x+1)^2}{12}\)(c) y=\(\frac{(4x-2)^2}{3}\)(d) y=\(\frac{(2x-1)^2}{3}\)The question was asked during an interview.I would like to ask this question from Methods of Solving First Order & First Degree Differential Equations topic in division Differential Equations of Mathematics – Class 12
Answer» Correct answer is (d) y=\(\frac{(2x-1)^2}{3}\) EXPLANATION: Given that, \(\frac{dy}{dx}+8x=16x^2+4\) \(\frac{dy}{dx}=16x^2-8x+4\) \(\frac{dy}{dx}=(4x-2)^2\) Separating the VARIABLES, we get dy=(4x-2)^2 dx Integrating both sides, we get \(\INT dy=\int (4x-2)^2 \,dx\) y=\(\frac{(4x-2)^2}{12}+C\) y=\(\frac{(2x-1)^2}{3}+C\) –(1) Given that, y=1/3 when x=1 Therefore, EQUATION (1) becomes, \(\frac{1}{3}=\frac{(2(1)-1)^2}{3}+C\) \(C=\frac{1}{3}-\frac{1}{3}\)=0 Hence, the PARTICULAR solution for the given differential solution is y=\(\frac{(2x-1)^2}{3}\).

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