Question

Solve the following equations
(a) tydt + (t + 1)dy = 0,
(b) y dy dt + t = 1,
(c) ty0 + y = y 2 , with the initial condition, y(1) = 1 2

Answer 1

Answer:

A) In y = - [ x-In (x+1) ] + c

B) $\frac{y^2}{2} = t - \frac{t^2}{2} + c$

Step-by-step explanation:

A) tydt + ( t + 1 ) dy = 0

dy/y = - ( $\frac{tdt}{t+1}$ )  we have to integrate both sides of the equation

In y = - [ x- In (x +1) ] + c

B) y dy/dt  + t = 1

we can express the equation as :

y dy = ( 1 - t ) dt

when we integrate the equation we have

$\frac{y^2}{2} = t - \frac{t^2}{2} + c$