Question

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 1

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=\(\frac{(2x-1)^2}{3}\).