Question

For
`\rm x \in \mathbb{R}`, let the function y(x) be the solution of the differential equation


`\rm (dy)/(dx) + 12y = \cos \bigg( (\pi)/(12)x \bigg ) , \: \: \: \: y(0) = 0 \\`
Then, which of the following statements is/are TRUE?

(A) y(x) is an increasing function

(B) y(x) is a decreasing function

(C) There exists a real number β such that the line y = β intersects the curve y = y(x) at infinitely many points

(D) y(x) is a periodic function
​

Answer 1

In the differential equation

`(dy)/(dx) + 12y = \cos\left((\pi x)/(12)\right)`

multiply on both sides by the integrating factor

`\mu = \exp\left(\displaystyle\int12\,dx\right) = e^(12x)`

Then the left side condenses to the derivative of a product.

`e^(12x) (dy)/(dx) + 12 e^(12x) y = e^(12x) \cos\left((\pi x)/(12)\right)`

`(d)/(dx)\left[e^(12x)y\right] = e^(12x)\cos\left((\pi x)/(12)\right)`

Integrate both sides with respect to
`x`, and use the initial condition
`y(0)=0` to solve for the constant
`C`.

`\displaystyle \int (d)/(dx) \left[e^(12x)y\right] \, dx = \int e^(12x) \cos\left((\pi x)/(12)\right) \, dx`

As an alternative to integration by parts, recall

`e^(ix) = \cos(x) + i \sin(x)`

Now

`e^(12x) \cos\left((\pi x)/(12)\right) = e^(12x) \mathrm{Re}\left(e^(i\pi x/12)\right) = \mathrm{Re}\left(e^((12+i\pi/12)x)\right)`

`\displaystyle \int \mathrm{Re}\left(e^((12+i\pi/12)x)\right) \, dx = \mathrm{Re}\left(\int e^((12+i\pi/12)x) \, dx\right)`

`\displaystyle. ~~~~~~~~ = \mathrm{Re}\left(\frac1{12+i\frac\pi{12}} e^((12+i\pi/12)x)\right) + C`

`\displaystyle. ~~~~~~~~ = \mathrm{Re}\left(\frac{12 - i\frac\pi{12}}{12^2 + (\pi^2)/(12^2)} e^(12x) \left(\cos\left((\pi x)/(12)\right) + i \sin\left((\pi x)/(12)\right)\right)\right) + C`

`\displaystyle. ~~~~~~~~ = (12)/(12^2 + (\pi^2)/(12^2)) e^(12x) \cos\left((\pi x)/(12)\right) + \frac\pi{12} e^(12x) \sin\left((\pi x)/(12)\right) + C`

`\displaystyle. ~~~~~~~~ = \frac1{12(12^4+\pi^2)} e^(12x) \left(12^4 \cos\left((\pi x)/(12)\right) + \pi (12^4+\pi^2) \sin\left((\pi x)/(12)\right)\right) + C`

Solve for
`y`.

`\displaystyle e^(12x) y = \frac1{12(12^4+\pi^2)} e^(12x) \left(12^4 \cos\left((\pi x)/(12)\right) + \pi (12^4+\pi^2) \sin\left((\pi x)/(12)\right)\right) + C`

`\displaystyle y = \frac1{12(12^4+\pi^2)} \left(12^4 \cos\left((\pi x)/(12)\right) + \pi (12^4+\pi^2) \sin\left((\pi x)/(12)\right)\right) + C`

Solve for
`C`.

`y(0)=0 \implies 0 = \frac1{12(12^4+\pi^2)} \left(12^4 + 0\right) + C \implies C = -(12^3)/(12^4+\pi^2)`

So, the particular solution to the initial value problem is

`\displaystyle y = \frac1{12(12^4+\pi^2)} \left(12^4 \cos\left((\pi x)/(12)\right) + \pi (12^4+\pi^2) \sin\left((\pi x)/(12)\right)\right) - (12^3)/(12^4+\pi^2)`

Recall that

`R\cos(\alpha-\beta) = R\cos(\alpha)\cos(\beta) + R\sin(\alpha)\sin(\beta)`

Let
`\alpha=(\pi x)/(12)`. Then

`\begin{cases} R\cos(\beta) = 12^4 \\ R\sin(\beta) = 12^4\pi+\pi^3 \end{cases} \\\\ \implies \begin{cases} (R\cos(\beta))^2 + (R\sin(\beta))^2 = R^2 = 12^8 + (12^4\pi + \pi^3)^2 \\ (R\sin(\beta))/(R\cos(\beta))=\tan(\beta)=\pi+(\pi^3)/(12^4)\end{cases}`

Whatever
`R` and
`\beta` may actually be, the point here is that we can condense
`y` into a single cosine expression, so choice (D) is correct, since
`\cos(x)` is periodic. This also means choice (C) is also correct, since
`\beta=\cos(x)\implies\beta=\cos(x+2n\pi)` for infinitely many integers
`n`. This simultaneously eliminates (A) and (B).