Form ODE by Eliminating a and b (Arbitrary Constants)

Form ODE by Eliminating a and b: In this article, let’s learn how to form an ordinary differential equation (ODE) by eliminating a and b which are arbitrary constants.

Lets solve some problems.

Solved Problems

$\boxed{\color{blue}\textbf{Q}1:}$ Eliminate the arbitrary constants a, b and c to from an ordinary differential equation.

1. y = ax+b
2. y = ax²+bx+c
3. y = 2cx+c²

$\boxed{\color{red}\textbf{Answer:}}$

Part (1) Given y = ax+b

Differentiating both sides with respect to x, we get that

$\dfrac{dy}{dx}$ = a

So, $\dfrac{d^2y}{dx^2}$ = 0.

Part (2) Given y = ax²+bx+c

Answer is, $\dfrac{d^3y}{dx^3}$ = 0.

Part (3) Here, y = 2cx+c² ….(i)

Differentiate w.r.t x, we obtain that

$\dfrac{dy}{dx}$ = 2c ⇒ c = $\dfrac{1}{2} \dfrac{dy}{dx}$

Putting this value in the original Equation (i), we get that

$y=\dfrac{dy}{dx}+\dfrac{1}{4}\left( \dfrac{dy}{dx}\right)^2$.

This is our desired ordinary differential equation after eliminating the arbitrary constant c.

$\boxed{\color{blue}\textbf{Q}2:}$ Eliminate a and b to form an ODE from y = e^-x(acosx +bsinx).

$\boxed{\color{red}\textbf{Answer:}}$

Given that y = e^-x(acosx +bsinx) …(ii)

Differentiating w.r.t x, we have

$\dfrac{dy}{dx}=-e^{-x}(a \cos x+b \sin x)$ $+e^{-x}(-a\sin x +b \cos x)$

Using Equation (ii) ⇒ $\dfrac{dy}{dx}=-y$ $+e^{-x}(-a\sin x +b \cos x)$…(iii)

Again differentiating w.r.t x, it follows that

$\dfrac{d^2y}{dx^2}=-\dfrac{dy}{dx}$ $-e^{-x}(-a\sin x +b \cos x)$ $+e^{-x}(-a\cos x -b \sin x)$

⇒ $\dfrac{d^2y}{dx^2}+\dfrac{dy}{dx}$ $=-\left(\dfrac{dy}{dx}+y \right)$ $-y$, by Equation (iii)

⇒ $\dfrac{d^2y}{dx^2}+2\dfrac{dy}{dx}+2y=0$.

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