The general solution of the differential equation \(y\,dx-(x+2y^2)\,dy = 0\)
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(x = 2y^2+ Cy\), Where C is constant of integration
The equation is not linear in y, but it is linear when x is taken as a function of y:
\(y\,dx = (x + 2y^2)\,dy\)
\(\displaystyle \frac{dx}{dy} = \frac{x + 2y^2}{y}\)
\(\displaystyle \frac{dx}{dy} - \frac{1}{y}\,x = 2y\)
This is a linear equation in x of the form \(\frac{dx}{dy} + Px = Q\), with \(P = -\frac{1}{y}\) and \(Q = 2y\).
\(\displaystyle \text{I.F.} = e^{\int -\frac{1}{y}\,dy} = e^{-\log_e y} = \frac{1}{y}\)
The solution is \(x \cdot \text{I.F.} = \int Q \cdot \text{I.F.}\,dy + C\):
\(\displaystyle \frac{x}{y} = \int 2y\cdot\frac{1}{y}\,dy = 2y + C\)
Multiplying by y:
\(x = 2y^2 + Cy\)
NCERT (2026-27 edition):
Linear Differential Equations
