- #1

- 5

- 0

## Main Question or Discussion Point

solve the initial value problem:

y

y(0)=1, y

y(x)=?

The way I tried to achieve is to solve the corresponding homo equation first:

y

y

Then try to find y

Any idea? Or I made some mistake?

Meric.

y

^{''}+8.4y^{'}+17.64y=e^{-4.2x}y(0)=1, y

^{'}(0)=1y(x)=?

The way I tried to achieve is to solve the corresponding homo equation first:

y

^{''}+8.4y^{'}+17.64y=0, which gives y_{c};y

_{c}=c_{1}e^{-4.2x}+c_{2}e^{-4.2x}xThen try to find y

_{p}, generally I would assume a y_{p}=Ae^{-4.2x}, but from the y_{c}got above, clearly y_{p}=Ae^{-4.2x}or y_{p}=Ae^{-4.2x}x is not good. If I add one more 'x' in y_{p}assumption,in which it seems A has to be 0, which is not right either....Any idea? Or I made some mistake?

Meric.