Question

Solution of differential equation t = 1 + (ty) $\frac{dy}{dt}$ + $\left( \frac{dy}{dt} \right)^{2}$ + ...  is –

• A. y = ±$\sqrt{(\log t)^{2} + c}$
• B. ty = t^{y + c}
• C. y = log t + c
• D. y = (log t)² + c

Answer 1

Answer: (A)

y = ±$\sqrt{(\log t)^{2} + c}$

Answer 2

The given equation is

t = 1 + (ty) $\left( \frac{dy}{dt} \right)$ + $\frac{(ty)^{2}}{2!}$ $\left( \frac{dy}{dt} \right)^{2}$+ ...  Ž t = $e^{ty\left( \frac{dy}{dt} \right)}$ Ž log t = ty $\frac{dy}{dt}$

Ž y dy = $\frac{\log t}{t}$ dt

On integration

$\frac{y^{2}}{2}$= $\frac{(\log t)^{2}}{2}$ + k

Ž y = ± $\sqrt{(\log t)^{2} + 2k}$

Ž y = ± $\sqrt{(\log t)^{2} + c}$