Question

Solution of differential equation $\frac{dt}{dx} = \frac{t\left( \frac{d}{dx}(g(x)) \right) - t^{2}}{g(x)}$is –

• A. t = $\frac{g(x) + c}{x}$
• B. t = $\frac{g(x)}{x}$ + c
• C. t = $\frac{g(x)}{x + c}$
• D. t = g(x) + x + c

Answer 1

Answer: (C)

t = $\frac{g(x)}{x + c}$

Answer 2

$\frac{dt}{dx}$ – t $\frac{g'(x)}{g(x)}$ = – $\frac{t^{2}}{g(x)}$

Ž – $\frac{1}{t^{2}}$ $\frac{dt}{dx}$ + $\frac{g'(x)}{g(x)}$ = $\frac{1}{g(x)}$... (1)

Let z = $\frac{1}{t}$ Ž – $\frac{1}{t^{2}}$ $\frac{dt}{dx}$ = $\frac{dz}{dx}$From (i)

$\frac{dz}{dx}$ + $\frac{g'(x)}{g(x)}$ z = $\frac{1}{g(x)}$

On comparing with $\frac{dz}{dx}$ + Pz = Q, we get

P = $\frac{g'(x)}{g(x)}$, Q =$\frac{1}{g(x)}$

\ IF = $e^{\int_{}^{}{\frac{g'(x)}{g(x)}dx}}$ = e ^{log [g(x)]} = g(x)

6tThus complete solution is

z . g(x) = $\int_{}^{}{g(x)}$. $\frac{1}{g(x)}$ dx + c Ž $\frac{1}{t}$g(x) = x + c Ž $\frac{g(x)}{x + c}$ = t