Question

Form the differential equation corresponding to $y=a\cos (nx+b),(a,b)$.

Answer 1

[It contain two constant

Then, the required differential equation in second order form]

$\begin{array}{c}\Rightarrow \frac{y}{a}=cos(nx+b)\\ \Rightarrow {\cos }^{-1}\left(\frac{y}{a}\right)=(nx+b)\\ \Rightarrow {\cos }^{-1}\left(\frac{y}{a}\right)-na=b\\ differentiatew.r.tx\\ \Rightarrow \frac{d\left[{\cos }^{-1}\left(\frac{y}{a}\right)\right]}{dx}-\frac{d\left(nx\right)}{dx}=\frac{db}{dx}\\ \Rightarrow \frac{-1}{\sqrt{1-{\left(\frac{y}{a}\right)}^2}}\times \frac{1}{a}\left(\frac{dy}{dx}\right)-n=0\\ \Rightarrow \frac{-\frac{1}{a}\left(\frac{dy}{dx}\right)}{\sqrt{1-{\left(\frac{y}{a}\right)}^2}}=n\\ \Rightarrow \frac{-\left(\frac{dy}{dx}\right)}{n}=a\sqrt{1-\frac{y^2}{a^2}}\\ squaringonbothsides\\ \Rightarrow \frac{{\left(\frac{dy}{dx}\right)}^2}{n^2}=a^2-y^2\\ againdifferentiatew.r.ttox\\ \Rightarrow \frac{1}{n^2}\frac{{\left(\frac{dy}{dx}\right)}^2}{dx}=\frac{d\left(a^2\right)}{dx}-\frac{d\left(y^2\right)}{dx}\\ \Rightarrow \frac{2}{n^2}\cdot \left(\frac{dy}{dx}\right)\left(\frac{d^{2}y}{d{x}^2}\right)=0-2y\frac{dy}{dx}\\ \Rightarrow \frac{\left(\frac{d^{2}y}{d{x}^2}\right)}{n^2}=-y\\ \therefore \Rightarrow \frac{\left(\frac{d^{2}y}{d{x}^2}\right)}{n^2}+y=0\\ Itisrequireddifferentialequation\end{array}$