Question

Solve the given differential equation by using an appropriate substitution. The DE is a Bernoulli equation. t2 (dy/dt) + y2 = ty

Answer 1

Answer:

Step-by-step explanation:

$t^2 \dfrac{dy}{dt}+ y^2 = ty \\ \\ t^2 \dfrac{dy}{dt}-ty = -y^2$

$\text{by dividing both sides by }{y^2},\text{ we have:}$

$\dfrac{t^2}{y^2}\dfrac{dy}{dt}- \dfrac{ty}{y^2}= -1$

$\dfrac{t^2}{y^2}\dfrac{dy}{dt}- t\dfrac{1}{y}= -1$

$\text{by dividing both sides by }t^2; \text{we have;}$

$\dfrac{1}{y^2}\dfrac{dy}{dt}- \dfrac{1}{t}\dfrac{1}{y}= -\dfrac{1}{t^2}--- (i)$

$Let \ v = \dfrac{1}{y}$

$\dfrac{dv}{dt}= -\dfrac{1}{y^2}\dfrac{dy}{dt}$

$-\dfrac{dv}{dt}= \dfrac{1}{y^2}\dfrac{dy}{dt}$

$\text{replace in equation (1); we have}$

$- \dfrac{dv}{dt}-\dfrac{1}{t}v = -\dfrac{1}{t^2}$

$- \Big ( \dfrac{dv}{dt}+ \dfrac{1}{t}v \Big) = -\dfrac{1}{t^2}$

$\dfrac{dv}{dt}+\dfrac{1}{t}v = \dfrac{1}{t^2}$

$\text{relating above with} \dfrac{dv}{dt}+ P(t) v = Q(t) \text{, we have:}$

$P(t) = \dfrac{1}{t} \ \ \ , \ \ \ \ Q(t) = \dfrac{1}{t^2}$

$\mu (t) = e^{\int P(t) \ dt }= e ^{\int \dfrac{1}{t} \ dt}= e^{In|t|}= t$

$v\mu(t) = \int \mu(t) Q(t) dt + C$

$vt = \int (t) (\dfrac{1}{t^2})dt + C$

$vt = \int \dfrac{1}{t}dt + C$

$vt = Int+C$

$v = \dfrac{In \ t + C}{t}$

$\text{substitute } v = \dfrac{1}{y}}, we \ get}$

$\dfrac{1}{y}=\dfrac{In \ t +C}{t}$

$\mathbf{y = \dfrac{t}{In \ t + C}}$