Question

1.  dy/dt + ty=y ; y(1)=3

Answer 1

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

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

$$\int\dfrac{dy}{y}=\int(1-t)dt$$

$$\ln |y|=(t-\dfrac{t^2}{2})+\ln C$$

$$y(t)=Ce^{(t-{t^2 \over 2})}$$

$$y(1)=3=>3=Ce^{(1-{1^2 \over 2})}$$

$$C=3e^{-{1 \over 2}}$$

$$y(t)=3e^{(t-{t^2 \over 2}-{1 \over 2})}$$

$$y(t)=3e^{-{(t-1)^2 \over 2}}$$