Question

Please solve for both parts ​

Answer 1

(a) The differential equation

`y' + \frac14 y = 3 + 2 \cos(2x)`

is linear, so we can use the integrating factor method. We have I.F.

`\mu = \displaystyle \exp\left(\int \frac{dx}4\right) = e^(x/4)`

so that multiplying both sides by
`\mu` gives

`e^(x/4) y' + \frac14 e^(x/4) y = 3e^(x/4) + 2 e^(x/4) \cos(2x)`

`\left(e^(x/4) y\right)' = 3e^(x/4) + 2 e^(x/4) \cos(2x)`

Integrate both sides. (Integrate by parts twice on the right side; I'll omit the details.)

`e^(x/4) y = 12 e^(x/4) + \frac8{65} e^(x/4) (8\sin(2x) + \cos(2x)) + C`

Solve for
`y`.

`y = 12 + \frac8{65} (\sin(2x) + \cos(2x)) + Ce^(-x/4)`

Given that
`y(0)=0`, we find

`0 = 12 + \frac8{65} (\sin(0) + \cos(0)) + Ce^0 \implies C = -(788)/(65)`

and the particular solution to the initial value problem is

`\boxed{y = 12 + \frac8{65} (\sin(2x) + \cos(2x)) - (788)/(65) e^(-x/4)}`

As
`x` gets large, the exponential term will converge to 0. We have

`\sin(2x) + \cos(2x) = \sqrt2 \sin\left(2x + \frac\pi4\right)`

which means the trigonometric terms will oscillate between
`\pm\sqrt2`. So overall, the solution will oscillate between
`12\pm\sqrt2` for large
`x`.

(b) We want the smallest
`x` such that
`y=12`, i.e.

`0 = \frac8{65} (\sin(2x) + \cos(2x)) - (788)/(65) e^(-x/4)`

`(788)/(65) e^(-x/4) = (8\sqrt2)/(65) \sin\left(2x + \frac\pi4\right)`

`(197)/(\sqrt2) e^(-x/4) = \sin\left(2x + \frac\pi4\right)`

Using a calculator, the smallest solution seems to be around
`\boxed{x\approx21.909}`