solve the following initial value problems:
dy/dt cos t + y sin t = 1 where y(Pi/4) = sqrt(2)
(1 + x^2) dy/dx = y where y(0) = 1
t* dy/dt - y = t^3 sin t where y(Pi) = 0
Divide through by $\displaystyle \cos t$ to get $\displaystyle \frac{\,dy}{\,dt}+y\tan t=\sec t$
The integrating factor is $\displaystyle \varrho\!\left(t\right)=\sec t$ (Verify)
Thus, the DE becomes $\displaystyle \frac{\,d}{\,dt}\left[y\sec t\right]=\sec^2 t$
Can you take it from here??
Separating the variables, you have $\displaystyle \frac{\,dy}{y}=\frac{\,dx}{1+x^2}$(1 + x^2) dy/dx = y where y(0) = 1
Can you take it from here?
Divide through by $\displaystyle t$ to get $\displaystyle \frac{\,dy}{\,dt}-\frac{1}{t}y=t^2\sin t$t* dy/dt - y = t^3 sin t where y(Pi) = 0
The integrating factor is $\displaystyle \varrho\!\left(t\right)=\frac{1}{t}$
Thus, the DE becomes $\displaystyle \frac{\,d}{\,dt}\left[y\frac{1}{t}\right]=t\sin t$
Can you take it from here?
Does this make sense?