Question

X'+7x=5cos(2t), x(0)=0

Answer 1

The ODE is linear. Multiplying both sides by
`e^(7t)` allows you to condense the left hand side as the derivative of a product.


`x'+7x=5\cos2t`

`e^(7t)x'+7e^(7t)x=5e^(7t)\cos2t`

`(\mathrm d)/(\mathrm dt)\left[e^(7t)x\right]=5e^(7t)\cos2t`

Integrating both sides with respect to
`t` yields


`e^(7t)x=5\displaystyle\int e^(7t)\cos2t\,\mathrm dt`

The integral can be done by parts. You should get


`e^(7t)x=\frac5{53}e^(7t)(7\cos2t+2\sin2t)+C`

`x=\frac5{53}(7\cos2t+2\sin2t)+Ce^(-7t)`

With the given initial condition, you have


`0=\frac5{53}(7+0)+C\implies C=-(35)/(53)`

So the particular solution to the ODE is


`x=\frac5{53}(7\cos2t+2\sin2t)-(35)/(53)e^(-7t)`