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

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- Jan 31st 2009, 10:24 PMrazorfeverinitial value problems
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 - Jan 31st 2009, 10:39 PMChris L T521
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??

Quote:

(1 + x^2) dy/dx = y where y(0) = 1

Can you take it from here?

Quote:

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?