Question

Please solve for both parts ​

Answer 1

(a) The differential equation

$y' + \dfrac14 y = 3 + 2 \cos(2x)$

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

$e^{x/4} y' + \dfrac14 e^{x/4} y = 3e^{x/4} + 2 e^{x/4} \cos(2x)$

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

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

$e^{x/4} y = 12 e^{x/4} + \dfrac8{65} e^{x/4} (8\sin(2x) + \cos(2x)) + C$

Solve for $y$.

$y = 12 + \dfrac8{65} (\sin(2x) + \cos(2x)) + Ce^{-x/4}$

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

$0 = 12 + \dfrac8{65} (\sin(0) + \cos(0)) + Ce^0 \implies C = -\dfrac{788}{65}$

and the particular solution to the initial value problem is

$\boxed{y = 12 + \dfrac8{65} (\sin(2x) + \cos(2x)) - \dfrac{788}{65} e^{-x/4}}$

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

$\sin(2x) + \cos(2x) = \sqrt2 \sin\left(2x + \dfrac\pi4\right)$

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.

$0 = \dfrac8{65} (\sin(2x) + \cos(2x)) - \dfrac{788}{65} e^{-x/4}$

$\dfrac{788}{65} e^{-x/4} = \dfrac{8\sqrt2}{65} \sin\left(2x + \dfrac\pi4\right)$

$\dfrac{197}{\sqrt2} e^{-x/4} = \sin\left(2x + \dfrac\pi4\right)$

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