Question

Solution of the differential equation \(\frac{dy}{dx} = (4x+2y+1)^2\) is ______

(a) \(\frac{1}{2\sqrt{2}} tan^{-1}⁡(\frac{4x+2y+1}{\sqrt{2}})=x+c  \)

(b) \(\frac{1}{\sqrt{2}} cot^{-1}⁡(4x+2y+1)=x+c  \)

(c) \(\frac{1}{\sqrt{2}} tan^{-1}⁡⁡(\frac{4x+2y+1}{\sqrt{2}})=c  \)

(d) cot^-1⁡⁡⁡(4x+2y+1)=x+c

This question was addressed to me in an international level competition.

I would like to ask this question from Separable and Homogeneous Equations topic in section Ordinary Differential Equations – First Order & First Degree of Engineering Mathematics

Answer 1

Correct choice is (a) \(\frac{1}{2\sqrt{2}} tan^{-1}⁡(\frac{4x+2y+1}{\sqrt{2}})=x+c  \)

For explanation I would say: \(\frac{dy}{dx} = (4x+2y+1)^2\)

here we use substitution for \( 4x+2y+1 = t→4 + 2\frac{dy}{dx} = \frac{dt}{dx} \rightarrow \frac{dy}{dx} = \frac{1}{2}  \frac{dt}{dx} – 2\)

\(\frac{1}{2}  \frac{dt}{dx} – 2 = t^2\)

\(\frac{dt}{dx}=2t^2+4\)

separating the variable and integrating

\(\int \frac{1}{2t^2+4} \,dt = \int dx\)

\(\frac{1}{2\sqrt{2}} tan^{-1}⁡ ⁡\frac{t}{\sqrt{2}} = x + c\)

\(\frac{1}{2\sqrt{2}} tan^{-1}⁡(\frac{4x+2y+1}{\sqrt{2}})=x+c  \)  is the solution.