Question

Find the general solution of the differential equation \(\frac{dy}{dx}=\frac{2+x^3}{4-y^3}\).

(a) x^3-y^3-4y+C=0

(b) x^4+8x+y^4-16y+C=0

(c) 2x+y^4-4y+C=0

(d) x^3+2x+C=0

I had been asked this question in an interview for job.

This question is from Methods of Solving First Order & First Degree Differential Equations in portion Differential Equations of Mathematics – Class 12

Answer 1

Right option is (b) x^4+8x+y^4-16y+C=0

The explanation: Given that, \(\frac{dy}{dx}=\frac{2+x^3}{4-y^3}\)

Separating the variables, we get

(4-y^3)dy=(2+x^3)dx

Integrating both sides, we get

\(\int 4-y^3 \,dy=\int 2+x^3 \,dx\)

\(4y-\frac{y^4}{4}=2x+\frac{x^4}{4}+C_1\)

x^4+8x+y^4-16y+C=0 (where 4C1=C)