Question

Form the differential equation by eliminating the arbitrary constant if $y=A{{e}^{2x}}+B{{e}^{-2x}}.$

Answer 1

To solve the given question, we will first find out what a differential equation is and what the arbitrary constants in any equation are. After doing this, we will differentiate y with respect to x. From here, we will get an equation in x, y, and derivative of y. Again, we will differentiate the obtained equation with respect to x. This will give as another equation in the second derivative of y, x, and y. From here, we will derive the relation between y and the second derivative of y.

Complete step-by-step answer:
Before solving the question, we must know about the differential equations and arbitrary constants. A differential equation is an equation with one or more derivative of a function. The derivative of the function is given by $\dfrac{dy}{dx}.$ In other words, it is defined as the equation that contains derivatives of one or more dependent variables with respect to one or more independent variables. An arbitrary constant is a constant to which various values can be assigned but which remains unaffected by the change in the values of the variables of the equation. In our case, A and B are arbitrary constants. The equation is given in the question of which we have to find the differential equation is
$y=A{{e}^{2x}}+B{{e}^{-2x}}.....\left( i \right)$
Now, we will differentiate both sides of (i) with respect to x. Thus, we will get,
$\Rightarrow \dfrac{dy}{dx}=\dfrac{d}{dx}\left[ A{{e}^{2x}} \right]+\dfrac{d}{dx}\left[ B{{e}^{-2x}} \right]$
$\Rightarrow \dfrac{dy}{dx}=A\dfrac{d}{dx}\left[ {{e}^{2x}} \right]+B\dfrac{d}{dx}\left[ {{e}^{-2x}} \right]$
Now, the differentiation of ${{e}^{ax}}$ is $a{{e}^{ax}},$ so we will get,
$\Rightarrow \dfrac{dy}{dx}=2A{{e}^{2x}}-2B{{e}^{-2x}}......\left( ii \right)$
Now, we will differentiate (ii) again with respect to x. Thus, we will get,
$\Rightarrow \dfrac{d}{dx}\left( \dfrac{dy}{dx} \right)=\dfrac{d}{dx}\left[ 2A{{e}^{2x}}-2B{{e}^{-2x}} \right]$
$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left[ 2A{{e}^{2x}} \right]-\dfrac{d}{dx}\left[ 2B{{e}^{-2x}} \right]$
$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=2A\dfrac{d}{dx}\left[ {{e}^{2x}} \right]-2B\dfrac{d}{dx}\left[ {{e}^{-2x}} \right]$
We know that $\dfrac{d}{dx}{{e}^{ax}}=a{{e}^{ax}},$ so we get,
$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=2A\left( 2{{e}^{2x}} \right)-2B\left( -2{{e}^{-2x}} \right)$
$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=4A{{e}^{2x}}+4B{{e}^{-2x}}$
$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=4\left( A{{e}^{2x}}+B{{e}^{-2x}} \right)......\left( iii \right)$
From (i) and (iii), we will get,
$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=4y$
$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}-4y=0.....\left( iv \right)$
Thus, (iv) is the required differential function of the given equation.

Note: The ‘2’ given in the power of e is not an arbitrary constant because if we change the power of 2 to 3, then we will get another differential equation. Similarly, for each value of the power of e, we will get a unique differential equation. Thus, power can change the behavior of the differential equation. Hence, it is not an arbitrary constant.