Question

Consider the differential equation $\frac{dy}{dx}=\frac{y^3}{2\left(x{y}^2-x^2\right)}$ such that $y(1)=1$, then the particular solution of the equation is

• A. $y^2=e^{\frac{y^2}{x}-1}$
• B. $y^2=e^{1-\frac{y^2}{x}}$
• C. $y=e^{1-\frac{y^2}{x}}$
• D. $y^2=e^{\frac{y^2}{x}}$

Answer 1

Answer: (A)

$\frac{dx}{dy}=\frac{2\left(x{y}^2-x^2\right)}{y^3}\Rightarrow \frac{1dx}{x^{2}dy}=2\left(\frac{1}{xy}-\frac{1}{y^3}\right)$
$\text{ Put, }\frac{1}{x}=t$
$\frac{dt}{dy}=2\left(\frac{1}{y^3}-\frac{t}{y}\right)$
$\frac{dt}{dy}+\frac{2t}{y}=\frac{2}{y^3}\Rightarrow t\cdot y^2=2\ln y+\ln c$
$\frac{y^2}{x}=\ln c^2$
$c{y}^2=e^{\frac{y^2}{x}}$
$u\mathrm{sing}y(1)=1\text{, we get }c=e$
$\Rightarrow y^2=e^{\frac{y^2}{x}-1}$