Question

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

Answer 1

Explanation:

`t^2 (dy)/(dt)+ y^2 = ty \\ \\ t^2 (dy)/(dt)-ty = -y^2`

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

`(t^2)/(y^2)(dy)/(dt)- (ty)/(y^2)= -1`

`(t^2)/(y^2)(dy)/(dt)- t(1)/(y)= -1`

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

`(1)/(y^2)(dy)/(dt)- (1)/(t)(1)/(y)= -(1)/(t^2)--- (i)`

`Let \ v = (1)/(y)`

`(dv)/(dt)= -(1)/(y^2)(dy)/(dt)`

`-(dv)/(dt)= (1)/(y^2)(dy)/(dt)`

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

`- (dv)/(dt)-(1)/(t)v = -(1)/(t^2)`

`- \Big ( (dv)/(dt)+ (1)/(t)v \Big) = -(1)/(t^2)`

`(dv)/(dt)+(1)/(t)v = (1)/(t^2)`

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

`P(t) = (1)/(t) \ \ \ , \ \ \ \ Q(t) = (1)/(t^2)`

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

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

`vt = \int (t) ((1)/(t^2))dt + C`

`vt = \int (1)/(t)dt + C`

`vt = Int+C`

`v = (In \ t + C)/(t)`

`\text{substitute } v = (1)/(y)}, we \ get}`

`(1)/(y)=(In \ t +C)/(t)`

`\mathbf{y = (t)/(In \ t + C)}`