Question

The solution of the differential equation x dx + y dy = x² y dy − y² x dx, is
(a) x² − 1 = C (1 + y²)
(b) x² + 1 = C (1 − y²)
(c) x³ − 1 = C (1 + y³)
(d) x³ + 1 = C (1 − y³)

Answer 1

(a) x² − 1 = C (1 + y²)


We have,
x dx + y dy = x²y dy − y²x dx

$$\begin{gathered} \Rightarrow \left(x+x{y}^2\right)dx=\left(x^{2}y-y\right)dy \\ \Rightarrow \frac{x}{\left(x^2-1\right)}dx=\frac{y}{\left(1+y^2\right)}dy \\ \Rightarrow \frac{2x}{2\left(x^2-1\right)}dx=\frac{2y}{2\left(1+y^2\right)}dy \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \frac{1}{2}\int \frac{2y}{\left(1+y^2\right)}dy=\frac{1}{2}\int \frac{2x}{\left(x^2-1\right)}dx \\ \Rightarrow \frac{1}{2}\log \left|\left(1+y^2\right)\right|=\frac{1}{2}\log \left|\left(x^2-1\right)\right|-\frac{1}{2}\log \left|C\right| \\ \Rightarrow \log \left|\left(1+y^2\right)\right|=\log \left|\left(x^2-1\right)\right|-\log \left|C\right| \\ \Rightarrow \log \left|\left(1+y^2\right)\right|=\log \left|\left(\frac{x^2-1}{C}\right)\right| \\ \Rightarrow 1+y^2=\frac{x^2-1}{C} \\ \Rightarrow C\left(1+y^2\right)=x^2-1 \end{gathered}$$